Is a group of order $2^kp$ not simple, where $p$ is a prime and $k$ is an positive integer?

Question

Is a group of order $2^kp$ not simple, where $p$ is a prime and $k$ is an positive integer?

I did this for the groups of order $2^k 3$. Here the intersection of two distinct Sylow $2$-subgroups (if number of sylow $2$-subgroup is more than $1$) is a nontrivial proper normal subgroup of the group. So group is not simple.

But if $p$ is any prime greater than $3$ then is this always not simple?

Please give some hint to solve this?

For $\;p=3\;$ I solved like this: https://math.stackexchange.com/a/2525990/934187

Answer 1

I think this can be done using just group actions and Sylow's theorems.

Suppose that $G$ is simple of order $2^kp$ with $p$ prime and $k>1$. Let $\Omega$ be the set of Sylow $2$-subgroups of $G$. Then $|\Omega|=p$, and we consider the conjugation action of $G$ on $\Omega$. Then the point stabilizers are the Sylow $2$-subgroups - let $S$ be one of these. Since $G$ is simple, its action on $\Omega$ is faithful.

If the Sylow $2$-subgroups have pairwise trivial intersection (or, equivalently, if all nontrivial elements of $S$ fix exactly one point of $\Omega$) then, by a standard counting argument, there are only $p-1$ elements of $G$ that do not lie in Sylow $2$-subgroup of $G$, and so $G$ must have a normal subgroup of order $p$, contradiction.

So some element of $S$ fixes more than one point of $\Omega$. Choose a subgroup $T < S$ such that $|{\rm Fix}(T)|$ is maximal subject to $|{\rm Fix}(T) > 1$, and such that $T$ is the full stabilizer of its fixed point set. Recall that elements of $N_G(T)$ fix the set ${\rm Fix}(T)$ (i.e. they permute its elements).

Then any element in $N_S(T) \setminus T$ fixes a unique point of ${\rm Fix}(T)$ (i.e. $S$). Since $T$ fixes more than one point, it is contained in some other Sylow $2$-subgroup $S' \ne S$ of $G$. Then $T < N_{S'}(T)$ and an element of $N_{S'}(T) \setminus T$ also fixes a unique point (i.e. $S'$) of ${\rm Fix}(T)$, and so this element does not fix $S$.

So $N_G(T)$ has no fixed points. But then $N_G(T)$ does not lie in a Sylow $2$-subgroup of $G$, so its order must be divisible by $p$. But now we have a $p$-cycle normalizing $T$ and hence fixing the set ${\rm Fix}(T)$, so $|{\rm Fix}(T)| \ge p$ and $T$ fixes all points of $\Omega$, contradicting the faithful action of $G$ on $\Omega$.

Answer 2

As mentioned in the comments, this is a special case of Burnside's $p^aq^b$-theorem, which states that if $p$ and $q$ are primes, every group of order $p^a q^b$ is solvable.

Before Burnside, special cases of this result were proven by Frobenius (1895, groups of order $p^aq$) and Jordan (1898, groups of order $p^a q^2$).

Here are the steps used in the proof (due to Frobenius) that a group order $p^a q$ is not simple. I think you can find this in a few textbooks as well.

Let $G$ be a finite group of order $p^a q$, where $p \neq q$ are primes. If $G$ has only one $p$-Sylow it is normal, so we are done.

Otherwise there are a total of $q$ Sylow $p$-subgroups. Pick $p$-Sylows $P_1 \neq P_2$ such that the intersection $D = P_1 \cap P_2$ has the largest possible order.

Steps for the proof:

1. Using the fact that $D \lneqq N_{P_i}(D)$, prove that $N_G(D)$ cannot be a $p$-group.

2. Thus $D$ is normalized by an element of order $q$. Conclude that $D$ is contained in every Sylow $p$-subgroup.

3. Hence $D$ is equal to the intersection of all $p$-Sylows, so $D \trianglelefteq G$ and we are done if $D \neq 1$.

4. Suppose then that $D = 1$. Then the $p$-Sylows have pairwise trivial intersection; by counting elements prove that $G$ contains a normal $q$-Sylow.

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References: (The result of Frobenius is §6, p.186 in [1].)

[1] F. G. Frobenius, Über endliche Gruppen, Sitzungsberichte der Königl. Preuß. Akad. der Wissenschaften (Berlin) (1895), 163-194.

[2] C. Jordan, Sur les groupes d'ordre $p^m q^2$, J. de Math., (5), t. IV (1898), 21-26.