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I have been asked to give all the non-isomorphic, abelian groups with order $12$ & order $15$ respectively.

The answer reads

Order $12$: $\Bbb Z_{12}$ and $\Bbb Z_6 \times\Bbb Z_2$

Order $15$: $\Bbb Z_{15}$

I don't really understand what makes it so clear to see that such groups are non-isomorphic. For example:

Q1) What property makes $\Bbb Z_6 \times \Bbb Z_2$ non isomorphic, but not, say, $\Bbb Z_4 \times\Bbb Z_3$?

Q2) Why is it that for order $15$, we can't have $\Bbb Z_3 \times\Bbb Z_5$? What makes this product isomorphic?

Any help appreciated!

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  • $\begingroup$ The underlying fact is that $4$ and $3$ are coprime while $6$ and $2$ are not. So for example $(1,1)$ generates $\mathbb{Z_4\times Z_3}$ but not $\mathbb{Z_6\times Z_2}$ (because of coprime-ness), while in $\mathbb{Z_6\times Z_2}$ the largest possible order of an element is $6$ (because of non-coprime-ness). $\endgroup$
    – user1729
    Commented May 29, 2021 at 14:23
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    $\begingroup$ A group can't be isomorphic by itself, the same way a number can't be equal by itself. Two groups can be isomorphic, just like two numbers can be equal. "Non-isomorphic" in this context just tells you to skip groups which are isomorphic to those you already mentioned. $\mathbb Z_4\times\mathbb Z_3$ is isomorphic to $\mathbb Z_{12}$. $\endgroup$ Commented May 29, 2021 at 14:29

1 Answer 1

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If $m,n\in\Bbb N\setminus\{1\}$ are coprime, then $\Bbb Z_{mn}\simeq\Bbb Z_m\times\Bbb Z_n$. Therefore, $\Bbb Z_4\times\Bbb Z_3\simeq\Bbb Z_{12}$ and $\Bbb Z_5\times\Bbb Z_3\simeq\Bbb Z_{15}$.

But $\Bbb Z_{12}$ has an element of order $12$, whereas $\Bbb Z_6\times\Bbb Z_2$ has no such element. Therefore $\Bbb Z_{12}\not\simeq\Bbb Z_6\times\Bbb Z_2$.

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