Group Theory/Printable version

Group Theory

The current, editable version of this book is available in Wikibooks, the open-content textbooks collection, at
https://en.wikibooks.org/wiki/Group_Theory

Permission is granted to copy, distribute, and/or modify this document under the terms of the Creative Commons Attribution-ShareAlike 3.0 License.

Groups, subgroups and constructions

Definition (group):

A group is a set $G$ together with three operations, namely

1. a nullary operation $e$ called the identity,
2. a unary operation $~^{-1}$ called inversion
3. and a binary operation $*$ called the group operation (also often referred to simply as the "operation" of the group)

such that the following axioms are satisfied for all elements $x,y,z$ of $G$:

1. $e*x=x$ (identity axiom)
2. $x*x^{-1}=e$ (inverse axiom)
3. $(x*y)*z=x*(y*z)$ (associativity axiom)

Definition (abelian group):

An abelian group is a group $G$, whose operation (which is usually denoted by $+$) is commutative.

Proposition (elementary rules for computing in groups):

Let $G$ be a group with group operation $*$ and let $x,y,z\in G$. Then we have the following rules of computation:

1. $x*y=x*z\Rightarrow y=z$ and $y*x=z*x\Rightarrow y=z$ ("cancellation laws")
2. $x^{-1}*x=e$
3. $x*e=x$

Proof: First note that when $x*y=x*z$, then we may multiply by $x^{-1}$ on the left using $*$ in order to get $y=z$ by using the identity axiom and $x^{-1}*x=e$ to get left cancellation, and right cancellation is proved similarly. Then, we note that by multiplying the inverse axiom $x*x^{-1}=e$ by $x^{-1}$ on the left and using the identity axiom, we get $x^{-1}*x*x^{-1}=x^{-1}$ and hence, since $e*x^{-1}=x^{-1}$ by the identity axiom, we may apply cancellation to infer $x^{-1}*x=e$, proving 2. Finally, $x*e=x*(x^{-1}*x)=(x*x^{-1})*x=e*x=x$. $\Box$

Henceforth, we shall sometimes refer to the group operation of a group simply as the "operation".

Definition (opposite group):

Let $(G,*)$ be a group, where $*$ denotes the operation of $G$. Then the opposite group $G^{o}$ of $G$ is defined as the group that has the same underlying set as $G$, but whose group operation $*'$ is given by $a*'b=b*a$ for $a,b\in G$ (inversion and identity are carried over from $G$).

Proposition (opposite groups are groups):

Whenever $G$ is a group, $G^{o}$ is also a group.

Proof: We have to check that the axioms for the given operations are satisfied. First note that $x*'x^{-1}=x^{-1}*x=e$ and that $e*'x=x*e=x$ by the rules that were derived above. Finally, $(x*'y)*'z=z*(y*x)=(z*y)*x=x*'(y*'z)$. $\Box$

Definition (subgroup):

Let $G$ be a group. A subset $H\subseteq G$ is called a subgroup of $G$ iff together with the group law of $G$ it is itself a group (in particular, for all $h,h'\in H$ we must have $h*h'\in H$.

Note: Often, the explicit notation for the group operation is omitted and the product of two elements is denoted solely by juxtaposition.

Subgroups with the inclusion map $\iota :H\to G$ represent subobjects of a group.

Proposition (subgroup criterion):

Let $G$ be a group and $H\subseteq G$ a subset. Then $H$ is a subgroup of $G$ if and only if $a\in H\wedge b\in H\Rightarrow ab^{-1}\in H$.

Proof: Suppose first that $H$ is a subgroup. Then the condition follows immediately from $H$ being closed under inversion and the group law. On the other hand, if the condition is satisfied, setting $a=1$ shows that $H$ is closed under inversion, and setting $b=c^{-1}$ shows that it is closed under the group law. Finally, setting $b=a$ shows that $H$ contains the identity. Associativity and the law that the identity must satisfy are automatically inherited from the group operation of $G$. $\Box$

Exercises

1. Make explicit the proof of right-cancellation ("right-cancellation" means $y*x=z*x\Rightarrow y=z$).
2. Let $G$ be a group, and let $H,J\leq G$ be subgroups such that neither $H\subseteq J$ nor $J\subseteq H$. Prove that $H\cup J$ is not a subgroup of $G$.
3. Let $G=\{1,-1\}$ together with the operation $1*1=1=-1*-1$, $1*-1=-1*1=-1$.
   1. Prove in detail that $G$, together with the operation $*$, is a group.
   2. Prove that in $G\times G$, there exists a subgroup which is not equal to $H\times L$ with subgroups $H,L\leq G$.

Representations

Definition (representation):

Let $G$ be a group, ${\mathcal {C}}$ a category, $A$ an object of ${\mathcal {C}}$. Then a representation (also called action) of $G$ on $A$ by automorphisms of ${\mathcal {C}}$ is a group homomorphism

$G\to \operatorname {Aut} (A)$.

Example (symmetric group acting on a product set):

Let $S$ be a set and let $S^{n}$ be the product of $n$ copies of $S$. The symmetric group acts on $S^{n}$ via

$\sigma (x_{1},\ldots ,x_{n}):=(x_{\sigma (1)},\ldots ,x_{\sigma (n)})$.

Note that in this notation, we identified the element of $G$ with the automorphism of $A$ to which the element is sent via the homomorphism of the representation. This convention is followed throughout group theory and will be understood by every mathematician.

Definition (equivalence of representations):

Let $G$ be a group, ${\mathcal {C}}$ a category, $A,B$ objects of ${\mathcal {C}}$ and $\rho :G\to A$, $\psi :G\to B$ two representations of $G$ on $A$ resp. $B$. Then an equivalence of representations is an isomorphism $f:A\to B$ such that

$\forall g\in G:g\circ f=f\circ g$.

