normal subgroup – Mathtuition88

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 $q\leq p-1<p$ .