Is $\mathbb{Z}/3\mathbb{Z} \times \mathbb{Z}/3\mathbb{Z}\hspace{0.5mm}$ isomorphic to $\hspace{0.5mm}\mathbb{Z}/9\mathbb{Z}$?

Question

I kind of know that the answer is no (It is no, right?) through brute force trying out, and I remember my professor emphatically saying that $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}\hspace{0.5mm}$ is not isomorphic to $\hspace{0.5mm}\mathbb{Z}/4\mathbb{Z}$, and the same logic should apply to the question at hand. However, I still don't quite understand the systematic reasoning here, it's all still a bit trial and error at this point for me. Can someone shed some light on this?

Answer 1

Hint:

There is no element of order $9$ in $\mathbb{Z}/3\mathbb{Z}\times\mathbb{Z}/3\mathbb{Z}$.

Answer 2

No. You can see this by considering the orders of elements: As a cyclic group, $\mathbb{Z}/9\mathbb{Z}$ has an element of order $9$ (for instance its generator). But, $\mathbb{Z}/3\mathbb{Z} \times \mathbb{Z}/3\mathbb{Z}$ only has elements of order $3$.

Answer 3

There are many quick ways of seeing this; however, if you want some light shed, then what you should be wondering is when $\mathbb{Z/mZ} \times \mathbb{Z/nZ}$ isomorphic to $\mathbb{Z/mnZ}$. This is precisely when $gcd(m,n) = 1$. You can construct an isomorphic explicitly by defining $f: \mathbb{Z/mZ} \times \mathbb{Z/nZ} \to \mathbb{Z/mnZ}$ by $f(1,1) \to 1$ and this should tell you what the rest of the map is.