Proposition (inverse of equivalence of representations is equivalence of representations):

Let $G$ be a group, ${\mathcal {C}}$ a category, $A,B$ objects of ${\mathcal {C}}$ and $\rho :G\to A$, $\psi :G\to B$ two representations of $G$ on $A$ resp. $B$. Let $f:A\to B$ be an equivalence of representations. Then $f^{-1}:B\to A$ is also an equivalence of representations.

Proof: We have

$g\circ f^{-1}=f^{-1}\circ g\Leftrightarrow f\circ g\circ f^{-1}=g\Leftrightarrow g=g$,

since $f$ is an equivalence of representations. $\Box$

Cosets and Lagrange's theorem

Definition (left coset):

Let $G$ be a group, and $H\leq G$ a subgroup, and $g\in G$. Then the left coset of $H$ represented by $g$ is the set

$gH:=\{gh|h\in H\}$.

Right cosets are defined in an analogous fashion:

Definition (right coset):

Let $G$ be a group, and $H\leq G$ a subgroup, and $g\in G$. Then the right coset of $H$ represented by $g$ is the set

$Hg:=\{hg|h\in H\}$.

For both of these, we have the following proposition:

Proposition (being in the same left coset is an equivalence relation):

Let $G$ be a group, and define a relation on $G$ by

$g\sim g':\Leftrightarrow \exists g''\in G:g\in g''H\wedge g'\in g''H$.

Then $\sim$ is an equivalence relation, and we also have the formula

$g\sim g'\Leftrightarrow g\in g'H$.

Proof: We first prove the alternative formula for being in the same coset. If $g\in g''H$ and $g'\in g''H$, we find from the latter equation an $h\in H$ so that $g'=g''h$, so that $g''=g'h^{-1}$ and hence $g\in g'h^{-1}H=g'H$, the latter identity because $H$ is a group and in particular closed under inversion. On the other hand, if $g\in g'H$, then $g=g'h$ for some $h\in H$, and hence $g'\in g'H$ (because the identity is in $H$) and then $g'\in g'H=gh^{-1}H=gH$.

Hence, the two formulae for the relations coincide, and it remains to check that we're dealing with an equivalence relation. Indeed, suppose $g\sim g'$ and $g'\sim g''$. Then there exist $h\in H$ so that $g=g'h$ and $h'\in H$ so that $g'=g''h'$. Then $g=g''h'h$, so that $g\in g''H$, ie. $g\in g''H$, proving transitivity. Reflexivity follows since the identity is in $H$, and symmetry follows because $g\in g'H$ implies $g=g'h$ and hence $g'\in g'H=ghH=gH$. $\Box$

Analogously, we have the following proposition:

Proposition (being in the same right coset is an equivalence relation):

Let $G$ be a group, and define a relation on $G$ by

$g\sim 'g':\Leftrightarrow \exists g''\in G:g\in Hg''\wedge g'\in Hg''$.

Then $\sim$ is an equivalence relation, and we also have the formula

$g\sim 'g'\Leftrightarrow g\in Hg'$.

Proof: Consider the opposite group $G^{o}$ of $G$. There is a bijection $G\to G^{o}$, given by $g\mapsto g$, under which two elements belonging to the same right coset of $H\leq G$ correspond to elements belonging to the same left coset of $H^{o}\leq G^{o}$. But the relation defined by the latter was seen to be an equivalence relation. $\Box$

Definition (index):

Let $G$ be a group, and let $H\leq G$ be a subgroup. Then the index of $H$ is defined to be the number

$[G:H]:=|\{gH|g\in G\}|$.

That is, the index is precisely the number of left cosets.

Proposition (Lagrange's theorem):

Let $G$ be a group, and let $H\leq G$ be a finite subgroup. Then

$[G:H]=|G|/|H|$.

In particular, the order of $H$ divides the order of $G$.

Proof: We have seen that being in the same left coset is an equivalence relation, so that the equivalence classes partition $G$. Moreover, every equivalence class (ie. coset) has the same cardinality as $H$ via the bijection $H\to gH,h\mapsto gh$. $\Box$

Proposition (number of right cosets equals number of left cosets):

Let $G$ be a group, and $H\leq G$ a subgroup. Then the number of right cosets of $H$ equals the number of left cosets of $H$.

Proof: By Lagrange's theorem, the number of left cosets equals $|G|/|H|$. But we may consider the opposite group $G^{o}$ of $G$. Its left cosets are almost exactly the right cosets of $G$; only the orders of the products are interchanged. But in particular, the number of left cosets of $G^{o}$, which, by Lagrange's theorem equals $|G^{o}|/|H^{o}|=|G|/|H|$, is equal to $[G:H]$, which is what we wanted to prove. $\Box$

Hence, we may also use the notation $[G:H]$ for the number of right cosets.

Proposition (degree formula):

Let $G$ be a group, and let $L\leq H\leq G$. Then

$[G:L]=[G:H][H:L]$.

Proof: We may partition $H$ into a family of left cosets $(h_{\lambda }L)_{\lambda \in \Lambda }$, where for all $\lambda \in \Lambda$ we have $h_{\lambda }\in H$. Moreover, $G$ may be partitioned into a family $(g_{\alpha }H)_{\alpha \in A}$ of left cosets of $H$. Then $(g_{\alpha }h_{\lambda }L)_{(\alpha ,\lambda )\in A\times \Lambda }$ is a family of left cosets of $L$ that partitions $G$ (since each element $j\in G$ is in one left coset $g_{\alpha }H$ of $H$, and then $g_{\alpha }^{-1}j\in H$ is in a unique coset $h_{\lambda }L$, and then $g_{\alpha }h_{\lambda }L$ is the unique coset in which $j$ is), and the cardinality of this family, which is $|A\times \Lambda |=|A||\Lambda |$, is the number of left cosets of $L$ in $G$. $\Box$

Exercises

1. Prove that $g\in g'H\Leftrightarrow gH=g'H$, thus establishing another formula for the equivalence relation of being in the same coset.
2. Formulate Lagrange's theorem for right cosets, without using index notation.

