## Advanced Method for Proving Normal Subgroup

For beginners in Group Theory, the basic method to prove that a subgroup $H$ is normal in a group $G$ is to show that “left coset = right coset”, i.e. $gH=Hg$ for all $g\in G$. Variations of this method include showing that $ghg^{-1}\in H$, $gHg^{-1}=H$, and so on.

This basic method is good for proving basic questions, for example a subgroup of index two is always normal. However, for more advanced questions, the basic method unfortunately seldom works.

A more sophisticated advanced approach to showing that a group is normal, is to show that it is a kernel of a homomorphism, and thus normal. Thus one often has to construct a certain homomorphism and show that the kernel is the desired subgroup.

Example: Let $H$ be a subgroup of a finite group $G$ and $[G:H]=p$, where $p$ is the smallest prime divisor of $|G|$. Show that $H$ is normal in $G$.

The result above is sometimes called “Strong Cayley Theorem”.

Proof: Let $G$ act on $G/H$ by left translation.

$G\times G/H\to G/H$, $(g,xH)\to gxH$.

This is a group action since $1\cdot xH=xH$, and $g_1(g_2\cdot xH)=g_1g_2xH=(g_1g_2)\cdot xH$.

This action induces a homomorphism $\sigma:G\to S_{G/H}\cong S_p$. Let $g\in\ker\sigma$. $\sigma(g)(xH)=xH$ for all $xH\in G/H$, i.e. $gxH=xH$ for all $x\in G$. In particular when $x=1$, $gH=H$. This means that $g\in H$. So we have $\ker\sigma\subseteq H$.

Suppose to the contrary $\ker\sigma\neq H$, i.e. $[H:\ker\sigma]>1$. Let $q$ be a prime divisor of $[H:\ker\sigma]$.

We also have

$[G:\ker\sigma]=[G:H][H:\ker\sigma]=p[H:\ker\sigma]$

By the First Isomorphism Theorem, $G/ker\sigma\cong\text{Im}\ \sigma\leq S_p$. By Lagrange’s Theorem, $[G:\ker\sigma]\mid p!$, i.e. $p[H:\ker\sigma]\mid p!$. This implies $[H:\ker\sigma]\mid(p-1)!$. Finally, $q\mid(p-1)!$ implies $q\leq p-1.

However, $q\mid [H:\ker\sigma]$ implies $q\mid[G:\ker\sigma]=\frac{|G|}{|\ker\sigma|}$ which implies $q\mid|G|$.

This is a contradiction that $p$ is the smallest prime divisor of $|G|$. Thus, $H=\ker\sigma$ and therefore $H$ is a normal subgroup.

This proof is pretty amazing, and hard to think of without any hints.