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Let $p$ be a prime and $G$ be a group of order $p^2$. Show that $Z(G)\neq 1$.

Is there a proof of this nice fact that doesn’t use the class equation?

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Probably the easiest proof to understand uses the class equation (counting elements in conjugacy classes) or group actions (orbit-stabilizer). Just in case you don’t understand one of the usual proofs already, I would suggest that you do so. The class equation is very useful in many other proofs in elementary group theory as well. I think you shouldn’t attempt to avoid it, especially if you are just learning about group theory.

In any case, as requested.. here is a horrible proof that doesn’t use the class equation. Let me emphasize that proving this fact in this way is pretty silly, and you probably shouldn’t spend time on it if you don’t understand the class equation or group actions.

Suppose that $G$ is a group, $|G| = p^2$ where $p$ is a prime. We prove that $G$ must be abelian, so $Z(G) = G$ and in particular $Z(G) \neq 1$. We may assume that every nonidentity element of $G$ has order $p$, because otherwise $G$ would be cyclic. Let $x \in G$ be a nonidentity element and $H = \langle x \rangle$.

I claim that $H$ must be a normal subgroup. If not, there exists $g \in G$ such that $gHg^{-1} \neq H$. In this case $gHg^{-1}$ and $H$ both have order $p$ and intersect trivially, so $gHg^{-1}H$ has order $p^2$. Thus $G = gHg^{-1}H$, and in particular $g = ghg^{-1}h’$ for some $h, h’ \in H$. But this forces $g \in H$ and thus $H = gHg^{-1}$, a contradiction.

Now pick $y \not\in H$. By the same argument, $K = \langle y \rangle$ is also a normal subgroup. Since $H \cap K = 1$, it follows that $G \cong H \times K \cong \mathbb{Z}/p\mathbb{Z}\times\mathbb{Z}/p\mathbb{Z}$ and $G$ is abelian.

You could try to prove that, up to isomorphism, there are only two groups of order $p^2$, both of which are abelian (they are $\mathbb{Z}/p^2\mathbb{Z}$ and $(\mathbb{Z}/p\mathbb{Z})^2$). Usually, this is proven using Cauchy’s theorem. On wikipedia, there is a proof of Cauchy’s theorem not using the class equation. It does use, however, the orbit-stabilizer theorem and thus assumes some knowledge of group actions.

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