Group Theory - Prime Index

Question

The index $(G : H)$ of a subgroup H of G is the number of cosets of H. Let H be a normal subgroup with (G : H) = p, where p is a prime, and let a be an element of G that is not in H. Show that for every $x ∈ G$, there is some integer n such that $xa^n$ $∈ H$

I don't really know how to start this proof. I don't have any theorems that involve prime indexes and haven't really worked with indexes at all in the first place. Any advice or solution would be appreciated!

Answer 1

We have that

$$H\lhd G\;\text{and}\;[G:H]=\left|G/H\right|=p\implies\color{red}{\forall\,x\in G\setminus H}\;,\;\;(xH)^p:=x^pH=H\iff x^p\in H$$

and thus we have that (observe that $\;x\notin H\iff x^{-1}\notin H\;\ldots$) :

$$\forall\;a\in G\setminus H\;:\;\;G/H=\langle x^{-1}H\rangle=\langle aH\rangle\implies \exists n\in\Bbb Z\;\;s.t.\;\; x^{-1}H=a^nH\iff xa^n\in H$$

Answer 2

If $a\in G$ but $a\notin H$ then the coset $aH$ is a non-identity element of the quotient group $G/H$.

But this quotient has order $p$ hence is cyclic, generated by $aH$ (since $aH$ must have order $p$ by Lagrange).

So consider the coset $xH$ for some $x\in G$. It must have an inverse $x^{-1} H$ in the quotient group. We must have that $x^{-1}H = (aH)^k$ for some $k$ and so $(xa^k)H = H$. Thus $xa^k\in H$.