# Group of order prime squared

You are given the following information: $$G - {\text{ is a group}},\,\,\,\left| G \right| = {p^2},\,\,\,p - {\text{ is prime}}{\text{.}}$$Prove that all nontrivial subgroups of $G$ are cyclic. What is the maximum number of different nontrivial subgroups $G$ might have at most?

I tried to tackle this problem in the following way. First, i refer to Lagrange's theorem to claim that the orders of the elements of $G$ can be $1$, $p$, or ${p^2}$. Then i consider the case, when $$\exists a \in G,\,\,\left| a \right| = {p^2} \Rightarrow A = \left\langle a \right\rangle ,\,\,A = G$$which shows that if $G$ has an element $a$ of order ${p^2}$, then this element generates a subgroup $A$ that will have all elements from $G$ and so it must be cyclic. Otherwise, there exists an element $b \in G,\,\,\left| b \right| = p$ and it generates a subgroup $B < G,\,\,\left| B \right| = p$. Any group of prime order is cyclic. Therefore, all subgroups of $G$ are cyclic.

Should we treat a subgroup with order same as $G$ as a nontrivial subgroup, since its clear that any such subgroup contains same elements as $G$. Now, about the different nontrivial number of subgroups. I assume that from showing that a subgroup of order ${p^2}$ or $p$ is cyclic, we conclude that $G$ is also cyclic and this means we have only $2$ different nontrivial subgroups (at most). Am i correct?

• I would usually not consider the entire group to be a trivial subgroup, but it seems from the context that whoever made the question does (I would have phrased it as proper instead, as certainly the trivial subgroup is cyclic). – Tobias Kildetoft Jan 25 '16 at 19:27
• @DietrichBurde should we consider both cases separately? If so, how it would affect maximum number of different nontrivial subgroups $G$ might have at most? – user1812 Jan 25 '16 at 19:33
• the exercice could have considered any group with $pq$ elements the product of two primes – reuns Jan 25 '16 at 19:36
• @user1812 Yes, you can count the maximal number of distinct subgroups of $C_{p^2}$, and the ones of $C_p\times C_p$. What do you get ? – Dietrich Burde Jan 25 '16 at 19:40
• @user1952009 That would make the second part considerably more complicated. – Tobias Kildetoft Jan 25 '16 at 19:41

The order of a non trivial subgroup $H$ of $G$ divides $p^2$ so it is $1$ or $p$ every group of order $p$ is cyclic and isomorphic to $\frac {\bf Z}{p\bf Z}$.
The intersection of two distinct subgroups of order $p$ is the neutral element. This implies that if $n$ is the maximal number of distinct subgroup of order $p$, $n(p-1)+1=p^2$ so $n=p+1$.
$H < G \implies$|H|divides $|G|=p^2$ thus $|H|=p$ ($1$ and $p^2$) are trivial.
Every group with $p$ elements is cyclic, because every element that is not 1 has order $p$ and thus generates the whole group.