Normal subgroups and the Noether isomorphism theorems

Definition (normal subgroup):

Let $G$ be a group. A subgroup $H\leq G$ is called a normal subgroup if and only if for each $g\in G$ we have $gH=Hg$.

Proposition (elements of disjoint normal subgroups commute):

Let $G$ be a group, and let $G_{1},G_{2}\trianglelefteq G$ so that $G_{1}\cap G_{2}=\{1\}$. Then $[G_{1},G_{2}]=1$.

Proof: Let $a\in G_{1}$ and $b\in G_{2}$. Since $G_{1}$ is a normal subgroup, $bG_{1}=G_{1}b$, so that there exists $\alpha \in G_{1}$ such that $ab=b\alpha$. Similarly, since $G_{2}\trianglelefteq G$, there exists $\beta \in G_{2}$ so that $b\alpha =\alpha \beta$. Then $ab=\alpha \beta$, hence $\alpha ^{-1}a=\beta b^{-1}$ and since $G_{1}\cap G_{2}=\{1\}$ we get $\alpha =a$, so that $ab=ba$ and since $a\in G_{1}$ and $b\in G_{2}$ were arbitrary, $[G_{1},G_{2}]=1$. $\Box$

Proposition (characterisation of direct products within groups):

Let $G$ be a group, and let $G_{1},\ldots ,G_{n}\leq G$ be subgroups of $G$. Consider the group $H:=\langle G_{1},\ldots ,G_{n}\rangle$ generated by these groups. The following are equivalent:

1. The function $\Phi :G_{1}\times \cdots \times G_{n}\to H,(g_{1},\ldots ,g_{n})\mapsto g_{1}\cdots g_{n}$ is an isomorphism
2. For $j\in [n]$ we have $G_{j}\trianglelefteq G$ and $G_{j}\cap \langle G_{1},\ldots ,G_{j-1},G_{j+1},\ldots ,G_{n}\rangle =\{1\}$
3. Each element in $H$ can be uniquely written as a product $g_{1}g_{2}\cdots g_{n}$, where for $j\in [n]$ we have $g_{j}\in G_{j}$

Proof: Certainly 1. $\Rightarrow$ 2., since for all $j\in [n]$, the subgroup $G_{j}$ of $G$ corresponds via $\Phi$ to the subgroup $\{(g_{k})_{1\leq k\leq n}|\forall k\in [n]:k\neq j\Rightarrow g_{k}=1\}$, which is certainly normal. For 2. $\Rightarrow$3., observe that any element of $H$ may be written as a product $g_{1}g_{2}\cdots g_{m}$, where $m\in \mathbb {N}$ and each $g_{j}$ is an element of some $G_{j}$. But by 2., the $G_{j}$ are pairwise disjoint, so that any elements in distinct $G_{j}$ commute. Hence, we may sort the product $g_{1}g_{2}\cdots g_{m}$ so that the first few entries are in $G_{1}$, the following entries are in $G_{2}$ and so on. The products of the entries that are contained within $G_{j}$ then form the element as required by the decomposition in 3. Finally, for 3. $\Rightarrow$ 1., observe that 3. implies that the given function is bijective. But it is also a homomorphism, because 3. immediately implies that the $G_{j}$ are disjoint, and we may use that elements of disjoint normal subgroups commute to obtain that $\Phi$ indeed commutes with the respective group laws. $\Box$

Definition (subgroup product):

Let $G$ be a group, and let $L,H\leq G$ be subgroups. Then the subgroup product of $L$ and $H$ is defined to be

$LH:=\{lh|l\in L,h\in H\}$.

In general, the subgroup product is not a subgroup. However, if one of the subgroups involved in the product is a normal subgroup, then it is:

Proposition (subgroup product of subgroup and normal subgroup is subgroup):

Let $G$ be a group, and let $L,H\leq G$ be subgroups such that one of $L,H$ is normal. Then $LH$ and $HL$ are subgroups of $G$.

Proof: Without loss of generality assume that $L$ is normal. Let $lh,jg\in LH$, where $l,j\in L$ and $h,g\in H$. Then

$lh(jg)^{-1}=lj'hg^{-1}\in LH$

for some $j'\in L$ because $L$ is normal (here we applied the subgroup criterion). Similarly, $HL$ is a subgroup. $\Box$

Proposition (product of normal subgroups is normal):

Let $G$ be a group, and let $L,H\triangleleft G$ be normal subgroups of $G$. Then $LH\triangleleft G$, ie. $LH$ is a normal subgroup of $G$, and the same holds for $HL$.

Proof: By symmetry, it suffices to prove that $LH$ is a normal subgroup. Indeed, let $g\in G$ and $lh\in LH$, where $l\in L$ and $h\in H$. Then

$g^{-1}lhg=g^{-1}lgh'=g^{-1}l'h'=l'h'$

for some $l'\in L$, $h'\in H$, since $L,H$ are normal. $\Box$

Exercises

1. Prove that the intersection of normal subgroups is again normal.
2. Let $G$ be a group, and let $H_{1},\ldots ,H_{n}\trianglelefteq G$ such that $[G:H_{1}],\ldots ,[G:H_{n}]$ are pairwise coprime. Prove that $G/(H_{1}\cap \cdots \cap H_{n})\cong G/H_{1}\times \cdots \times G/H_{n}$.

Cardinality identities for finite representations

Definition (permutation representation):

Let $G$ be a group and let $X$ be a set. A permutation representation of $G$ on $S$ is a representation $\rho :G\to \operatorname {Aut} (S)$, where the automorphisms of $S$ are taken in the category of sets (that is, they are just bijections from $S$ to itself).

