# Group Theory/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:
* 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
* For $j\in [n]$ we have $G_{j}\trianglelefteq G$ and $G_{j}\cap \langle G_{1},\ldots ,G_{j-1},G_{j+1},\ldots ,G_{n}\rangle =\{1\}$
* 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

* Prove that the intersection of normal subgroups is again normal.
* 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 $G/(H_{1}\cap \cdots \cap H_{n})\cong G/H_{1}\times \cdots \times G/H_{n}$.