Normal Sylow Subgroups of Solvable Groups

Question

If $G$ is square free such that $|G|=p_1\cdots p_n$, where $p_n>p_{n-1}>\cdots$. Then we can use the N/C Theorem, along with some induction, to show that the Sylow $p_n$-subgroup of $G$, $P_n$, is normal in $G$.

Suppose $G$ isn't square free, and let's assume $G$ is solvable. Suppose $p_k$ is the largest prime divisor of $G$ with exponent $1$. Could we necessarily say $P_k\trianglelefteq G$?

Answer 1

The answer is no, as noted in a comment by Steve D. For example, consider $G = A_4$, which has order $2^2 \cdot 3$ but does not have a normal Sylow $3$-subgroup.

In fact a finite solvable group does not need to have any normal Sylow subgroups, consider for example $G = S_4$.

But the answer is yes for some classes of solvable groups. For example besides groups of squarefree order, more generally the property you ask about holds for supersolvable finite groups.

That is, if $G$ is a finite supersolvable group and $p$ is the largest prime divisor of $|G|$, then $G$ has a normal $p$-Sylow.

Answer 2

$\DeclareMathOperator{\Aut}{Aut}\newcommand{\Span}[1]{\left\langle #1 \right\rangle}$One of the possible generalizations of the $A_{4}$ example is the following.

Let $p$ be any odd prime.

Let $k$ be the order of $2$ modulo $p$, so that $p \mid 2^{k} - 1$.

Let $N$ be an elementary abelian group of order $2^{k}$. Then $N$ has an automorphism $\alpha$ of order $p$. This is because $N$ can be regarded as the additive group of the field $F$ with $2^{k}$ element. The multiplicative group of $F$ is cyclic of order $2^{k} - 1$, generated by an element $b$, say. Multiplication by $b$ yields an automorphism $\beta$ of $N$ of order $2^{k} - 1$, and then one can take $$ \alpha = \beta^{(2^{k} - 1)/p}. $$

Let $P = \Span{a}$ be a cyclic group of order $p$

The homomorphism $P \to \Aut(N)$ determined by $a \mapsto \alpha$ yields a semidirect product $$ G = N \rtimes P $$ which is soluble, and whose order $p \cdot 2^{k}$ is not squarefree as $k > 1$.

Since $N$ is normal in $G$, but $G$ is non-abelian, we have that $P$ is not normal in $G$.

The case of $A_{4}$ occurs when $p = 3$, and $k = 2$.