Definition (pointwise stabilizer):

Let $G$ be a group, let ${\mathcal {A}}$ be an algebraic variety and let $A$ be an instance of ${\mathcal {A}}$. Suppose that $\rho :G\to \operatorname {Aut} (A)$ is a representation in the category defined by ${\mathcal {A}}$. Let $S\subseteq A$. Then the pointwise stabilizer of $S$ is given by

$G_{S}:=\{g\in G|\forall a\in S:ga=a\}$.

Proposition (transitive permutation representation is equivalent to right multiplication on quotient by stabilizer):

Let $G$ be a group, let $X$ be a set and suppose that we have a permutation representation $\pi :G\to \operatorname {Sym} (X)$ which is transitive. Let $x\in X$ be arbitrary and let $G_{x}\leq G$ be the pointwise stabilizer of $x$. Consider the action $\rho :G\to \operatorname {Sym} (G/G_{x})$ by left multiplication, where $G/G_{x}$ is the set of left cosets of $G_{x}$ (which is in fact never a normal subgroup in this situation, unless the action is trivial, because $G_{gx}=g^{-1}G_{x}g$). Then there exists a $G$-isomorphism from $X$ to $G/G_{x}$.

Proof: We define $f:G/G_{x}\to X$ as follows: $gG_{x}$ shall be mapped to $gx$. First, we show that this map is well-defined. Indeed, suppose that we take $h\in G_{x}$. Then $ghG_{x}$ is mapped to $ghx=x$. Then we note that the map is surjective by transitivity. Finally, it is also injective, because whenever $gx=hx$, we have $h^{-1}gx=x$ by applying $h^{-1}$ to both sides and using a property of a group action, and thus $h^{-1}g\in G_{x}$, that is to say $gG_{x}=hG_{x}$. That $f(ghG_{x})=g(hx)$ follows immediately from the definition, so that we do have an isomorphism of representations. $\Box$

We are now in a position to derive some standard formulae for permutation representations.

Theorem (orbit-stabilizer theorem):

Let $G$ be a group, and let $\pi :G\to \operatorname {Sym} (X)$ be a permutation representation on a set $X$. Then

$|Gx|=[G:G_{x}]$.

Proof: $G$ acts transitively on $Gx$. The above $G$-isomorphism between $Gx$ and $G/G_{x}$ is bijective as an isomorphism in the category of sets. But the notation $[G:G_{x}]$ stood for $|G/G_{x}|$. $\Box$

Theorem (class equation):

Let $G$ be a finite group and let $\pi :G\to \operatorname {Sym} (X)$ be a permutation representation on the finite set $X$. Then

$|X|=\sum _{j=1}^{n}[G:G_{x_{j}}]$,

where $Gx_{1},\ldots ,Gx_{n}$ are the orbits of $G$, and $Gx_{k}\neq Gx_{j}$ for $k\neq j$. (We also say that $x_{1},\ldots ,x_{n}$ are a system of representatives for the orbits of $G$.)

Proof: $G$ acts transitively on each orbits, and the orbits partition $X$. Hence, by the orbit-stabilizer theorem,

$|X|=\sum _{j=1}^{n}|Gx_{j}|=\sum _{j=1}^{n}[G:G_{x_{j}}]$. $\Box$

Definition (fixed point set):

Let $G$ be a group that acts on a set $X$, and let $S\subseteq G$ be a subset of $G$. Then the fixed point set of $S$ is defined to be

$\operatorname {Fix} (S):=\{x\in X|\forall s\in S:sx=x$.

Free products and amalgamated sums

Definition (reduced word):

Let $S$ be any set, and define the set $S^{-1}$ to be the set of formal inverses to the elements of $S$; that is, $S^{-1}:=\{s^{-1}|s\in S\}$, so that $S\cap S^{-1}=\emptyset$; for example, we could define $s^{-1}:=\{\emptyset ,s\}$. Let $\varepsilon$ denote the empty tuple. Then a reduced word over $S$ is either

1. the empty tuple $\varepsilon$, or
2. a finite tuple $(r_{1},\ldots ,r_{n})$ of elements of $S\cup S^{-1}$ such that whenever $r_{j-1},r_{j}$ are two adjacent elements, then neither $r_{j-1}\in S$ and $r_{j-1}=r_{j}^{-1}$ nor $r_{j-1}\in S^{-1}$ and $r_{j-1}^{-1}=r_{j}$.

Definition (empty word):

The empty tuple $\varepsilon =()$ is also called the empty word

Proposition (reduction of tuples to reduced words):

Let $S$ be any set, and let $S^{-1}$ be the set of formal inverses. Suppose that $(r_{1},\ldots ,r_{n})$ is any tuple (not necessarily a reduced word). Then in finitely many steps, one may obtain a reduced word from $(r_{1},\ldots ,r_{n})$ by removing adjacent elements $r_{j-1},r_{j}$ such that either $r_{j-1}\in S$ and $r_{j-1}=r_{j}^{-1}$ or $r_{j-1}\in S^{-1}$ and $r_{j-1}^{-1}=r_{j}$.

Proof: This follows immediately since the length of the tuple $(r_{1},\ldots ,r_{n})$ is an integer, which is reduced by 2 whenever adjacent elements that contradict the definition of a reduced word are eliminated. Doing this elimination repeatedly until it is no longer possible will hence lead to a reduced word in a finite number of steps. $\Box$

Note that when $n$ is odd, then the reduced word obtained in this way will not be the empty tuple. Otherwise, the empty tuple may result.

Definition (free group):

Let $S$ be any set. Then the free group over $S$ is defined to be the group $F\langle S\rangle$ whose elements are the reduced words over $S$ and whose group operation is given by first concatenation and then reduction to a reduced word.

