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?


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?

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.