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I want to find an example of a finite group $G$, which has two non-conjugate $\pi$-subgroups of Hall.

From the Hall's theorem I know that I have to search in the class of non-solvable groups. So my $G$ has order at least $60$ and it is divisible for at least three distinct prime numbers. My idea was to find a group with two subgroups $H_1$ and $H_2$ of the proper order to be $\pi$ -subgroups of Hall and such that one of them was normal in $G$ and the other was not. But I got stuck.

I would be very thankful if someone could help me finding this example.

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    $\begingroup$ Your idea can unfortunately not work. If a group has a normal Hall subgroup $N$ and $H$ is any subgroup whose order divides that of $N$ then $H\leq N$ (this can be proven in a fairly elementary way by considering the product of the subgroups). $\endgroup$
    – Tobias Kildetoft
    Commented Feb 22, 2016 at 9:19

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${\rm PSL}(2,13)$ has two subgroups of order $12$, one dihedral, and the other isomorphic to $A_4$.

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    $\begingroup$ Thanks for the answer. Is there a simple way to prove this statement? $\endgroup$
    – GGG
    Commented Feb 22, 2016 at 9:48

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