Proposition (the free group is a group):

Let $S$ be a set. Then $F\langle S\rangle$ is a group.

Proof: The empty tuple $\varepsilon$ serves as an identity. Associativity holds because if $(r_{1},\ldots ,r_{n}),(t_{1},\ldots ,t_{m}),(u_{1},\ldots ,u_{k})$ are three reduced words, then

Finally, whenever $(r_{1},\ldots ,r_{n})$ is a reduced word, we claim by induction on $n$ that it has an inverse. Certainly the empty word has Indeed, suppose that $r_{n}\in S$; then $(r_{1},\ldots ,r_{n})(r_{n}^{-1})=(r_{1},\ldots ,r_{n-1})$, which has an inverse $(t_{1},\ldots ,t_{m})$ by the induction hypothesis, so that by associativity $(r_{n}^{-1})(t_{1},\ldots ,t_{m})$ is an inverse of $(r_{1},\ldots ,r_{n})$. $\Box$

Exercises

1. Prove that when $S$ is a set such that $|S|\geq 2$, then $F\langle S\rangle$ is not an abelian group.

Abelian groups and the Grothendieck group of a monoid

Definition (abelian):

Let $G$ be a group. We call $G$ an abelian group if and only if for all $g,h\in G$, we have $gh=hg$ (where we denote the group operation by juxtaposition).

Definition (cyclic group):

A cyclic group is a group that is generated by a single of its elements, ie. $G=\langle g\rangle$ for a certain $g\in G$.

Proposition (cyclic group is abelian):

Let $G$ be a cyclic group. Then $G$ is abelian.

Proof: Indeed, write any two elements $h,j\in G$ as $h=g^{n}$, $j=g^{m}$, where $g\in G$ is such that $G=\langle g\rangle$. Then $hg=g^{n}g^{m}=g^{m}g^{n}=gh$, using associativity. $\Box$

The action by conjugation and p-groups

Definition (global stabilizer):

Let $G$ be a group that acts on $A$, where $A$ belongs to some algebraic variety ${\mathcal {A}}$. Let $S\subseteq A$ be a subset. Then the global stabilizer of $S$ is the set

$G(S):=\{g\in G|gS\subseteq S\}$,

where the notation $gS$ stands for the set $\{gs|s\in S\}$.

Definition (p-group):

Let $p\in \mathbb {N}$ be a prime number. Then a $p$-group is a group of order $p^{k}$ for some $k\in \mathbb {N}$.

Proposition (cardinality of fixed point set of a p-group equals cardinality of set mod p):

Let $G$ be a $p$-group that acts on a set $X$. Then

$|\operatorname {Fix} (G)|\equiv |X|\mod p$.

Proof: By the class equation,

$|X|=\sum _{k=1}^{n}[G:G_{x_{k}}]$,

where for each orbit of the action of $G$ on $X$ we pick one representative $x_{k}$ of that orbit. Since $G$ is a $p$-group, whenever $[G:G_{x_{k}}]$ is not $1$, it is divisible by $p$ by Lagrange's theorem. Hence, by taking the above equation $\mod p$, we get

$|X|\equiv \sum _{j=1}^{l}1\equiv l\mod p$,

where $l$ is the number of those $x_{k}$ for which $[G:G_{x_{k}}]=1$. But $[G:G_{x_{k}}]=1$ means precisely that the orbit of $x_{k}$ is trivial, that is, that $x_{k}$ is fixed by all of $G$. $\Box$

Proposition (p-groups have nontrivial center):

Let $G$ be a $p$-group. Then $Z(G)\neq \{e\}$, where $e\in G$ denotes the identity.

Proof: $G$ acts on itself via conjugation. Furthermore,

$h\in Z(G)\Leftrightarrow \forall g\in G:g^{-1}hg=h$,

so that $Z(G)$ is precisely the fixed point set of $G$ under that action. But since the cardinality of the fixed point set of a p-group equals the cardinality of the whole set mod p, we get that

$0\equiv |G|\equiv |\operatorname {Fix} (G)|\equiv |Z(G)|\mod p$,

which would be impossible if $|Z(G)|=1$. $\Box$

Simple groups and Sylow's theorem

Definition (Sylow p-subgroup):

Let $G$ be a group and let $p\in \mathbb {N}$ be a prime number such that $p||G|$. Then a Sylow $p$-subgroup of $G$ is a subgroup $P\leq G$ such that $|P|=p^{k}$, where $k$ is maximal such that $p^{k}||G|$.

Theorem (Cauchy's theorem):

Let $G$ be a group whose order $|G|$ is divisible by a prime number $p\in \mathbb {N}$. Then $G$ contains an element of order $p$.

Proof: $G$ acts on itself via conjugation. Let $x_{1},\ldots ,x_{n}\in G$ be a system of representatives of cojugacy classes. The class equation yields

$|G|=\sum _{k=1}^{n}[G:G_{x_{k}}]$.

Either, there exists $k\in \{1,\ldots ,n\}$ such that $[G:G_{x_{k}}]$ is both not $1$ and not divisible by $p$, in which case we may conclude by induction on the group order, noting that $p$ divides $|G_{x_{k}}|$ and $|G_{x_{k}}|<|G|$, or for all $k\in \{1,\ldots ,n\}$ the number $[G:G_{x_{k}}]$ is either $1$ or divisible by $p$; but in this case, by taking the class equation $\mod p$, we obtain that $Z(G)$ is nontrivial and moreover that its order is divisible by $p$. Hence, it suffices to consider the case where $G$ is an abelian group. Take then any element $h\in G$. If $h$ has order divisible by $p$, raising $h$ to a sufficiently high power will produce an element of order $p$. Otherwise, the order of $G/\langle h\rangle$ is divisible by $p$, and by induction we find an element $g+\langle h\rangle \in G/\langle h\rangle$ whose order is divisible by $p$. Then the order of $g$ will also be divisible by $p$, because otherwise, passing to the quotient, $(g+\langle h\rangle )^{k}=e+\langle h\rangle$ for some $k$ not divisible by $p$. $\Box$

