# $\Bbb Q_p^×/(\Bbb Q_p^×)^p$ is isomorphic to $\Bbb Z/p\Bbb Z×\Bbb Z/p\Bbb Z$

I would like to prove $$\Bbb Q_p^×/（\Bbb Q_p^×）^p$$ is isomorphic to $$\Bbb Z/p\Bbb Z×\Bbb Z/p\Bbb Z$$ as a group.

I know $$\Bbb Q_p^×$$ is isomorphic to $$\Bbb Z×\Bbb Z_p×\Bbb Z/（p-1）\Bbb Z$$ using formal logarithm.

I tried to find group homomorphism $$\Bbb Q_p^×$$$$\Bbb Z/p\Bbb Z×\Bbb Z/p\Bbb Z$$ of which kenerl is $$（\Bbb Q_p^×）^p$$, but I couldn't.

• Thank you!, but is $G^p$ isomorphic to $\Bbb pZ×\Bbb pZ_p×\Bbb Z/（p-1）\Bbb Z$? Commented Mar 12, 2021 at 13:27
• This is false when $p = 2$: the group $\mathbf Q_2^\times/(\mathbf Q_2^\times)^2$ has order $8$, not $4$.
– KCd
Commented Mar 12, 2021 at 20:07

Is it clear to you that (Hensel lemma) for $$p$$ odd: $$\Bbb{Q}_p^\times = p^\Bbb{Z}\times (1+p\Bbb{Z}_p)\times \langle \zeta_{p-1}\rangle$$
Then $$(1+p\Bbb{Z}_p)^p=1+p^2\Bbb{Z}_p$$ because it contains $$(1+cp^k)^p\equiv 1+ c p^{k+1}\bmod p^{k+2}$$ so any $$1+p^2 a$$ is of the form $$\prod_{k\ge 1} (1+c_k p^k)^p$$ with $$c_k\in 0 \ldots p-1$$.
Whence $$\Bbb{Q}_p^\times/(\Bbb{Q}_p^\times)^p\cong p^\Bbb{Z}/p^\Bbb{pZ}\times (1+p\Bbb{Z}_p)/(1+p^2\Bbb{Z}_p)\cong \Bbb{F}_p\times \Bbb{F}_p$$
For $$p=2$$ it is the same except that $$(1+2\Bbb{Z}_2)^2=1+8\Bbb{Z}_2$$ so $$\Bbb{Q}_2^\times/(\Bbb{Q}_2^\times)^2\cong \Bbb{F}_2\times \Bbb{F}_2\times \Bbb{F}_2$$.