# Subgroup with finite index contains a normal group [duplicate]

Let H be a subgroup of finite index of an infinite group G. Prove that G has a normal subgroup of finite index which is contained in H.

I am not sure how to start on this problem, and would appreciate any suggestions.

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• If $[G:H] = n,$ try to define a group homomorphism $\phi: G \to S_{n},$ using the existence of $H.$ – Geoff Robinson Nov 17 '14 at 16:30

Let $\;X:=H/G:=$ the set of left cosets of $\;H\;$ in $\;G\;$ , and define an action

$$G\times X\to X\;,\;\;g'\cdot(gH):=(g'g)H\;,\;\;\forall\;g',g\in G$$

Show the above is indeed an action, and it thus determines a groups homomorphism $\;\phi:G\to Sym_X\cong S_n\;$ , with $\;n=[G:H]\;$ , given by $\;\phi(g')(gH):=(g'g)H\;$ .

Prove now that

$$\ker\phi\le H\;\;\text{and}\;\;G/\ker\phi\cong T\le S_n$$

so that $\;N:=\ker\phi\;$ is the wanted normal subgroup.

BTW, $\;N\;$ as above is called the core of $\;H\;$, and it is characterized for being the maximal normal subgroup of $\;G\;$ contained in H.

Extra exercise: prove that

$$N=\ker\phi=\bigcap_{g\in G}H^g\;,\;\;\text{with}\;\;H^g:=g^{-1}Hg$$