# Group Theory/Normal subgroups and the Noether isomorphism theorems

Definition (normal subgroup):

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

Proposition (elements of disjoint normal subgroups commute):

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

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

Proposition (characterisation of direct products within groups):

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

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

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

Definition (subgroup product):

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

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

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

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

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

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

Proposition (product of normal subgroups is normal):

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

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

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

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

## Exercises

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