Isomorphism question about groups I know that the group $\mathbb{Z}/9\mathbb{Z}$ is not isomorphic to the group $\mathbb{Z}/3\mathbb{Z}\times\mathbb{Z}/3\mathbb{Z}$, but I just do not know how to prove this. 
 A: Hint: the group $\mathbb{Z}/9\mathbb{Z}$ is cyclic. Is $\mathbb{Z}/3\mathbb{Z}\times\mathbb{Z}/3\mathbb{Z}$ cyclic?
A: $$\mathbb Z_9=\{\bar{0},\bar{1},\bar{2},\bar{3},\bar{4},\bar{5},\bar{6},\bar{7},\bar{8},\bar{9}\}=\langle \bar{1} \rangle$$ and $$\mathbb Z_3\times\mathbb Z_3=\{(x,y)\mid x,y\in\mathbb Z_3\}$$ By checking, we find that the equation $3*x=0$ has $3$ solutions in $\mathbb Z_9$ while the same equation has $9$ solutions in the other group.
A: Suppose, seeking a contradiction, that $\varphi:\mathbb{Z}_9 \rightarrow \mathbb{Z}_3 \times \mathbb{Z}_3$ is an isomorphism.
Suppose $\varphi(1)=(a,b) \in \mathbb{Z}_3 \times \mathbb{Z}_3$.  Then
\begin{align*}
\varphi(3) &= \varphi(1+1+1) \\
&= \varphi(1)+\varphi(1)+\varphi(1) & \text{since } \varphi \text{ is an homomorphism} \\
&= (a,b)+(a,b)+(a,b) & \text{since } \varphi(1)=(a,b) \\
&= (a+a+a,b+b+b) \\
&= (0,0) & \text{since } a \in \mathbb{Z}_3 \text{ and } b \in \mathbb{Z}_3.
\end{align*}
Similarly, we can show $\varphi(-3)=(0,0)$, contradicting that $\varphi$ is a bijection.
A: Hint: $\mathbb{Z}_9$ has only one subgroup of order three, namely $\{0,3,6\}$.
