Using generators to check for a normal subgroup

[gauss mouse]

Homework Statement

Perhaps I should say first that this question stems from an attempt to show that in the group
[itex] \langle x,y|x^7=y^3=1,yxy^{-1}=x^2 \rangle,\ \langle x \rangle [/itex] is a normal subgroup.

Let [itex] G [/itex] be a group with a subgroup [itex] H [/itex]. Let [itex] G [/itex] be generated by [itex]A\subseteq G [/itex]. Suppose that [itex] H [/itex] is closed under conjugation by elements from [itex] A [/itex] in the sense that [itex] aha^{-1}\in H [/itex] for any [itex] a\in A [/itex] and any [itex] h\in H [/itex]. Is it then true that [itex] H [/itex] is a normal subgroup of [itex] G [/itex]?

The Attempt at a Solution

I know that [itex] G [/itex] is the set of all products of elements from [itex] A\cup A^{-1} [/itex] where [itex] A^{-1}:=\{a^{-1}|a\in A\}. [/itex] Since we are assuming that [itex] aha^{-1} \in H[/itex] for any [itex] a\in A [/itex] and any [itex] h\in H [/itex], it will be enough to show that [itex] a^{-1}ha\in H[/itex] for any [itex] a\in A [/itex] and any [itex] h\in H [/itex]. However I have not been able to show this.

 

[Nino MMP]

I have tried to prove this by contradiction, but in my attempts I get stuck in a case where I don't know if it is possible to find an element of H which does not satisfy the condition. If my attempt is wrong, or if I am missing something, please let me know. Any help will be appreciated. Thank you.