Theorem (Sylow's theorem):

Let $G$ be a finite group, such that $|G|=p^{k}m$ with $p\nmid m$. Then the following hold:

1. $G$ has a Sylow subgroup
2. The action of $G$ by conjugation on the Sylow subgroups is transitive
3. If $n_{p}$ is the number of Sylow $p$-subgroups, then $n_{p}|m$ and $n_{p}\equiv 1\mod p$
4. Every $p$-group of $G$ is contained within some Sylow $p$-group of $G$

Definition (simple group):

A group $G$ is a simple group if $G$ and $\{e\}$ are the only normal subgroups of $G$ (where $e\in G$ denotes the identity).

Subnormal subgroups and series

{{definition|subnormal subgroup|Let $G$ be a group. A subgroup $H\in G$ is called subnormal subgroup if and only if there exists

Definition (subnormal series):

Let $G$ be a group. Then a subnormal series is a finite family of subgroups $H_{0},H_{1},\ldots ,H_{n}\leq G$ such that

$\{e\}=H_{0}\triangleleft H_{1}\triangleleft \cdots \triangleleft H_{n}=G$,

where $e\in G$ is the identity.

Definition (composition series):

Let $G$ be a group. A composition series of $G$ is a subnormal series

$\{e\}=H_{0}\triangleleft H_{1}\triangleleft \cdots \triangleleft H_{n}=G$

of $G$ such that for all $k\in \{1,\ldots ,n\}$ the quotient group $H_{k}/H_{k-1}$ is simple.

Theorem (Schreier refinement theorem):

Let $G$ be a group, and let

$\{e\}=H_{0}\triangleleft H_{1}\triangleleft \cdots \triangleleft H_{n}=G$

be a subnormal series of $G$.

Commutators, solvable and nilpotent groups

Definition (commutator):

Let $G$ be a group and let $a,b\in G$. Then the commutator of $a$ and $b$ is defined to be the element

$[a,b]:=a^{-1}b^{-1}ab\in G$.

{{definition|commutator

Proposition (commutators form a subgroup):

Let $G$ be a group. Then the set $\{[a,b]|a,b\in G\}$ forms a subgroup of $G$.

Proof: By the subgroup criterion, it is sufficient to show that for $a,b,c,d\in G$, the element $[a,b][c,d]^{-1}$ is of the form $[e,f]$ for suitable $e,f\in G$. Indeed,

Failed to parse (syntax error): {\displaystyle \begin{equation*} t \end{equation*}} $\Box$

Definition (commutator subgroup):

Let $G$ be a group. Then the commutator subgroup of $G$ is defined to be $[G,G]$.

Definition (perfect):

A group $G$ is called perfect if and only if $[G,G]=G$.

Proposition (subdirect normal product of perfect groups is direct):

Let $H_{1},\ldots ,H_{n}$ be perfect groups, and let

$N\trianglelefteq H_{1}\times \cdots \times H_{n}$

be a subdirect product which is simultaneously a normal subgroup of their outer direct product. Then in fact $N=H_{1}\times \cdots \times H_{n}$.

Proof: It suffices to show that whenever $k\in \{1,\ldots ,n\}$ and $g,h\in H_{k}$, then

$(1,\ldots ,{\overset {\overset {k{\text{-th entry}}}{\downarrow }}{g^{-1}h^{-1}gh}},1\ldots ,1)\in N$,

since $H_{k}$ is a perfect group. Thus, let $h\in H_{k}$ be arbitrary, and pick $(h_{1},\ldots ,h_{n})\in N$, where $h_{j}\in H_{j}$ for all $j\in \{1,\ldots ,n\}$, such that $h_{k}=h$. Since $N$ is a subgroup and normal, the element

$(1,\ldots ,1,{\overset {\overset {k{\text{-th entry}}}{\downarrow }}{g^{-1}}},1,\ldots ,1)(h_{1}^{-1},\ldots ,h_{n}^{-1})(1,\ldots ,1,{\overset {\overset {k{\text{-th entry}}}{\downarrow }}{g}},1,\ldots ,1)(h_{1},\ldots ,h_{n})=(1,\ldots ,{\overset {\overset {k{\text{-th entry}}}{\downarrow }}{g^{-1}h^{-1}gh}},1\ldots ,1)$

is in $N$. $\Box$

Definition (solvable):

Proposition (group is solvable iff maximal normal subgroup is solvable):

Let $G$ be a group, and let $H\triangleleft G$ be a maximal normal subgroup. Then $G$ is solvable if and only if $H$ is solvable.

Characteristic subgroups

Definition (characteristic subgroup):

Let $G$ be a group. A characteristic subgroup of $G$ is a subgroup $H\leq G$ such that $\varphi (H)=H$ for all $\varphi \in \operatorname {Aut} _{\textbf {Grp}}(G)$.

Proposition (characteristic subgroups are normal):

Any characteristic subgroup of a group $G$ is a normal subgroup of $G$.

Proof: This follows since the map $j\mapsto g^{-1}jg$ is a group automorphism of $G$. $\Box$

Definition (characteristically simple):

A group $G$ is called characteristically simple if and only if its only characteristic subgroups are $G$ and $\{e\}$, where $e$ denotes the identity of $G$.

Proposition (characteristically simple groups):

Let $G$ be a characteristically simple finite group, and let $H$ be any of its minimal normal subgroups. Then $G$ is isomorphic to a product of copies of $H$, that is, $G\cong \prod _{\alpha \in A}H$, where $A$ is an index set (of finite cardinality).

