# Thread: Isomorphism between A4 and Z2 X S3

1. ## Isomorphism between A4 and Z2 X S3

Show that $A_4$ and $\mathbb{Z}_2$ X $S_3$ are not isomorphic.

Be grateful for help.

I note in passing that both groups have order 12 so in that sense they potentially could be isomorphic. Further, they are both non-abelian so we cannot construct an argument that one is abelian and the other is not.

For those who need an update or re-acquainting with $A_4$ I have attached the relevant pages of Gallian: Contemporary Abstract Algebra

For those who need an update or re-acquainting with $S_3$ I have attached the relevant pages of Fraleigh: A First Course in Abstract Algebra

Peter

2. ## Re: Isomorphism between A4 and Z2 X S3

Originally Posted by Bernhard

Show that $A_4$ and $\mathbb{Z}_2$ X $S_3$ are not isomorphic.

Be grateful for help.

I note in passing that both groups have order 12 so in that sense they potentially could be isomorphic. Further, they are both non-abelian so we cannot construct an argument that one is abelian and the other is not.

For those who need an update or re-acquainting with $A_4$ I have attached the relevant pages of Gallian: Contemporary Abstract Algebra

For those who need an update or re-acquainting with $S_3$ I have attached the relevant pages of Fraleigh: A First Course in Abstract Algebra

Peter
Hint: Count subgroups of order 3.

3. ## Re: Isomorphism between A4 and Z2 X S3

Originally Posted by Bernhard

Show that $A_4$ and $\mathbb{Z}_2$ X $S_3$ are not isomorphic.

Be grateful for help.

I note in passing that both groups have order 12 so in that sense they potentially could be isomorphic. Further, they are both non-abelian so we cannot construct an argument that one is abelian and the other is not.

For those who need an update or re-acquainting with $A_4$ I have attached the relevant pages of Gallian: Contemporary Abstract Algebra

For those who need an update or re-acquainting with $S_3$ I have attached the relevant pages of Fraleigh: A First Course in Abstract Algebra

Peter
We know that there exists $H\subseteq S_3$ of order $3$ (e.g $\langle (1,2,3)\rangle$) and so $\mathbb{Z}_2\times H$ is a subgroup of $\mathbb{Z}_2\times S_3$ of order six. Yet, you should know (as the classic minimal counterexample to the converse of Lagrange's theorem) that $A_4$ has no subgroup of order six.

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# group Z2 x S3

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