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I was asked to prove: no group can have a minimal normal subgroup isomorphic to a $\mathrm{Syl}_2(A_7)$.

I think I should find some property that $\mathrm{Syl}_2(A_7)$ has but not a minimal normal subgroup. So then I thought that it a $\mathrm{Syl}_2(A_7)$ always has characteristic subgroup, i.e. exist $G\ char\ \mathrm{Syl}_2(A_7)$, then the image of G under isomorphism will contradict minimality of "group can have a minimal normal subgroup isomorphic to a $\mathrm{Syl}_2(A_7)$". But I get stuck then about whether it is true that $\mathrm{Syl}_2(A_7)$ always has characteristic subgroup.

Maybe this is not the correct way to prove it. Can anyone help me?

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It is indeed a fine way to prove it. Note that $Syl_2(A_7)$ is a 2-group, so it has non-trivial center. All you then need to show is that it is not abelian. For this, try to find two non-commuting elements whose orders are powers of 2, such that their product also has this property. (Alternatively, it is not too hard to see that this Sylow-subgroup is in fact isomorphic to the dihedral group of order 8) –  Tobias Kildetoft Dec 20 '10 at 13:28
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up vote 3 down vote accepted

Minimal normal subgroups are direct products of isomorphic simple groups. If they have order a power of a prime, like 8, then they are abelian of exponent p. The dihedral group of order 8 is not abelian.

Your idea works. Just take the derived subgroup of the dihedral group; it has order 2 and is normal in any group containing the dihedral subgroup as a normal subgroup.

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