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Thread: Prove G is a solvable group

  1. #1
    r45
    r45 is offline
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    Prove G is a solvable group

    Hi there,

    I am trying to solve this problem:

    Suppose G is a finite group, and the order of G is $\displaystyle {p_1}^2 {p_2}^2$, where $\displaystyle p_1,p_2$ are primes such that $\displaystyle {p_1}^2 < p_2$. Show that G is a solvable group.

    I've tried various approaches through Sylow groups and normal subgroups etc but haven't been able to do it, can anyone lend a hand?

    Many thanks!
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  2. #2
    Member Black's Avatar
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    Let's look at the number of Sylow $\displaystyle p_2$-subgroups. By the Sylow theorems, since $\displaystyle n_{p_2}=kp_2+1$ and $\displaystyle n_{p_2}|p_1^2$, $\displaystyle n_{p_2}$ must be 1. Thus, the Sylow $\displaystyle p_2$-subgroup, $\displaystyle P$ is normal in $\displaystyle G$. Now consider

    $\displaystyle 1 \unlhd P \unlhd G$.

    Any group of order $\displaystyle p^2$ (for p prime) is abelian, and since $\displaystyle |G/P|=p_1^2,$ $\displaystyle G/P$ is abelian.
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