# Group Theory/Groups, subgroups and constructions

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

 To do:#Product subgroups renaming of this section to something more appropriate

Definition (group):

A group is a set ${\displaystyle G}$ together with three operations, namely

1. a nullary operation ${\displaystyle e}$ called the identity,
2. a unary operation ${\displaystyle ~^{-1}}$ called inversion
3. and a binary operation ${\displaystyle *}$ 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 ${\displaystyle x,y,z}$ of ${\displaystyle G}$:

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

Definition (abelian group):

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

Proposition (elementary rules for computing in groups):

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

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

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

Definition (opposite group):

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

Proposition (opposite groups are groups):

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

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

Definition (subgroup):

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

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 ${\displaystyle \iota :H\to G}$ represent subobjects of a group.

Proposition (subgroup criterion):

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

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

## Exercises

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