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Thread: Finite groups of even cardinality

  1. #1
    Forum Admin topsquark's Avatar
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    Finite groups of even cardinality

    Let (G,*) be a finite group with an even number of elements. Show that there must exist at least one element $\displaystyle a \neq e \in G$ such that $\displaystyle a^2=e$.

    I have no clue how to start and what ticks me off about that is that I think I've seen it before. (And I don't recall it was that difficult, either.) However my class notes (from years ago) doesn't contain the proof.

    -Dan
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    Consider pairing up elements with their inverses.
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by rgep
    Consider pairing up elements with their inverses.
    Got it. Let me formalize it then.

    Let (G,*) be a finite group with an even number of elements. Consider the subset of G: $\displaystyle A \equiv \{b|b^2=e, b \in G\} $ where e is the identity in G. It may be that A is empty. If not, the inverse of any element $\displaystyle b \in A$ is also in A and is distinct from b. So b and its inverse form a pair of elements in A and this is true of any element in A. Thus we know that A has no elements or has an even number of them.

    Now form the set G - A. (For others, I believe the notation is G\A?) We know that G - A is not empty because $\displaystyle e^2=e$. We also know that the set G - A has an even number of elements since both G and A do (calling 0 an even number). Because of this, the set G - A - {e} cannot be empty, since this would imply that G has an odd number of elements. Since any element $\displaystyle a \in G-A-{e} $ is not in A we know that $\displaystyle a^2=e$.

    Thus there is at least one element a in G that is not the identity that has the property $\displaystyle a^2=e$. (End of proof.)

    Wow. That took a lot more writing than I thought it would.

    Thanks for the tip!

    -Dan
    Last edited by topsquark; Feb 25th 2006 at 01:56 PM. Reason: Corrections
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    Quote Originally Posted by topsquark
    Got it. Let me formalize it then.

    Let (G,*) be a finite group with an even number of elements. Consider the subset of G: $\displaystyle A \equiv \{b|b^2=e, b \in G\} $
    Don't you need to add $\displaystyle b\ne e$ to the definition of $\displaystyle A$?

    where e is the identity in G. It may be that A is empty. If not, the inverse of any element $\displaystyle b \in A$ is also in A and is distinct from b.
    Take the group $\displaystyle \mathcal{G}=(\{e,b\},*)$, where $\displaystyle b=b^{-1}$. Then $\displaystyle b \in A$ but $\displaystyle b^{-1}$ is not distinct form $\displaystyle b$.

    So b and its inverse form a pair of elements in A and this is true of any element in A. Thus we know that A has no elements or has an even number of them.

    Now form the set G - A. (For others, I believe the notation is G\A?) We know that G - A is not empty because $\displaystyle e^2=e$. We also know that the set G - A has an even number of elements since both G and A do (calling 0 an even number). Because of this, the set G - A - {e} cannot be empty, since this would imply that G has an odd number of elements. Since any element $\displaystyle a \in G-A-{e} $ is not in A we know that $\displaystyle a^2=e$.

    Thus there is at least one element a in G that is not the identity that has the property $\displaystyle a^2=e$. (End of proof.)

    Wow. That took a lot more writing than I thought it would.

    Thanks for the tip!

    -Dan
    Last edited by CaptainBlack; Feb 26th 2006 at 04:32 AM.
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    Grand Panjandrum
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    Quote Originally Posted by topsquark
    Let (G,*) be a finite group with an even number of elements. Show that there must exist at least one element $\displaystyle a \neq e \in G$ such that $\displaystyle a^2=e$.

    I have no clue how to start and what ticks me off about that is that I think I've seen it before. (And I don't recall it was that difficult, either.) However my class notes (from years ago) doesn't contain the proof.

    -Dan
    If there is no such element every element $\displaystyle a$ except $\displaystyle e$ has a distinct
    partner $\displaystyle a^{-1}$, adding up to an even number of elements, but
    $\displaystyle e$ is un-partnered so there are an odd number of elements in the
    group, a contradiction.

    RonL
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    Forum Admin topsquark's Avatar
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    In regards to your second to last post on this thread, CaptainBlack, yes I forgot to include b<>e in the definition of A, thank you.

    For providing a proof MUCH shorter than mine...thppppt!

    -Dan
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