Consider the following question:
Show that a group $G$ of order $8p$ is solvable, for any prime $p$.
I am kind of stuck, but here are my first attempts:
I chose the series of subroups $G>H_8>H_4>H_2>H_1=\{e\}$ where $H_k$ has order $k$. All of these subgroups exist due to Sylow's Theorems. We have the quotients $H_8/H_4\cong H_4/H_2 \cong H_2/H_1 \cong \mathbb Z/2\mathbb Z$, so $G>H_8\rhd H_4\rhd H_2\rhd H_1=\{e\}$. Only the factor $G/H_8 \cong \mathbb Z/p\mathbb Z$ is causing me a headache, because although its of prime order (and hence abelian) I don't know whether $H_8$ is normal in $G$ (unless $p=2$).
If $p\ne 2$, then the number $k$ of Sylow $2$-subgroups is $1$ mod $2$. Since $k$ divides $|G|$ we might have $k=1$ or $k=p$. If we had $k=1$ then $H_8$ would be the only Sylow $2$-subgroup (and hence normal in $G$). But how could we show this?
Or is there even an easier way to approach this problem?
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edit: I am going to try a different approach: The case $p=2$ is clear. 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 (as Mariano pointed out).
In any other case we get a Sylow $p$-subgroup $H_p$ of order $p$, which is is normal in $G$. The quotient $G/H_p$ has order $8=2^3$; and prime power order implies solvable. So $G/H_p$ is solvable and $H_p$ is also solvable. And since the composition factors of $G$ are those of $H_p$ together with those of $G/H_p$, we conclude that $G$ is solvable iff $H_p$ and $G/H_p$ are solvable, which is the case. Hence $G$ is solvable if $|G|=8p$.
Best Answer
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.