Proof: Let $M$ be a subgroup of $G$ maximal subject to the following two conditions:

1. $M$ is the direct sum of images of $H$ under a $\varphi \in \operatorname {Aut} _{\textbf {Grp}}(G)$
2. $M$ is normal

Suppose that $M\neq G$. Note that the group $\langle \varphi (H)|\varphi \in \operatorname {Aut} _{\textbf {Grp}}(G)\rangle$ is characteristic, so that it equals all of $G$. Hence, we find $\varphi \in \operatorname {Aut} _{\textbf {Grp}}(G)$ such that $\varphi (H)$ is not a subgroup of $M$. Since $\varphi$ is an automorphism, $\varphi (H)\triangleleft G$, so that $M\cap \varphi (H)=\{e\}$. Since the product of normal subgroups is normal, we conclude that the product subgroup $M\varphi (H)\triangleleft G$ is a normal subgroup that is a direct product of homomorphic images of $H$ in $G$, in contradiction to the maximality of $M$ with these properties. Hence, $M=G$ and we are done. $\Box$

Proposition (minimal normal subgroups of a characteristically simple groups are simple):

Let $G$ be a characteristically simple group, and let $H\triangleleft G$ be a minimal normal subgroup of $G$. Then $H$ is simple.

Proof: $\Box$

Proposition (powers of characteristically simple groups are characteristically simple):

We conclude:

Theorem (structure theorem of finite, characteristically simple groups):

The finite, characteristically simple groups are precisely the powers of simple groups.

Proof: We have seen that each characteristically simple finite group is the direct product of copies of isomorphic images of any of its minimal normal subgroups, and that the latter are always simple in characteristically simple groups. We conclude that each finite, characteristically simple group is a power of simple groups. Conversely, let $H$ be a simple group, $n\in \mathbb {N}$, and set

$G:=H^{n}:=\overbrace {H\times \cdots \times H} ^{n{\text{ times}}}$. $\Box$

Exercises

1. Prove that all subgroups of $\mathbb {Z} _{6}$ are characteristic.
2. Let $H,L$ be two finite simple groups such that $|L|$ is divisible by a prime number $p$ that does not divide $|H|$. Use the structure theorem for characteristically simple groups to prove that $H\times L$ is not characteristically simple.
3. Prove that a subgroup of a characteristically simple group need not be characteristically simple.
4. Prove that the product of characteristically simple subgroups whose minimal normal subgroups are not isomorphic is not characteristically simple.

The symmetric group

Definition (symmetric group):

Let $X$ be a set. Then the symmetric group of $X$ is defined to be

$\operatorname {Sym} (X):=\operatorname {Aut} _{\textbf {Set}}(X)$;

that is, it is the set of all bijective functions from $X$ to itself with composition as operation.

Definition (permutation):

A permutation is, by definition, an element of $\operatorname {Sym} (X)$.

Proposition (symmetric group essentially depends only on the cardinality of the underlying set):

Let $X,Y$ be sets of the same cardinality. Then there exists a group isomorphism

$\operatorname {Sym} (X)\cong \operatorname {Sym} (Y)$.

Proof: Suppose that $\phi :X\to Y$ is a bijective function. Then the group isomorphism is given by

$\operatorname {Sym} (X)\ni f\mapsto \phi \circ f\circ \phi ^{-1}$;

indeed, an inverse is given by

$\operatorname {Sym} (Y)\ni g\mapsto \phi ^{-1}\circ g\circ \phi$. $\Box$

Definition (finite symmetric group):

Let $n\in \mathbb {N}$. Then the symmetric group of order $n$, denoted $S_{n}$, is defined to be

$S_{n}:=\operatorname {Sym} (\{1,\ldots ,n\})$.

Theorem (Cayley's theorem):

Let $G$ be a finite group, and set $n:=|G|\in \mathbb {N}$. Then there exists a subgroup of $S_{n}$ which is isomorphic to $G$.

Proof: $G$ acts transitively on itself by left multiplication in the category of sets. This means that we have a group homomorphism $\rho :G\to \operatorname {Aut} _{\textbf {Set}}(G)\cong S_{n}$. Moreover, this morphism is injective; indeed, only the identity element of $G$ induces the identity element in $S_{n}$. Hence, the claim follows from the first Noether isomorphism theorem. $\Box$

Definition (matrix representation of a permutation):

The representation of $S_{n}$ in the category of vector spaces over a field $\mathbb {F}$ given by

$\rho :S_{n}\to \operatorname {Aut} (\mathbb {F} ^{n}),\sigma \mapsto [(x_{1},\ldots ,x_{n})\mapsto (x_{\sigma (1)},\ldots ,x_{\sigma (n)})]$

is called the matrix representation of the permutations contained within $S_{n}$.

Definition (sign):

Let $\sigma \in S_{n}$ be a permutation. Then the sign of $\sigma$, written $\operatorname {sgn}(\sigma )$, is defined to be $(-1)^{k}$, where there exist $k$ transpositions $\tau _{1},\ldots ,\tau _{k}$ such that $\sigma =\tau _{1}\tau _{2}\cdots \tau _{k}$.

The following proposition shows that this notion is well-defined:

Proposition (equivalent characterisations of the sign of a permutation):

Definition (alternating group):

Let $n\in \mathbb {N}$. Then the alternating group, a subgroup of $S_{n}$, is defined to be

$A_{n}:=\ker \operatorname {sgn} =\{g\in S_{n}|\operatorname {sgn}(g)=1\}$.

Proposition (alternating group is maximal and normal in the symmetric group):

Let $n\in \mathbb {N}$. Then $A_{n}\triangleleft S_{n}$, and further $A_{n}$ is a maximal subgroup of $S_{n}$.

In particular, $A_{n}$ is a maximal normal subgroup in $S_{n}$ (ie. maximal among the normal subgroups).

