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Let $G$ be a finite non-solvable group. Does $G$ always have a maximal subgroup of even order which is not normal in G?

Attempt: As $G$ is non-solvable, $|G|$ is even and has an element of order $2$, say $a$. Thus there exists a maximal subgroup $M$ of $G$ containing $a$. Clearly $|M|$ is even. Now, if $G$ is simple, then we are done.

However, even if $G$ is not simple, I believe that still we can find a maximal subgroup of even order which is not normal in $G$. Any help is appreciated.

Yet another attempt: As $G$ is non-solvable, Sylow $2$-subgroups of $G$ are not normal in $G$. Consider a maximal subgroup $M$ of $G$ containing a Sylow $2$-subgroups of $G$. If $M$ is not normal in $G$, we are done. If $M$ is normal in $G$, then all Sylow $2$-subgroups of $G$ are contained in $M$. Again, I am stuck at this point.

Cross-posted in Math Stack Exchange: https://math.stackexchange.com/questions/5097690/existence-of-maximal-subgroups-of-even-order-which-are-not-normal

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    $\begingroup$ It's normal to wait a week before cross-posting from MSE. $\endgroup$ Commented Sep 24 at 9:31

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If the Sylow 2-subgroup of $G$ is normal, then $G$ is solvable, using the Feit-Thompson odd order theorem. Hence we only need to consider finite groups $G$ with a Sylow $2$-subgroup $S$ which is not normal.

Then $N_{G}(S)$ is a proper subgroup of $G$, so is contained in a maximal subgroup, say $M$, of $G$. By a standard argument, making use of the Frattini argument, $M= N_{G}(M),$ so that $M$ is a maximal subgroup of even order of $G$ which is not normal.

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  • $\begingroup$ I suppose that the right generalization is that for any prime $p$, a finite group $G$ which is not $p$-solvable has a maximal subgroup of order divisible by $p$ which is not normal (even if $G$ is not $p$-closed, there is such a a non-normal maximal subgroup). $\endgroup$ Commented Sep 24 at 10:13
  • $\begingroup$ Thank you for the answer. I have a further query or rather speculation which is generated by the construction of M: If T is the set of all involutions in G, can T be contained in M? My guess is NO. $\endgroup$ Commented Sep 24 at 13:03
  • $\begingroup$ No, I think there are examples where the maxima aubgroup $M$ contains all involutions of $G$. As an extremely example, take $G = {\rm SL}(2,q)$ where $q$ is odd and greater than $4$, and let $M$ be a maximal subgroup containing a Sylow 2-normalizer. Note that $G$ has only one involution, which is in $M$. $\endgroup$ Commented Sep 24 at 13:13
  • $\begingroup$ Ok. What I feel is: Given a finite non solvable group G, there exist two conjugate maximal subgroups M and N of G such that there exist two elements a in M-N, and b in N-M such that the subgroup generated by a and b is a proper subgroup of G. If G is simple, this is easy to prove. I am stuck with the non-simple, non-solvable case. $\endgroup$ Commented Sep 25 at 3:59
  • $\begingroup$ I will think about that, but it seems like a different question, and this thread may be long enough. $\endgroup$ Commented Sep 25 at 14:44

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