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Math Help - Centre of a group proof

  1. #1
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    Centre of a group proof

    Hey,

    I have this question which asks to prove that if a group has an element, a such that |a|=2, (exactly one order 2 element a) that this element is in the centre of the group.

    so ag=ga for all g in G

    aga-1=g, aa-1g=g so aa-1=e which is true since a has order 2.

    But I feel like this isn't enough at all and I cant really see where the "exactly 1 element with |a|=2" comes in,

    Does anyone have any ideas?
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  2. #2
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    Re: Centre of a group proof

    You're supposed to prove that ag=ga for all g\in G, not assume ag=ga and play around with it (a common mistake with many students).

    This is what you do. Let b=g^{-1}ag. Show that b order 2. But a is the unique element of order 2 in G. What does this tell you?
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  3. #3
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    Re: Centre of a group proof

    So you can say that since a is in the group then its conjugate must also be in the group, let b be its conjugate then

    bb=g-1agg-1ag= g-1aag=g-1g= e so b also had order 2

    Since a is the only element of order to b must be a

    Does that look alright?
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  4. #4
    Member Sylvia104's Avatar
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    Re: Centre of a group proof

    That's correct. One more line completes the proof.
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    Re: Centre of a group proof

    Thanks Sylvia!

    Do you mean like, for the final line, a conclusion like,

    Therefore as b = a, since the group has only one order 2 element,

    b = g-1ag
    gb=ag
    ga=ag

    and hence a is in the centre of G
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