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Math Help - Isomorphism between A4 and Z2 X S3

  1. #1
    Super Member Bernhard's Avatar
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    Isomorphism between A4 and Z2 X S3

    Can anyone please help with the following problem:

    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
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  2. #2
    MHF Contributor Swlabr's Avatar
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    Re: Isomorphism between A4 and Z2 X S3

    Quote Originally Posted by Bernhard View Post
    Can anyone please help with the following problem:

    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.
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  3. #3
    MHF Contributor Drexel28's Avatar
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    Re: Isomorphism between A4 and Z2 X S3

    Quote Originally Posted by Bernhard View Post
    Can anyone please help with the following problem:

    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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