Prove that every group of order $p^2$, where $p$ is prime, is an abelian group.
Solution
If the order of $G$ is $p^2$, then for the order of $Z(G)$ we have the following possibilities $p^2, p, 1$ (since $p$ is prime). It cannot hold $|Z(G)|=1$ , since we must have $p/|Z(G)|$ . This follows immediately from the class equation .
$$|G|= |Z(G)|+ \sum [G: CG(x_i)]$$
Solution
If the order of $G$ is $p^2$, then for the order of $Z(G)$ we have the following possibilities $p^2, p, 1$ (since $p$ is prime). It cannot hold $|Z(G)|=1$ , since we must have $p/|Z(G)|$ . This follows immediately from the class equation .
$$|G|= |Z(G)|+ \sum [G: CG(x_i)]$$
- If $|Z(G)|=p^2$ then we have that $G=Z(G)$ hence $G$ is an abelian.
- If $|Z(G)|=p$ the order of $G/Z(G)$ is equal to $p^2/p=p$ hence it is cyclic and again we have that it is abelian.
Hello Tolaso. This is a really nice exercise.
ReplyDeleteHere is a second solution :
Let \(\displaystyle{\left(G,\cdot\right)}\) be a group of order \(\displaystyle{|G|=p^2}\), where \(\displaystyle{p}\)
is a prime number. Let \(\displaystyle{x\,,y\in G}\) .Then, \(\displaystyle{\langle{x\rangle}\,,\langle{y\rangle}\leq G}\)
and according to \(\displaystyle{\rm{Lagrange's}}\) theorem, we get :
\(\displaystyle{o(x)\,,o(y)\mid p^2\implies o(x)\,,o(y)\in\left\{1,p,p^2\right\}}\) .
If \(\displaystyle{o(x)=1}\) or \(\displaystyle{o(y)=1}\), then \(\displaystyle{x=e}\) or \(\displaystyle{y=e}\)
and \(\displaystyle{x\cdot y=y\cdot x\in\left\{x,y\right\}}\) .
If \(\displaystyle{o(x)=p^2}\) or \(\displaystyle{o(y)=p^2}\), then \(\displaystyle{G=\langle{x\rangle}}\)
or \(\displaystyle{G=\langle{y\rangle}}\) and the group \(\displaystyle{\left(G,\cdot\right)}\) is abelian
as cyclic. Futhermore, \(\displaystyle{\left(G,\cdot\right)\simeq \left(\mathbb{Z}_{p^2},+\right)}\) .
Let now \(\displaystyle{o(x)=o(y)=p}\) . We have that
\(\displaystyle{o(H=\langle{x\rangle}\cap \langle{y\rangle})\mid o(x)=p\implies o(H)\in\left\{1,p\right\}}\) .
If \(\displaystyle{o(H)=p=o(x)}\), then, \(\displaystyle{H=\langle{x\rangle}\cap \langle{y\rangle}=\langle{x\rangle}}\)
and thus: \(\displaystyle{\langle{x\rangle}\subseteq \langle{y\rangle}}\), so :
\(\displaystyle{x\cdot y=y^{k}\cdot y=y^{k+1}=y^{1+k}=y\cdot y^{k}=y\cdot x}\) .
If \(\displaystyle{o(H)=1}\), then the subgroups \(\displaystyle{\langle{x\rangle}\,,\langle{y\rangle}}\)
of \(\displaystyle{G}\) are normal, \(\displaystyle{H=\langle{x\rangle}\cap \langle{y\rangle}=\left\{e\right\}}\)
and \(\displaystyle{o(\langle{x\rangle}\,\langle{y\rangle})=p^2=|G|}\), so :
\(\displaystyle{G=\langle{x\rangle}\,\langle{y\rangle}\simeq \left(\mathbb{Z}_{p}\times \mathbb{Z}_{p},+\right)}\)
and we deduce that \(\displaystyle{\left(G,\cdot\right)}\) is an abelian group.
In any case, \(\displaystyle{x\cdot y=y\cdot x}\) and the group \(\displaystyle{\left(G,\cdot\right)}\)
is abelian. Also,
\(\displaystyle{\left(G,\cdot\right)\simeq \left(\mathbb{Z}_{p^2},+\right)}\)
or
\(\displaystyle{\left(G,\cdot\right)\simeq \left(\mathbb{Z}_{p}\times \mathbb{Z}_{p},+\right)}\) .