Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

This question already has an answer here:

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

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


share|cite|improve this question

marked as duplicate by Derek Holt, M Turgeon, TMM, Joe Johnson 126, Sami Ben Romdhane Mar 23 '14 at 21:46

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

There's another abelian group. – user61527 Mar 23 '14 at 20:23
Do you have any means of constructing a large group out of smaller ones. – Mark Bennet Mar 23 '14 at 20:26

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

share|cite|improve this answer

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: $\;A_4\;,\;\;C_2\times S_3\;$

share|cite|improve this answer

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 $T$.

share|cite|improve this answer

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 $D_6$ has 3 elements of order $2$.

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

share|cite|improve this answer

Not the answer you're looking for? Browse other questions tagged or ask your own question.