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Thread: 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 $\displaystyle \mathbb{Z}_2$ but in the spirit of your question...no. If $\displaystyle G$ is a finite group and $\displaystyle |G|=21$ where $\displaystyle m>1$ then a simple application of Cauchy's theorem says that $\displaystyle G$ must have an element of order $\displaystyle 2$. More generally, Sylow's first theorem implies that if $\displaystyle G$ is a finite group and $\displaystyle p^m\mid |G|$ with $\displaystyle p$ prime then $\displaystyle G$ must contain a subgroup of order $\displaystyle p$. From this it's easy to see that the only groups you mention have order $\displaystyle 1$ or $\displaystyle p$ where $\displaystyle p$ is prime.
    Last edited by Drexel28; Apr 3rd 2011 at 04:47 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 $\displaystyle \mathbb{Z}_2$ but in the spirit of your question...no. If $\displaystyle G$ is a finite group and $\displaystyle |G|=2m$ where $\displaystyle m>0$ then a simple application of Cauchy's theorem says that $\displaystyle G$ must have an element of order $\displaystyle 2$. More generally, Sylow's first theorem implies that if $\displaystyle G$ is a finite group and $\displaystyle p^m\mid |G|$ with $\displaystyle p$ prime then $\displaystyle G$ must contain a subgroup of order $\displaystyle p$. From this it's easy to see that the only groups you mention have order $\displaystyle 1$ or $\displaystyle p$ where $\displaystyle p$ is prime.
    Thanks. "Cauchy" isn't until the next chapter, but I can follow the logic easily enough. Thanks!

    -Dan
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