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Math Help - G is a group with |G|=mp where p is a prime and 1<m<p. Prove that G is not simple.

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    Question G is a group with |G|=mp where p is a prime and 1<m<p. Prove that G is not simple.

    Suppose that G is a group with |G| = mp where p is a prime and 1 < m < p. Prove that G is not simple.
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    Senior Member Tinyboss's Avatar
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    How many Sylow p-subgroups can G have?
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    Quote Originally Posted by Tinyboss View Post
    How many Sylow p-subgroups can G have?
    My class has not covered Sylow p-subgroups.
    There is a Sylow p-subgroup of order p, right? Are there others or just one?
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    Senior Member Tinyboss's Avatar
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    There are always Sylow p-subgroups. In this case, it follows immediately from Sylow's theorems that there is a unique one. Unique Sylow subgroups are normal.

    But as you say, you haven't covered that yet, so I'm sure you're expected to make a more direct argument.
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    Since p|mp, there is a a subgroup H of order p. A (sub)group of order p is cyclic, thus H is Abelian thus normal. Since we have a normal subgroup H\ne\{e\} and H\ne G, G is not simple.

    Is this argument correct?
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    Senior Member Tinyboss's Avatar
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    Abelian subgroups need not be normal. But I think you're on the right track considering the cyclic subgroup generated by an order-p element.
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    Quote Originally Posted by Tinyboss View Post
    Abelian subgroups need not be normal. But I think you're on the right track considering the cyclic subgroup generated by an order-p element.
    Right. The subgroups of an Abelian subgroup H are normal in H but necessarily normal in the group G.


    I am stuck now.
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by JJMC89 View Post
    Suppose that G is a group with |G| = mp where p is a prime and 1 < m < p. Prove that G is not simple.

    Let H\leqslant G be such that |H|=p. It is trivial that there is a homomorphism \phi:G\to \text{Sym}\left(G/H\right) by having \phi_g(aH)=gaH. Moreover, one can prove that \ker\phi\subseteq H. Now, since p is prime and \ker\phi\leqslant H we must have that \ker\phi=\{e\} or \ker\phi=H. Suppose that \ker\phi=\{e\} then \text{im}(\phi) is a subgroup of \text{Sym}\left(G/H\right) of order mp and so mp\mid m! but since p is prime and m<p this is impossible. Thus, H=\ker\phi and so \{e\}\triangleleft H\triangleleft G so that G isn't simple.
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