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Math Help - Suppose that |G| = p^n and further suppose that G is simple. Prove that |G| = p.

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    Question Suppose that |G| = p^n and further suppose that G is simple. Prove that |G| = p.

    Suppose that |G| = p^n and further suppose that G is simple. Prove that |G| = p.
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  2. #2
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    Quote Originally Posted by joestevens View Post
    Suppose that |G| = p^n and further suppose that G is simple. Prove that |G| = p.

    1) Using the Class Equation, show that any finite p-group has non-trivial center,

    2) By induction on n, show that for any 0\leq k\leq n there exists a normal

    sbgp. N_k\leq G

    3) Solve your problem now.

    Tonio
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by joestevens View Post
    Suppose that |G| = p^n and further suppose that G is simple. Prove that |G| = p.
    Or, use the fact that if a p-group has order p^n then any subgroup of order p^{n-1} must be normal.
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    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by Drexel28 View Post
    Or, use the fact that if a p-group has order p^n then any subgroup of order p^{n-1} must be normal.
    You would also need to prove subgroups of order p^{n-1} exist. This is not obvious (you need that all the composition factors in your composition series are of order p, or something like it).
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    Quote Originally Posted by tonio View Post
    1) Using the Class Equation, show that any finite p-group has non-trivial center,
    done.

    Quote Originally Posted by tonio View Post
    2) By induction on n, show that for any 0\leq k\leq n there exists a normal sbgp. N_k\leq G
    done.

    Quote Originally Posted by tonio View Post
    3) Solve your problem now.

    Tonio
    n must be 1. then the only normal subgroups have order p and 1, which are G and \{e\}.
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Swlabr View Post
    You would also need to prove subgroups of order p^{n-1} exist. This is not obvious (you need that all the composition factors in your composition series are of order p, or something like it).
    I'm sorry, isn't the fact that if p^k\mid |G| where |G|<\infty then \exists H\leqslant G such that |H|=p^k commonly known?
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    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by Drexel28 View Post
    I'm sorry, isn't the fact that if p^k\mid |G| where |G|<\infty then \exists H\leqslant G such that |H|=p^k commonly known?
    Well, yes, but so is the fact that the centre is non-trivial. As the OP doesn't appear to know that the centre is non-trivial, we must assume they don't know that either! My point was, if one is going to prove one of `maximal subgroups have order p', `Z(G) is non-trivial', one would prove the latter.

    Unless, of course, one doesn't know the class formula...

    (I've used `one' as if I'd used `you' the last line is obviously false, because I'm pretty sure you know the class formula...so I was trying to not insult you, not trying to sound posh...)
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