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Let $G$ be a group of order $p^2$ and put $\mathcal A=\{U\leq G, \#U=p\}$.

What is $\#\mathcal A$?

If $G$ is cyclic, then $G$ is generated by some element $x$ of order $p^2$. It seems like there is only one order $p$ subgroup here.

Question 1 How do I prove this rigorously?

If $G$ is not cyclic, then there is no element of order $p^2$. By Lagrange's theorem, any nontrivial element $g\in G$ then has order $p$, hence generates an order $p$ subgroup $\langle g \rangle\in \mathcal A$.

Question 2 Which of these groups $\langle g\rangle$ coincide?

Question 3 Are there any more order $p$ subgroups?

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Since we know

$$G\cong C_{p^2}\;\;\;\text{or}\;\;\;G\cong C_p\times C_p$$

there are not many options: in the first case $\; |A|=1\;$ as any cyclic group of finite order has one single subgroup of any order dividing the group's.

In the second case: since we can consider $\;G\;$ a a vector space of dimension two over the prime field of characteristic $\;p\,,\,\,\Bbb F_p\;$, the wanted number equals the number of different subspaces of dimension one this space has. Can you take it from here?

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  • $\begingroup$ Why does a cyclic group of finite order have exactly one subgroup of any order dividing the group order? $\endgroup$ – MyNameIs Jul 20 '16 at 11:10
  • $\begingroup$ How do we know that $G\cong C_{p^2}$ or $G\cong C_p\times C_p$? Sure, those two are possible, but why no other group? $\endgroup$ – MyNameIs Jul 20 '16 at 11:11
  • $\begingroup$ @MyNameIs Both questions are pretty elementary and should be known before this or together with this question's level, yet we can resume: first question: prove first that for every divisor of the order there's a subgroup of that order, and then use the fact that if $\;G=\langle x\rangle\;$ is of order $\;n\in\Bbb N\;$ , then we get that $\;G=\langle x^k\rangle\iff gcd(k,n)=1\;$ . For the second question: we may need the basic fact that a finite $\;p\,-$ group always has a non-trivial center, and also that $\;G/Z\;$ cannot be cyclic and non-trivial... $\endgroup$ – DonAntonio Jul 20 '16 at 11:16
  • $\begingroup$ thank you for the explanations. I like the idea of going to vector spaces in the second case. $\endgroup$ – MyNameIs Jul 20 '16 at 17:30

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