Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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
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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.