# $\mathbb{Z}_{p^2}$ and $\mathbb{Z}_p\times \mathbb{Z}_p$ [duplicate]

I try to prove if de groups $$\mathbb{Z}_{p^{2}}$$ and $$\mathbb{Z}_p\times \mathbb{Z}_p$$ are isomorphic. I was using the fact that $$\mathbb{Z}_ {mn}$$ is isomorphic to $$\mathbb{Z}_m \times \mathbb{Z}_n$$ if and only if $$(m, n) = 1$$ taking $$m = n = p$$,in this case as $$p$$ is a prime greater than 1, then $$(p, p) = p \neq 1$$ and so I don't have the isomorphism, but I don't know if this is true.

• The cyclic group $C_n$ has an element of order $n$. Does $C_p\times C_p$ has an element of order $p^2$? Mar 3, 2021 at 4:22

I claim that $$\mathbb{Z}/p^2$$ is not isomorphic to $$\mathbb{Z}/p\times \mathbb{Z}/p$$. By definition, the former is the cyclic group of order $$p^2$$, which implies that it has a generating element of order $$p^2$$. Any element of $$\mathbb{Z}/p\times \mathbb{Z}/p$$ is of form $$(a,b)$$, with $$a,b\in \mathbb{Z}/p$$. But observe that $$p(a,b)=(pa, pb)=(0,0)$$, so the order of any such $$(a,b)$$ is at most $$p$$. Since isomorphisms preserve order, no such isomorphism is possible.

• In this case, is the notation $\mathbb{Z} / p$. Is the same that $\mathbb{Z}_{p}$, that is, are the integers modulo p? Mar 3, 2021 at 4:45
• @randal Yes; this is the case. Mar 3, 2021 at 4:45
• ok, I understand now. Mar 3, 2021 at 4:51

It is. The statement $$\Bbb Z_m\times\Bbb Z_n\cong\Bbb Z_{mn}$$ is equivalent to $$(m,n)=1$$.

• I meant it's true that the OP doesn't have an isomorphism. The OP asked if that was true.
– user403337
Mar 3, 2021 at 4:40
• There are loads of times when the statements of the problems on this site are not terribly precise. @peterag messy, convoluted, near nonsense. I'm used to it. But I believe that has something to do with (trying to do) mathematics. Sometimes they get cleaned up; alot of times they don't.
– user403337
Mar 3, 2021 at 4:43