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Math Help - Order of the Centre of a Subgroup

  1. #1
    Member Haven's Avatar
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    Order of the Centre of a Subgroup

    I've been racking my brain over this problem:

    Let G be a group of order p^n
    Prove the Center of G cannot have order p^{n-1}

    Naturally, I assume for a contradiction that |Z(G)| = p^{n-1}
    By Lagrange's Theorem, there are \frac{|G|}{|Z(G)|} = \frac{p^n}{p^{n-1}} = p distinct left cosets of Z(G)

    Let a \in G.
    If a \in Z(G) then aZ(G) = Z(G)

    This is as far as I get and I get the contraditiction. Any help would be greatly appreciated
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Haven View Post
    I've been racking my brain over this problem:

    Let G be a group of order p^n
    Prove the Center of G cannot have order p^{n-1}

    Naturally, I assume for a contradiction that |Z(G)| = p^{n-1}
    By Lagrange's Theorem, there are \frac{|G|}{|Z(G)|} = \frac{p^n}{p^{n-1}} = p distinct left cosets of Z(G)

    Let a \in G.
    If a \in Z(G) then aZ(G) = Z(G)

    This is as far as I get and I get the contraditiction. Any help would be greatly appreciated
    What have you learned up to in class. That will dictate members' answers.
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  3. #3
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    Quote Originally Posted by Haven View Post
    I've been racking my brain over this problem:

    Let G be a group of order p^n
    Prove the Center of G cannot have order p^{n-1}

    Naturally, I assume for a contradiction that |Z(G)| = p^{n-1}
    By Lagrange's Theorem, there are \frac{|G|}{|Z(G)|} = \frac{p^n}{p^{n-1}} = p distinct left cosets of Z(G)

    Let a \in G.
    If a \in Z(G) then aZ(G) = Z(G)

    This is as far as I get and I get the contraditiction. Any help would be greatly appreciated
    Hint.

    |G/Z(G)| = p (a prime number) ----> G/Z(G) is cyclic ------> G is abelian------>G/Z(G)=1. Contradiction?
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by aliceinwonderland View Post
    Hint.

    |G/Z(G)| = p (a prime number) ----> G/Z(G) is cyclic ------> G is abelian------>G/Z(G)=1. Contradiction?
    I wanted to do that!!

    Haha! I'm just kidding!
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  5. #5
    Member Haven's Avatar
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    Quote Originally Posted by Drexel28 View Post
    What have you learned up to in class. That will dictate members' answers.
    I have not learned Quotient Groups yet, we're only allowed to use left cosets and Lagrange's Theorem.
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