# All subgroups of a group with order the square of a prime

So I have a group of order $p^2$ (where $p$ is a prime number) and I'm wondering how many subgroups it can have. By Lagrange's theorem I know that if a subgroup exists its order has to divide the order of the group i.e. $p^2$ in other words it has to be of order $1$, $p$ or $p^2$. Of order $1$ we have only the trivial group and of order $p^2$ the group itself while the existence of subgroups of order $p$ is enstablished by Cauchy's theorem but how many of them are there?

I tried to reduce the problem to a combinatorial one however I'm not too familiar with this branch. I reasoned as it follows:

i) I have to choose $p$ element from a set which has $p^2$

ii) the unit must be in the subgroup so we have only to choose $p-1$ elements from a set which has $p^2-1$

iii) for every element the inverse must be in the subgroup so we have only to choose $\frac {p-1}2$ elements (if $p\neq 2$) from a set which has $p^2-1$

iv) for every two elements their composition must be in the subgroup however I don't know how to use this fact and so I don't know how to end the problem. Maybe using the criterion for subgroups can shorten the computation but I'm not sure how to use it.

Tell me if my reasoning is correct and how I should end this exercise

• Possible duplicate of Number of subgroups of groups with prime power order – Arnaud D. Jan 15 '17 at 11:55
• @ArnaudD. sorry but I think my question is different, I wrote my approach and asked if it's correct or not not only to find the number of subgroups – Renato Faraone Jan 15 '17 at 11:57
• @RenatoFaraone I wrote a complete hint for this question yet I deleted it as I read you only want to know whether your reasoning is correct. I think it is...but I doubt whether it'll take you very far away. Better, try to think in the two possibilities for group: both are abelian, but one is the cyclic group $\;C_{p^2}\;$ (and here we have no problem as there's one single subgroup of every order dividing $\;p^2\;$), and the other one is the elementary abelian $\;C_p\times C_p\;$ , which can be made into a linear space and then things get much easier... – DonAntonio Jan 15 '17 at 12:03