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


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.

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.

Author: mathtuition88

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