Group isomorphic to $\Bbb Z/3\Bbb Z\times \Bbb Z/3\Bbb Z$

Question

Is a group with $9$ elements such that all elements (excepted the natural element) are of order $3$ is isomorphic to $\Bbb Z/3\Bbb Z\times \Bbb Z/3\Bbb Z$ ?

Answer 1

Yes: since a group of order $p^2$ ($p$ prime) is abelian, we can use the structure theorem for finite abelian groups. A group of order $9$ is either cyclic, isomorphic to $\mathbf Z/9\mathbf Z$ or isomorphic to $\mathbf Z/3\mathbf Z\times\mathbf Z/3\mathbf Z $ .

Answer 2

If a group $G$ has order $9$ and is not cyclic then, by Lagrange's theorem all of its elements $\neq e$ are fo order $3$. Now take $\alpha \in G$ and $\beta \in G - \langle\alpha\rangle$. Then we would have $$G = \{e, \alpha, \alpha^2,\beta,\alpha\beta,\alpha^2\beta,\beta^2,\alpha\beta^2,\alpha^2\beta^2\}$$

Then

$$\begin{cases}|G| = 9\\G=\langle\alpha,\beta\rangle\\\alpha^3=e\\\beta^3=e\end{cases}$$

Find the product $\beta\alpha$.

And work on the cases $\beta\alpha= \alpha\beta$ (take $G = \mathbb{Z}/3\mathbb{Z} \times \mathbb{Z}/3\mathbb{Z}$), $\beta\alpha = \alpha^2\beta$, $\beta\alpha = \alpha\beta^2$ and $\beta\alpha^2\beta^2$, there is a contradiction for the last three cases.

Case $\beta\alpha = \alpha^2\beta^2$

$$(\alpha\beta)^2 = \alpha\beta\alpha\beta = \alpha \alpha^2\beta ^2\beta = e$$