$G$ is a non-cyclic group of order $p^2$. Then $\langle g \rangle \cap \langle k \rangle = \{e\}$

Question

I need to prove the following result:

Let $G$ be a group of order $p^2$. Then $G \simeq \mathbb Z_{p^2}$ or $G \simeq \mathbb Z_p \times \mathbb Z_p$.

and the results I have at my disposal are the following:

1. Lagrange's Theorem and it's usual corollaries (order of element divides order of the group, etc.)
2. The centre $Z(G)$ of a $p$-group $G$ has order at least $p$.
3. Let $H, K \leq G$. Then $G \simeq H \times K$ if:
   • $H \cap K = \{e\}$,
   • $\forall h\in H, \forall g\in G, gh = hg$, and
   • $G = HK = \{hk : h \in H \wedge k \in K\}$.
4. $\displaystyle |HK| = \frac{|H||K|}{|H\cap K|}.$

My attempt was the following.

Suppose $G$ is not cyclic, that is, there is no element $g\in G$ with order $p^2$. Thus any $g \in G$ has $|g| < p^2$, which by a corollary to Lagrange's theorem must divide $p^2$, that is, $|g| = 1$ or $p$. If $g \neq e$, then $|g| \neq 1$, therefore we must have $|g| = p$. In particular, take $g \in Z(G)$ with $g \neq e$ (we know we can take such a $g$ since $|Z(G)| \geq 2$ by result 2 above), and consider the cyclic subgroup $H = \langle g \rangle \leq G$ generated by $g$. Now take some $k \in G\setminus H$, and let $K = \langle k \rangle \leq G$.

All the elements of $K$ commute with those of $H$ (since $H$ is generated by $g\in Z(G)$), and $|HK| = |H| |K| / |H \cap K| = (p \cdot p)/ 1 = p^2 = |G|$, so $HK = G$.

All that remains to show is that $H \cap K = \{e\}$, but I don't know how to do it. Is there an easier way to go about the proof? I appreciate any assistance.

Answer 1

Note that $H \cap K$ is a subgroup since $H$ and $K$ are both subgroups and intersection of subgroups is subgroups.

The order of $H \cap K$ must be smaller than $p$, because there is one element in $K$ that is not in $H$, by construction.

The order must divide $p^2$ by Lagrange.

Therefore, the order must be $1$.

Therefore, $H \cap K = \{e\}$.