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Hello Mathematics Community. I am unsure about how to solve this problem involving the number of subgroups in an abelian group.

How many subgroups of order $p^2$ does the abelian group $\mathbb{Z}_p \oplus \mathbb{Z}_p$ have?

I have first tried to use the First Isomorphism Theorem, but I do not think it helped. Then I considered the following:

The order of our group is $|\mathbb{Z}_p \oplus \mathbb{Z}_p|=p^2$. By the structure theorem of finite abelian groups, then $\mathbb{Z}_p \oplus \mathbb{Z}_p \cong \mathbb{Z}_{p^2}$. Now I am stuck and do not know how to proceed, or whether this is the right direction to solving the problem.

As always, any help is greatly appreciated; thanks in advance.

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    $\begingroup$ You already say that the order of the group is $p^2$. So you have a bag with $X=p^2$ items. You take $X$ items and put them in another bag. How many ways to do this? $\endgroup$ – ypercubeᵀᴹ Nov 7 '14 at 15:32
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    $\begingroup$ Note that (if I understand what you mean by $\oplus$) $\mathbb{Z}_{p} \oplus \mathbb{Z}_{p}$ is NOT isomorphic to $\mathbb{Z}_{p^{2}}$ by the structure theorem. In fact, the structure theorem says the opposite. Consider the easy counterexample of $\mathbb{Z}_{2} \oplus \mathbb{Z}_{2} \cong V_{4}$, which is not isomorphic to $\mathbb{Z}_{4}$. $\endgroup$ – Alex Wertheim Nov 7 '14 at 15:33
  • $\begingroup$ Note that there are two groups of order $p^2$ - one is cyclic and has elements of order $p^2$ - that's $\mathbb Z_{p^2}$ and the other is not cyclic and all elements other than the identity have order $p$ - think about the groups you know of order $4$ - this is $\mathbb Z_p \oplus \mathbb Z_p$ - the structure theorem doesn't tell you they are the same. $\endgroup$ – Mark Bennet Nov 7 '14 at 15:36
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The big group has, as you pointed out, only $p^2$ elements!

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    $\begingroup$ So the fact that the big group ($\mathbb{Z}_p \oplus \mathbb{Z}_p$) only has $p^2$ elements implies that there is only one such subgroup of that order? $\endgroup$ – Jamil_V Nov 7 '14 at 15:37
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    $\begingroup$ Yes, the only $p^2$-element subgroup is the whole group. Any proper subgroup would have fewer than $p^2$ elements (the only possibilities are $1$ and $p$, but we don't need to know that). Zero machinery needed. $\endgroup$ – André Nicolas Nov 7 '14 at 15:38
  • $\begingroup$ Ok, I think I am understanding now. Is it because every group is also a subgroup of itself? $\endgroup$ – Jamil_V Nov 7 '14 at 15:43
  • $\begingroup$ Yes, a group is a subgroup of itself. And if you knoe that $G$ hs $n$ elements, and $H$ is a subgroup of $G$ with $n$ elements, then $H$ must be all of $G$. $\endgroup$ – André Nicolas Nov 7 '14 at 16:46

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