# Help with isomorphic groups?

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

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!

• 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). Commented May 29, 2021 at 14:23
• 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 $\mathbb Z_{12}$. Commented May 29, 2021 at 14:29

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 $$\Bbb Z_{12}\not\simeq\Bbb Z_6\times\Bbb Z_2$$.