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Math Help - Subgroups of a group of even order

  1. #1
    Forum Admin topsquark's Avatar
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    Subgroups of a group of even order

    My question is this: Can we construct a group of even order that has no subgroups? I am tolerably certain that I have heard of this as a theorem, but I may have only heard it in reference to cyclic groups where it is obvious that a subgroup of order 2 exists.

    Thanks!
    -Dan
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by topsquark View Post
    My question is this: Can we construct a group of even order that has no subgroups? I am tolerably certain that I have heard of this as a theorem, but I may have only heard it in reference to cyclic groups where it is obvious that a subgroup of order 2 exists.

    Thanks!
    -Dan
    I assume that you mean no non-trivial proper subgroups. The answer is yes \mathbb{Z}_2 but in the spirit of your question...no. If G is a finite group and |G|=21 where m>1 then a simple application of Cauchy's theorem says that G must have an element of order 2. More generally, Sylow's first theorem implies that if G is a finite group and p^m\mid |G| with p prime then G must contain a subgroup of order p. From this it's easy to see that the only groups you mention have order 1 or p where p is prime.
    Last edited by Drexel28; April 3rd 2011 at 05:47 PM.
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Drexel28 View Post
    I assume that you mean no non-trivial proper subgroups. The answer is yes \mathbb{Z}_2 but in the spirit of your question...no. If G is a finite group and |G|=2m where m>0 then a simple application of Cauchy's theorem says that G must have an element of order 2. More generally, Sylow's first theorem implies that if G is a finite group and p^m\mid |G| with p prime then G must contain a subgroup of order p. From this it's easy to see that the only groups you mention have order 1 or p where p is prime.
    Thanks. "Cauchy" isn't until the next chapter, but I can follow the logic easily enough. Thanks!

    -Dan
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