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Math Help - commutes

  1. #1
    Senior Member sfspitfire23's Avatar
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    commutes

    If a group is non-abelian, then the identity is the only thing that commutes with everything right?
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  2. #2
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    Quote Originally Posted by sfspitfire23 View Post
    If a group is non-abelian, then the identity is the only thing that commutes with everything right?
    Not necessarily. Plz read this
    Center (group theory) - Wikipedia, the free encyclopedia

    Group is abelian IFF Z(G)=G
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  3. #3
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by sfspitfire23 View Post
    If a group is non-abelian, then the identity is the only thing that commutes with everything right?
    For example, G = S_4 \times C_2. The C_2 will commute with everything.

    Also, every group of order p^n has a non-trivial centre, p prime, for instance in D_8 the rotation by 180 degree commutes with everything and is non-trivial.
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  4. #4
    Senior Member sfspitfire23's Avatar
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    In a non-abelian group that is generated by <a>, will the only thing that commutes be e or will what you guys said still hold?
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  5. #5
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    Quote Originally Posted by sfspitfire23 View Post
    In a non-abelian group that is generated by <a>, will the only thing that commutes be e or will what you guys said still hold?

    Any cyclic group will always be abelian. This follows from associative property.
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  6. #6
    Member alunw's Avatar
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    A group generated by a single element is cyclic and (therefore) abelian. So there is no such thing as the non-abelian group generated by <a> , whatever a may be.
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  7. #7
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by sfspitfire23 View Post
    In a non-abelian group that is generated by <a>, will the only thing that commutes be e or will what you guys said still hold?
    Let G=<a>. Then for g,h \in G arbitrary, g=a^n, h=a^m, m,n \in \mathbb{Z}. Can you show that gh=hg?

    EDIT: 3 replies in 2 minutes? That must be some kind of record!
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  8. #8
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    Quote Originally Posted by Swlabr View Post
    Let G=<a>. Then for g,h \in G arbitrary, g=a^n, h=a^m, m,n \in \mathbb{Z}. Can you show that gh=hg?
    oops....too many helpers
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  9. #9
    Member alunw's Avatar
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    Since the centre of a group is a normal subgroup, you can be sure that in a non-abelian simple group the identity is the only thing that commutes with everything. The smallest such group is the alternating group on 5 symbols.
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