Let $G$ be a group of order $2^2 \cdot 5 \cdot 23^r$ where $r \gt 0$
a) Show that $G$ is not simple.
b) Show that $G$ is solvable.
My attempt:
a) Assume that:
$S$ is the set of sylow-$23$-subgroups and $n_{23}$ is the number of conjugates of $S$
By Sylow's 3rd theorem, we have
$n_{23} | 2^2 \cdot 5$ and $n_{23}\equiv 1 \pmod {23}$
Which implies that $n_{23}=1,\ 2^a,\ 5,\ 2^b \cdot 5$ where $a,b\leq 2$, and $n_{23}=1+23k$
Since $23>2^a,\ 5,\ 2^b.5$, if we assume that $n_{23}= 2^a,\ 5, $ or $2^b \cdot 5$, that would lead us to a contradiction. Hence, $n_{23}=1$, and then $S$ is normal.
As a result, $G$ is not simple.
b) $S$ is a $23$-group $\implies S$ is nilpotent $\implies S$ is solvable.
Since $S$ is normal, $G/S$ is a group. We have
$o(G/S)=23$ $\implies G/S$ is cyclic $\implies G/S$ is abelian $\implies G/S$ is solvable $\implies G$ is solvable.
Edit:
$S$ is a $23$-group $\implies S$ is nilpotent $\implies S$ is solvable.
Since $S$ is normal, $G/S$ is a group. We have
$o(G/S)=2^2 \cdot 5$
Suppose that $G/S=H$.
We have $o(H)=2^2 \cdot 5$
Suppose that
$R$ is the set of Sylow-${5}$-subgroups and $n_{5}$ is the number of conjugates of $R$
From the 3rd Sylow's theorem, we have
$n_5\equiv 1 \pmod 5$ and $n_5 | 2^2$
which implies that
$n_5=1+5k$ and $n_5=1,\ 2,$ or $2^2$.
Since $5>\gt 2,\ 2^2$, the only possible value of $n_5$ is $1$. So we can conclude that $R$ is normal in $H$.
$R$ is a $5$-group $\implies R$ is nilpotent $\implies R$ is solvable.
Since $R$ is normal, $H/R$ is a group. We have
$o(H/R)=4$ $\implies H/R$ is abelian $\implies H/R$ is solvable $\implies H=G/S$ is solvable $\implies G$ is solvable.
The Sylow $5$-subgroup of $G/S$ must be normal because $n_5(G/S)=1, 2, \text{or } 4$.