# 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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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 $\mathbb Z_{12}$ 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 $A_4\not\cong \mathbb Z_2 \times \mathbb S_3$.

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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.

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if i have two groups G and H, is GxH isomorphic to HxG in general? –  Paul Malinowski Feb 20 '14 at 5:28
Yes, the two groups are isomrphic. –  Wei Zhou Feb 20 '14 at 5:35
so groups that i came up with are Z12, A4, Z6xZ2. i cant think of a fourth one? I don't understand your example –  Paul Malinowski Feb 20 '14 at 5:42
$S_3$ is a non-abelian group of order 6. So $Z_2 \times S_3$ is of order 12, non-abelian. –  Wei Zhou Feb 20 '14 at 5:48
but A4 is also not abelian –  Paul Malinowski Feb 20 '14 at 5:51

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:

Check the SPOILERS for the problem you are stuck in.

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