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

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 $g\in A$. 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. Commented Mar 15, 2021 at 16:43
• Sorry to be that person, but for $p=2$, the index of the $p$-powers is $2^\color{red}{3}$. Commented Mar 15, 2021 at 17:04

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 $$g \in G$$.