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Math Help - Prove that if b is the only element of order 2 in G, then b E Z(G)

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    Prove that if b is the only element of order 2 in G, then b E Z(G)

    the center Z(G) is defined as Z(G) = {a E G | ax = xa for every x E G}
    Prove that if b is the only element of order 2 in G, then b E Z(G)
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  2. #2
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by rainyice View Post
    the center Z(G) is defined as Z(G) = {a E G | ax = xa for every x E G}
    Prove that if b is the only element of order 2 in G, then b E Z(G)
    This is actually quite a neat question. What you need to do it to prove that b satisfies this condition. I will just give you a hint though. Look in this thread of yours. You will use one of these three facts, along with the fact that if x^2=e then x=b. You will also have to fiddle around with the condition. ax=xa \Leftrightarrow \ldots
    Last edited by Swlabr; April 3rd 2010 at 12:30 AM. Reason: removed a full stop after \ldots as it looked silly...
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    Then G has only one subgroup of order 2, then <b> is normal in G since it doesn't have any conjugates. So then if x\neq e, then xbx^{-1}=b since it must be in <b> but it's not e. QED
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