## Cosets and normal subgroups

Just as in the case of abelian groups, each subgroup $H\leq G$ gives rise to a coset decomposition of $G,$ or, more precisely, to two decompositions, into left cosets i.e.

$\displaystyle \{gH\mid g\in G\},$

and right cosets, i.e.

$\displaystyle \{Hg\mid g\in G\}.$

It is proved by the same argument as in the case of abelian groups that one indeed has decompositions of $G$ in this way.
Unlike in the case of abelian groups, these two decompositions need not be the same: an example is given by $S_3\geq H\cong S_2.$
The following is an important and elementary implication of the fact that all the cosets of $H$ have the same cardianlity:

Let $|G|<\infty,$ and $H\leq G.$ Then $|H|$ divides $|G|.$

An extremely important case (and, in fact, the only case) when these two decompositions are the same is when $H=Ker(\phi),$ where $\phi: G\to F$ is a group homomorphism. (Just as in the case of abelian groups, one can show that $Ker (\phi)\leq G$ and $Im(\phi)\leq F.$)
We need to show that for any $g\in G$ and $k\in Ker(\phi)$ one has $gk=k'g$ for some $k'\in Ker(\phi).$ We compute $\phi(gk)=\phi(g)\cdot 1_F=\phi(k')\phi(g)=\phi(k'g).$ Hence $gK=Kg.$

A subgroup $H\leq G$ satisfying $gH=Hg$ for all $g\in G$ is called normal (notation: $H\unlhd G,$ or $H\lhd G,$ when $H\neq G.$)

Often it is more convenient to work with the equivalent condition for normality, namely that $gHg^{-1}=H$ for all $g\in G.$

Examples of normal subgroups.

• The subgroup generated by $\begin{pmatrix}1&2&3\\3&1&2\end{pmatrix}$ in $S_3$ is normal.
• $H\leq G=GL(n,\mathbb{F})$ defined by $H=\{x\in G\mid\det x=1\}$ is normal.
• $H\leq G$ of index 2 (i.e. the one that has exactly 2 left cosets in $G$) is normal.

The 1st is easy to establish directly (it also follows from the 3rd).The 2nd follows from the identity $\det x=\det (gxg^{-1}).$
The 3rd follows from the equality $G=H\sqcup gH=H\sqcup Hg',$ and an observation that $g\in Hg',$ so $Hg'=Hg.$

Cosets play an important role in constructing “easier” groups from “complicated” ones, by the quotient group construction. The difference with the case of abelian groups is that only normal subgroups can be kernels of homomorphism.

Let $H\unlhd G.$ Then $G/H:=\{gH\mid g\in G\}$ is a group with operation $(gH)(g'H)=(gg')H,$ and $g\mapsto gH$ a group homomorphism $\phi: G\to G/H$ with the kernel $Ker\phi=H.$

We need to use $gH=Hg$ to establish that the multiplication in $G/H$ is well-defined: $(gH)(g'H)=(gH)(Hg')=g(HH)g'=g(Hg')=(gg')H.$
Then $G/H$ is a group with the identity element $H,$ as associativity in $G/H$ follows from the associativity in $G,$ and $(gH)^{-1}=(g^{-1})H.$
It can be readily checked that $\phi$ is a homomorphism, and that $H= Ker\phi.$ Indeed, $\phi(gg')=(gg')H=(gH)(g'H).$ Also, $\phi(h)=hH=H$ for $h\in H,$ and so $H\leq Ker\phi.$ On the other hand if $g\in G-H$ then $\phi(g)=gH\neq H,$ and so $H\geq Ker\phi.$

Matrix groups
Subgroups of $GL_n(\mathbb{F}),$ the group, with respect to matrix multiplication, of invertible $n\times n$ matrices with entries in a field $\mathbb{F},$ provide a rich playground for various examples of normal subgroups and quotient groups. To this end, we will work with block matrices, i.e. we will partition our matrices into rectangular blocks of appropriate size, so that this block decomposition is preserved under the matrix multiplication. The simplest case is the following partition into 4 blocks (i.e. $2\times 2$ block matrices).

Given an $n\times n$ matrix $A$ and positive integer $p we can view $A$ as $A=\begin{pmatrix} A_{11}&A_{12}\\A_{21} & A_{22}\end{pmatrix},$ where $A_{11}$ is a $p\times p$ matrix, $A_{22}$ a $(n-p)\times (n-p)$ matrix (and certainly $A_{12}$ and $A_{21}^T$ being of size $p\times (n-p)$).

