Nonisomorphic groups of order 12.

Question

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? `

score 3 · Answer 1 · answered Feb 20 '14 at 04:32

3

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.

answered Feb 20 '14 at 04:32

Wei Zhou

1,651

if i have two groups G and H, is GxH isomorphic to HxG in general? – pmal Feb 20 '14 at 05:28
Yes, the two groups are isomrphic. – Wei Zhou Feb 20 '14 at 05:35
so groups that i came up with are Z12, A4, Z6xZ2. i cant think of a fourth one? I don't understand your example – pmal Feb 20 '14 at 05:42
$S_3$ 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 05:48
but A4 is also not abelian – pmal Feb 20 '14 at 05:51
Note $Z_2$ is in the center of $Z_2 \times S_3$. But the center of $A_4$ is trivial. So they are not isomorphic. – Wei Zhou Feb 20 '14 at 07:14

score 3 · Answer 2 · answered Feb 20 '14 at 04:40

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.

amWhy · Accepted Answer · 2016-10-10T11:24:31.830

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 $A_4\not\cong \mathbb Z_2 \times \mathbb S_3$.

Nonisomorphic groups of order 12.

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