Nonisomorphic groups of order 12. I'm trying to find 4 groups of order 12, none of which are isomorphic to each other.  Should I be trying external direct products? 
So far I have $A_4, \mathbb Z_{12},\,$ and $\,\mathbb Z_6\times \mathbb Z_2.\,$ How do I show all of these are non-isomorphic to each other? And how do I find a fourth?
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 A: Yes, external direct product is need. You can get $A_4$ and $Z_2\times S_3$ are not isomorphic. 
And you can get all groups in abelian case easily.
A: You can use the external direct product for this problem.
There are actually five non-isomorphic groups of order $12$, but you need to find four.  Here are few catches to determine the specific groups:


*

*Apply the fundamental theorem of finitely-generated abelian groups.  Be careful of which groups are isomorphic.

*Try to find other groups by constructing homomorphism and applying stabilizer theorem.



Check the SPOILERS for the problem you are stuck in.

A: The groups you've found so far:
$$ A_4, \;\mathbb Z_{12}, \mathbb Z_6\times \mathbb Z_2$$ are, indeed, non-isomorphic. Why?


*

*By the Fundamental Theorem of Finitely Generated Abelian groups, we
know $\mathbb Z_{12}$ and $\mathbb Z_6\times \mathbb Z_2 \cong
   \mathbb Z_2\times \mathbb Z_6$ are abelian and further, that $\mathbb
   Z_{12}\cong \mathbb Z_3 \times \mathbb Z_4$ and is cyclic, and not isomorphic to the abelian group $\mathbb
   Z_{2}\times \mathbb Z_6$.$\mathbb Z_{12} = \mathbb Z_{2\times 6} \not\cong \mathbb Z_2\times \mathbb Z_6$ because $\gcd(2, 6) = 2\neq 1$. Indeed, $\mathbb Z_{12}$ is cyclic, but $\mathbb Z_2\times \mathbb Z_6$ is not.

*Neither of these two non-isomorphic abelian groups is isomorphic to
$A_4$, since $A_4$ is not abelian.
Finally, for a fourth group of order $12$ which is not isomporphic to any of the above three groups, we have $\;\mathbb Z_2\times S_3$. This group is not abelian, and so not isomorphic to $\mathbb Z_{12},$ nor to $\mathbb Z_2\times \mathbb Z_6$. All that's left for you to justify is the fact that $A_4\not\cong \mathbb Z_2 \times \mathbb S_3$.
