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.
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5$\begingroup$ The cyclic group $C_n$ has an element of order $n$. Does $C_p\times C_p$ has an element of order $p^2$? $\endgroup$– user10354138Mar 3, 2021 at 4:22
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2 Answers
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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.
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$\begingroup$ In this case, is the notation $ \mathbb{Z} / p $. Is the same that $\mathbb{Z}_{p} $, that is, are the integers modulo p? $\endgroup$– randalMar 3, 2021 at 4:45
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It is. The statement $\Bbb Z_m\times\Bbb Z_n\cong\Bbb Z_{mn}$ is equivalent to $(m,n)=1$.
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$\begingroup$ I meant it's true that the OP doesn't have an isomorphism. The OP asked if that was true. $\endgroup$– user403337Mar 3, 2021 at 4:40
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1$\begingroup$ 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. $\endgroup$– user403337Mar 3, 2021 at 4:43