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Give examples of four groups of order 12 no two of which are isomorphic.

So far I've thought of $Z_{12}$ and $D_6$.

Thanks!

EgoKilla
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4 Answers4

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Consider $\mathbb{Z}_2\times \mathbb{Z}_2\times \mathbb{Z}_3$ and $A_4$.

wanderer
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Hints for you to prove:

The two abelian ones: $\;C_{12}\;,\;\;C_2\times C_6\cong C_2\times C_2\times C_3\;$ , and now two non-abelian ones which is easy to see aren't isomorphic: $\;A_4\;,\;\;C_2\times S_3\;$

DonAntonio
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There is also $T=C_3\rtimes C_4$, since ${\rm Aut}\, C_3\simeq C_2$ is isomorphic to a subgroup of $C_4$. In fact, the only non-abelian groups of order $12$ are $A_4$, $D_{12}$ and $T$.

Pedro
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Okay, here is my reasoning.

Take $Z_{12}$, $Z_2 x Z_2 x Z_3$, $D_6$ and $A_4$.

$Z_{12} \not\cong Z_2 x Z_2 x Z_3$ since $Z_{12}$ is cyclic.

$Z_{12} \not\cong D_6$ since $Z_{12}$ is cyclic.

$Z_{12} \not\cong A_4$ since $Z_{12}$ is cyclic.

$D_6 \not\cong A_4$ since $D_6$ has an element of order $6$.

$D_6 \not\cong Z_2 x Z_2 x Z_3$ since $D_6$ is non-Abelian and Z_2 x Z_2 x Z_3$ is an abelian group

$A_4 \not\cong Z_2 x Z_2 x Z_3$ since $A_4$ has elements of order $2$.

Matthew Les
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EgoKilla
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  • $\mathbb{Z}_2\times \mathbb{Z}_2\times \mathbb{Z}_3$ also has elements of order $2$, moreover exactly $3$ elements of order $2$ which is same as number of order $2$ elements in $A_4$. A possible correction is to claim $\mathbb{Z}_2\times \mathbb{Z}_2\times \mathbb{Z}_3$ has only one element of order $3$ whereas $A_4$ has $8$ elements of order $3$. – Aman Kushwaha Sep 09 '22 at 02:10