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The hint for this problem essentially lays out a strategy for the proof (which I will outline below) I would just like some help seeing how to show some of the properties listed. The hint says

If there exists an element of order $p^2$ then $G \cong \mathbb{Z}_{p^2}$ so assume that this is not the case and deduce that every nonidentity element must have order $p$. Then prove that for any two elements $g,h \in G$ either $\langle g \rangle = \langle h \rangle$ or $\langle g \rangle \cap \langle h \rangle = \{ e \}$. Then prove there exists two nonidentity elements so that $\langle g \rangle \cap \langle h \rangle = \{ e \}$ and appeal to proposition $3.7.1$.

For this problem I have so far:

Assume there does not exist an element $g \in G$ such that $\lvert g \rvert = p^2$. Then by Lagrange's Theorem we know that the order of any element of a group must divide the order of the group. Hence the only possible orders for elements of $g$ are $1,p$ or $p^2$ but by assumption no such element with order $p^2$ exists. Then the order of any element of $G$ is either $1$ or $p$. For any $g \in G$ if $\lvert g \rvert = 1$ then $g = e$ the identity element for $G$, by definition of the order of an element. Hence any nonidentity element must have order $p$.

Now for the next part: my original instinct was to show that if $\langle g \rangle$, $\langle h \rangle$ both had order $p$ then they were equivalent since they were both Sylow $p$-subgroups and by the Third Sylow Theorem we have $$n_p(G) \mid \frac{\lvert G \rvert}{p^2} = \frac{p^2}{p^2} = 1 \implies n_p(G) \mid 1$$ and $$n_p(G) \equiv 1 (\text{mod p})$$ Hence $n_p(G) = 1$. In other words, there's only $1$ sylow $p$-subgroup in $G$. Hence if any $2$ elements have order $p$ then the cyclic subgroups generated by them must be the same, namely the Sylow $p$-subgroup. However, in the next part we see this can't be the case since they ask us to show that any $2$ nonidentity elements (i.e., ones with order $p$) have trivial intersection (and thus cannot be the same subgroup).

Given this I would appreciate some guidance regarding how to prove the $2\text{nd}$ and $3\text{rd}$ parts of this problem.

EDIT

Right as I posted this I just realized that the cyclic subgroups generated by an element of order $p$ would not be a Sylow $p$-subgroup since the order of the subgroup is not maximal. In other words $\lvert \langle g \rangle \rvert = p$ Where only the cyclic subgroup generated by an element of order $p^2$ would generate a Sylow $p$-subgroup as $p^2$ is maximal in the order of $G$.

Shaun
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Numerical Disintegration
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  • The hint will go right through. A sledgehammer would be the structure theorem (idk if you know it). Oops you need that it's abelian. – calc ll Jul 16 '22 at 02:08
  • I would first use the lemma that the center of $G$ must be non-trivial, because $G$ is a $p$-group (follows from the class equation). Then use the fact that if $G/Z(G)$ is cyclic, we actually have $G=Z(G)$ (another standard exercise we have covered on this site already). Then you can start looking at the orders of the elements. Or invoke the structure theorem of finitely generated abelian groups. The latter is overkill for this though. – Jyrki Lahtonen Jul 16 '22 at 02:24
  • Once you get two distinct cyclic groups of order $p$, it should be their direct product. – calc ll Jul 16 '22 at 02:35
  • What's prop $3.7.1$? – calc ll Jul 16 '22 at 04:45
  • Can we say something like clearly if we have a group of order p2 then it is abelian. So, the isomorphic copy can be cyclic or not. Thus, we have 2 choices for the isomorphism – nozalp10 Jul 16 '22 at 09:19

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