Let us fix $n$ and $p.$
First, we need to convince ourselves (an easy exercise in matrix multiplication) that the following multiplication law works for two block matrices $A$ and $B$:
$\displaystyle AB=C=\begin{pmatrix} A_{11} B_{11}+A_{12}B_{21}& A_{11}B_{12}+A_{12}B_{22}\\ A_{21}B_{11}+A_{22}B_{21}&A_{21}B_{12}+A_{22}B_{22}\end{pmatrix}$
It should of course look very familiar, as this is exactly how two $2\times 2$ matrices are multiplied.
When $A_{21}=B_{21}=0,$ we see that the shape is preserved, i.e. $C_{21}=0.$ More precisely
$\displaystyle AB=C=\begin{pmatrix} A_{11} B_{11}& A_{11}B_{12}+A_{12}B_{22}\\ 0 &A_{22}B_{22}\end{pmatrix}$
We see that the blocks with indices 11 and 22 behave as if “nothing around them matters”, as if they sit in their own groups $G_p(\mathbb{F})$ and $G_{n-p}(\mathbb{F})$ (that is, assuming the matrices we talk about are invertible). We can formalise this observation as follows.

Let $H be a subgroup of $2\times 2$ block matrices, with diagonal blocks of sizes $p\times p$ and $(n-p)\times (n-p),$ and such that $A_{21}=0$ for all $A\in H.$ Then

1. the maps
$\phi_1 : H\to GL_p(\mathbb{F}):\quad A\mapsto A_{11}$ and
$\phi_2 : H\to GL_{n-p}(\mathbb{F}):\quad A\mapsto A_{22}$
are group homomorphisms.
2. $Ker(\phi_k)=\{A\in H\mid A_{kk}=I\}$ for $k=1,2.$

The reader is encouraged to provide a proof.
More generally, this result can be interpreted in terms of the actions of $H$ on the $n-$dimensional vectorspace $V$ over $\mathbb{F}.$

Exercise. Prove that the intersection of two normal subgroups of a group is a normal subgroup. Based on this, describe $Ker(\phi_1)\cap Ker(\phi_2),$ and derive that $\{A\in H\mid A_{11}=I, A_{22}=I\}$ is a normal subgroup of $H.$

Note that $Im(\phi_k)$ (where $k=1$ or $K=2$) cannot be obtained by simply setting the remaining off-diagonal block of $H$ to 0, and the remaining diagonal block to the identity matrix. For instance, let $H be generated by the matrix $X=\begin{pmatrix} \mathbf{i}&0\\ 0& -1\end{pmatrix},$ where $\mathbf{i}=\sqrt{-1}.$ Then $H$ is a cyclic group of order 4, and it satisfies the conditions of the above lemma with $n=2,$ $p=1.$ It is easy to see that $\phi_1$ is an isomorphism, and $\phi_2$ is a homomorphism with the kernel generated by $X^2.$

Let us denote $K_k=Ker(\phi_k),$ and $K_{12}=K_1\cap K_2.$
As $|H|=|K_k||Im(\phi_k)|,$ by the usual argument involving decomposition of $H$ into the cosets of $K_k,$ computing $|H|$ for $\mathbb{F}= \mathbb{Z}_q,$ for $q$ prime, can be accomplished by computing $|K_k|$ and $|Im(\phi_k)|$ separately. In turn, $|K_k|=|K_{12}||Im(\phi_{3-k}|,$ so we can simplify the task of computing $|K_k|,$ too.

To illustrate this principle, let us compute the order of
$\displaystyle H=\{A\in GL_n( \mathbb{Z}_q)\mid A_{21}=0.\}\qquad\qquad (1)$
First we derive the following:

$|GL_m( \mathbb{Z}_q)|=(q^m-1)(q^m-q)(q^m-q^2)\dots (q-1),$ where $q$ is prime.

Indeed, we can fill in the 1st row of an $m\times m$ matrix in $q^m-1$ way (everything goes, except all zeros). Fixing this 1st row vector $v_1,$ we can again fill in the 2nd row with entries of any vector $v_2,$ which is linearly independent (to get an invertible matrix) of $v_1,$ i.e. lies outside of the subspace $V_1$ spanned by $v_1.$ As $|V_1|=q,$ we get $q^m-q$ ways to choose $v_2.$ Now we fix $v_2,$ as well, and fill in the 3rd row with entries of any vector $v_3,$ which is linearly independent of $v_1$ and $v_2,$ i.e. lies outside of the subspace $V_2$ spanned by $v_1$ and $v_2.$ As $|V_2|=q^2,$ we get $q^m-q^2$ ways to choose $v_2.$
So when filling the $(\ell+1)$-th row, we have $q^m-q^\ell$ choices, giving us the formula above.

Now, we claim that $|K_{12}|=q^{p(n-p)}.$ Indeed, $K_{12}$ consists of $A$ such that $A_{21}=0,$ $A_{11}=I,$ $A_{22}=I,$ whereas we can fill the entries of $A_{12}$ without any strings attached – the result will be invertible. Thus we have
$|H|=(q^p-1)(q^p-q)\dots (q-1) q^{p(n-p)} (q^{n-p}-1)(q^{n-p}-q)\cdots (q-1).$

Excercise. Compute $|H'|$ for $H'=\{A\in H\mid \det A=1\}$ and $H$ as in (1).
(Hint: use the fact that $\det$ is a group homomorphism.)