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.

If $B$ is a normal subgroup of $A$ of index $n$ then for any $gB\in A/B, g^nB=1\, B$ ie. $g^n\in B$. When $B$ is not normal it stays true that $g^{n!}\in B$ for all $g\in A$. — reuns, Commented Mar 15, 2021 at 16:39
How are you typing your formulas? There's this weird space before and after the parentheses that seems to be part of the parenthesis, and results in bad spacing when rendered. — Arturo Magidin, Commented Mar 15, 2021 at 16:43
Sorry to be that person, but for $p=2$, the index of the $p$-powers is $2^\color{red}{3}$. — Torsten Schoeneberg, Commented Mar 15, 2021 at 17:04

WhatsUp · Accepted Answer · 2021-03-15 16:38:46Z

7

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 $g \in G$.

answered Mar 15, 2021 at 16:38

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