Proof: Note first that $A_{n}$ is normal as the kernel of a group homomorphism. We then have that $\operatorname {sgn}$ is a group homomorphism from $S_{n}$ to $\mathbb {Z} _{2}$, and by the first Noether isomorphism theorem, $Z_{2}\cong S_{n}/A_{n}$. In particular, there are only two cosets of $A_{n}$. Suppose that there existed a subgroup $A_{n}\lneq H\lneq S_{n}$. Then by the degree formula, we would have $[S_{n}:H][H:A_{n}]=2$, so that either $[S_{n}:H]=1$ or $[H:A_{n}]=1$. In both cases, one of the inclusions is not strict, a contradiction. $\Box$

Proposition (conjugation in the symmetric group is re-labeling):

Let $(a_{1}a_{2}\ldots a_{k})\in S_{n}$ be a cycle, and let $g\in S_{n}$ be any element. Then

$g^{-1}(a_{1}a_{2}\cdots a_{n})g=(g(a_{1})g(a_{2})\cdots g(a_{n})$.

Proposition (in degrees 5 or larger all three-cycles are conjugate in the alternating group):

Let $n\geq 5$, and let $(abc)$ and $(def)$ be any two three-cycles in $S_{n}$. Then there exists $g\in A_{n}$ such that $g(abc)g^{-1}=(def)$.

Proof: Since being in the same conjugacy class is an equivalence relation, assume $(abc)=(123)$. $\Box$

Theorem (in degrees 5 or larger the alternating group is simple):

Let $n\geq 5$. Then $A_{n}$ is a simple group.

Proposition (in degrees 5 or larger neither the alternating nor the symmetric group are solvable):

Let $n\geq 5$. Then $A_{n}$ and $S_{n}$ are both not solvable.

Proof: Since $A_{n}$ is a maximal normal subgroup of $S_{n}$, $\Box$

Groups with structure

Definition (group with structure):

Let ${\mathcal {C}}$ be a concrete category. Then the category of groups with structure is the subcategory of ${\mathcal {C}}$ which is defined as follows:

• Its objects are the objects $G$ of ${\mathcal {C}}$ such that the categorical product $G\times G$ exists (and, on a set level, equals the set-theoretic product of $G$ with itself, projections included) and whose underlying sets bear a group structure such that the group law is a morphism $G\times G\to G$ in ${\mathcal {C}}$ and inversion is a morphism $G\to G$ in ${\mathcal {C}}$.
• Its morphisms are morphisms $f:G\to G'$ in ${\mathcal {C}}$ that, on the set level, are also group homomorphisms.

Proposition (multipication by an element is an automorphism in the underlying concrete category for every group with structure in a category admitting enough constant morphisms):

Let $G$ be a group with structure whose additional structure is given by the concrete category ${\mathcal {C}}$, such that every constant morphism $G\to G$ is a morphism of ${\mathcal {C}}$. Further, suppose that $g\in G$. Then the function

$G\to G,h\mapsto gh$

is an automorphism of $G$ in the category ${\mathcal {C}}$.

Proof: If we show that the given function is a morphism, we've completed the proof, since an inverse is given by left multiplication by $g^{-1}$.

Thus, consider the morphism $G\to G\times G$ whose first component is given by the constant function associated to $g$ and whose second component is given by the identity on $G$: Postcomposing it with the group law yields the morphism in the theorem statement, which is hence a morphism of ${\mathcal {C}}$. $\Box$

Topological groups

Definition (topological group):

A topological group $G$ is a group whose underlying set is endowed with a topology such that

1. the group law is a continuous function $G\times G\to G$ and
2. inversion is a continuous function $G\to G$.

Thus, a topological group is a group with structure in the category of topological spaces.

Proposition (every topological group is a uniform space):

Let $G$ be a topological group, and let $(U_{\alpha })_{}$ be a neighbourhood system of its identity. Then the sets

${\displaystyle }$

form an entourage system whose induced topology is identical to the topology of $G$.

Proof: $\Box$

Theorem (Birkhoff‒Kakutani theorem):

Proposition (the connected component of the identity of a topological group is one of its normal subgroups):

Let $G$ be a topological group, and let $K\subseteq G$ be the connected component of its identity. Then $K\trianglelefteq G$.

Proof: $\Box$

Proposition (each locally compact topological group is the disjoint union of translates of one of its σ-compact open subgroups):

Let $G$ be a locally compact topological group. Then there exists a σ-compact open subgroup $H\leq G$, from which we may of course deduce that

$G=\bigsqcup _{g\in S}gH$,

where $S$ is a set that contains one element of each left coset of $H$ (the squared union symbol indicating that the union is disjoint). Moreover, each left coset $gH$ of $H$ is σ-compact and open.

Proof: We shall denote the identity of $G$ by $e$. Let $K$ be a compact neighbourhood of $e$. We set $L:=K\cup K^{-1}$. Since the image of a compact set via a continuous map is compact and the union of two compact sets is compact, $L$ is a compact neighbourhood of $e$. Moreover, induction, the fact that the product of two compact sets is compact and the fact that the image of a compact set via a continuous map is compact (applied to the continuous group law map) yield that all the sets $L^{2},L^{3},\ldots$ are compact. Yet, the group

$H:=\langle L\rangle \leq G$,

ie. the group generated by the elements of $L$, is the union of these sets, hence σ-compact.

It remains to show that $H$ is open. To this end, we may use that since $L$ is a neighbourhood of $e$, there exists an open set $V\subseteq L$ such that $e\in L$. Since multiplication by a group element is an isomorphism, the set $hV$ are open in $H$ whenever $h\in H$. Hence,

$H=\bigcup _{h\in H}hV$

is open.

Finally, $gH$ is open and σ-compact because multiplication by $g$ is an automorphism of $G$ in the category of topological spaces, whence it preserves openness and compactness. $\Box$