subgroup of $\Bbb Q_p^×$ of index p contains $(\Bbb Q_p^×)^p$ as subgroup

Question

Let $H$ be a subgroup of $\Bbb Q_p^×$ of index p. Then, I would like to prove $H$ contains $(\Bbb Q_p^×)^p$ as subgroup.

I know $(\Bbb Q_p^×)^p$ has index $p^2$ because $ \Bbb Q_p^×/(\Bbb Q_p^×)^p$ has order $p^2$.

I encountered this question when I was trying to count $\Bbb Q_p^×$'s abelian extension of order $p$.

Thank you in advance.

Answer 1

If $G$ is any group (written multiplicatively) and $H$ is any normal subgroup of $G$ of index $n$, then $H$ must contain $G^n$.

This is because the quotient group $G/H$ has order $n$ and hence the $n$-th power of any element of $G/H$ is the identity element.

Translating back to $G$, we see that $g^n \in H$ for any $g \in G$.