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Let $p$ be prime integer. I've read that every subgroup of $(\mathbb{Q}_p/\mathbb{Z}_p)^2$ has the form

\begin{equation} (\mathbb{Q}_p/\mathbb{Z}_p)^e\times U \end{equation}

with $0\le e \le 2$ and $U$ a finite group. My question is: why?

Some ideas:

$\mathbb{Q}_p/\mathbb{Z}_p$ is a $p$-divisible group that has only finite subgroups. One idea could be: consider $H$ a subgroup of $(\mathbb{Q}_p/\mathbb{Z}_p)^2$ and consider $D$ its maximal divisible subgroup. If $D\cong (\mathbb{Q}_p/\mathbb{Z}_p)^e$ and has finite index in $H$, then we're done since divisible subgroups are always direct summands. But I don't know if (or why) the hypotheses are true.

Fraz
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    The main step is to show that if $H $ is infinite then it contains a sequence $(a_n), p a_n = a_{n-1},a_1\ne 0,pa_1=0$. If it is infinite then take $f\in Aut(Q_p^2/Z_p^2)\cong GL_2(Z_p)$ such that $f(H)=Q_p/Z_p\times 0$. Then quotient $f(H)$ by $Q_p/Z_p\times 0$ to obtain a subgroup of $(Q_p/Z_p \times Q_p/Z_p)/(Q_p/Z_p \times 0)\cong Q_p/Z_p$. – reuns May 17 '20 at 19:31
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    I meant such that $f(a_n) = (p^{-n},0)$. The quotient $f(H)$ by $Q_p/Z_p\times 0$.. – reuns May 17 '20 at 23:20
  • The existence of the sequence is ok, if $H$ is infinite. Now I try to interpret your hint: consider $K$ the subgroup generated by the sequence $(a_n)$: this is isomorphic to $\mathbb{Q}_p/\mathbb{Z}_p$ via the map $a_n\mapsto p^{-n}$. This means that $K$ contins a copy of $\mathbb{Q}_p/\mathbb{Z}_p$ as direct summand, since $\mathbb{Q}_p/\mathbb{Z}_p$ is divisible. Hence I can work on $K/(\mathbb{Q}_p/\mathbb{Z}_p)$: if it is finite, I'm done. If it is infinite, I make the same reasoning done before. – Fraz May 18 '20 at 05:52
  • Edit: the existence of that sequence is not ok, I still can't manage to prove it precisely. – Fraz May 18 '20 at 06:41

1 Answers1

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By Goursat's lemma (see this link), any subgroup $H$ of $(\mathbb{Q}_p/\mathbb{Z}_p)^2$ corresponds to a quintuplet $(G_1,G_2,N_1,N_2,\phi)$ where $G_1$ and $G_2$ are subgroups of $\mathbb{Q}_p/\mathbb{Z}_p$, $N_1$ is a subgroup of $G_1$, $N_2$ is a subgroup of $G_2$ and $\phi:G_1/N_1\to G_2/N_2$ is an isomorphism. Explicitely, we have that

$H=\{(x,y)\in (\mathbb{Q}_p/\mathbb{Z}_p)^2 : x\in G_1, \ y\in G_1, \ \phi(x+N_1)=y+N_2\}$.

Let's consider all possible cases.

(1) Assume that $G_1$ and $G_2$ are both finite. Since $H$ is a subgroup of $G_1\times G_2$, we obtain that $H$ is finite.

(2) If one of $G_1$ and $G_2$ is finite and the other one is $\mathbb{Q}_p/\mathbb{Z}_p$, from the fact that $G_1/N_1\cong G_2/N_2$ we obtain that $N_1=G_1$ and $N_2=G_2$, since $\mathbb{Q}_p/\mathbb{Z}_p$ has no finite quotients other than $\{0\}$. Therefore, $H=G_1\times G_2$ has the required form.

(3) We're left with the case $G_1=G_2=\mathbb{Q}_p/\mathbb{Z}_p$.

(3a) If $N_1=N_2=\mathbb{Q}_p/\mathbb{Z}_p$, then $H=(\mathbb{Q}_p/\mathbb{Z}_p)^2$.

(3b) We're left with the case $N_1$ and $N_2$ finite. Then, there are $n_1,n_2\in\mathbb{N}$ such that $N_1=p^{-n_1}\mathbb{Z}_p/\mathbb{Z}_p$ and $N_n=p^{-n_2}\mathbb{Z}_p/\mathbb{Z}_p$. Since the multiplication by $p^{n_1}$ (resp. $p^{n_2}$) yields an isomorphism between $G_1/N_1$ and $\mathbb{Q}_p/\mathbb{Z}_p$ (resp. between $G_2/N_2$ and $\mathbb{Q}_p/\mathbb{Z}_p$), one can show that the isomorphism $\phi:G_1/N_1\to G_2/N_2$ is given by the multiplication by an element $z_\phi\in\mathbb{Q}_p$ with valuation $n_1-n_2$. Then one can easily show that the map $\Phi: \mathbb{Q}_p/\mathbb{Z}_p\times N_2\to H$ defined by $(a,b)\mapsto (a,z_\phi a+b)$ is an isomorphism, with inverse $(x,y)\mapsto(x,y-z_\phi x)$.

Fraz
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