It is false in general that "abelian intersect maximal implies normal": $A_5$ is maximal in $S_5$, the subgroup generated by $(1,2,3,4)$ is abelian, but the intersection of the two is nontrivial (contains $(1,3)(2,4)$) and not normal in $S_5$.

However, it *is* true that the intersection of a maximal subgroup and an **abelian normal subgroup** is normal.

**Proposition.** Let $G$ be a group, and $H$ a maximal subgroup of $G$. If $N$ is an abelian normal subgroup of $G$, then $N\cap H$ is normal in $G$.

*Proof.* If $N\subseteq H$, then $N\cap H = N\triangleleft G$ and we are done.

If $N$ is not contained in $H$, then maximality of $H$ and normality of $N$ imply that $HN=G$ (since $HN$ is a subgroup). Let $x\in H\cap N$ and $g\in G$. Then we can write $g = hn$ with $h\in H$ and $n\in N$. Then
$$gxg^{-1} = (hn)x(hn)^{-1} = h(nxn^{-1})h^{-1} = hxh^{-1}$$
with the last equality since $N$ is abelian. Now, $x,h\in H$, so $hxh^{-1}\in H$. And $x\in N$, so $hxh^{-1}\in N$. Thus, $hxh^{-1}\in H\cap N$, proving that $H\cap N$ is normal in $G$. **QED**

*Added.* More generally: note that $H\cap N$ is certainly normal in $H$. If $HN=G$, then you only need to show that $N$ normalizes $H\cap N$: then $(hn)x(hn)^{-1} = h(nxn^{-1})h^{-1}$, which will lie in $H\cap N$ if $nxn^{-1}\in H\cap N$. Therefore:

**Proposition.** Let $H$ be a subgroup of $G$ and let $N$ be a normal subgroup of $G$. Then $H\cap N\triangleleft HN$ if and only if $N\subseteq N_{HN}(N\cap H)$, where $N_{NH}(N\cap H)$ is the normalizer of $N\cap H$ in $NH$.

*Proof.* If $N\subseteq N_{NH}(N\cap H)$, then the argument proceeds as above. Conversely, if $N\cap H\triangleleft NH$, then $N\subseteq NH=N_{NH}(N\cap H)$. $\Box$

In particular, if $N$ is abelian then $N\subseteq C_{NH}(N\cap H)\subseteq N_{NH}(N\cap H)$; and if $H$ is maximal, then this gives the proposition above in the nontrivial case.

Now apply this to $H$ and the abelian normal subgroup $G^{(n-1)}$ to conclude that $H\cap G^{(n-1)}\triangleleft G$.

Now, you know that $G^{(n-1)}$ is normal in $G$ (because the derived terms are always normal in $G$), and has order $p$. Since a $p$-Sylow subgroup of $G$ has order $p$, then $G^{(n-1)}$ is a $p$-Sylow subgroup of $G$. Since it is normal in $G$, the $p$-Sylow subgroup of $G$ is normal in $G$. So $G$ is contained in the normalizer of the $p$-Sylow subgroup $G^{(n-1)}$ of $S_p$.

The number $n_p$ of $p$-Sylows divides $8$, so it is one of $1$, $2$, $4$ or $8$, and it is congruent to $1$ modulo $p$, so it is of the form $ap+1$ for some $a$.

$n_p$ cannot be $2$ for then $ap=1$, and $p\neq1$.

If $n_p$ is $4$, then $ap=3$ so $p=3$, and

if $n_p$ is $8$, then $ap=7$ and $p=7$.

It follows that for most primes we have $n_p=1$ so the $p$-Sylow is *normal*, and
you can start the composition series on the other end!

Notice that when $p=3$ or $p=7$ the group has order $24$ or $56$ and, since the smallest simple non-abelian group has order $60$, the group must also be solvable in these exceptional cases.

## Best Answer

Yes you can conclude that $G$ is not solvable, and many of your assumptions are unnecessary.

Let $p$ be prime, and let $G$ be a solvable subgroup of $S_p$ containing a $p$-cycle. Then $G$ is transitive and hence primitive, so its nontrivial normal subgroups are transitive.

Let $P$ be a minimal normal subgroup of $G$. Then $P$ is abelian and transitive, so $|P|=p$, and hence $G$ is contained in the normalizer of $P$ in $S_p$, which is a $2$-transitive group of order $p(p-1)$ in which no non-identity element fixes more than one point