# Group of order $8p$ is solvable, for any prime $p$

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 $|G|=8p$.

• This is direct by Philip Hall's theorem. Dec 4, 2014 at 3:31

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.

• So (for most primes) we have a Sylow $p$-subgroup $H$ of order $p$ which is normal in $G$, but how do we know that the quotient $G/H$ is simple? Apr 7, 2013 at 2:37
• It isn't. It can't be: it has order $8$! Apr 7, 2013 at 2:41
• But for a group to be solvable, don't we need the factors/quotients to be simple? Apr 7, 2013 at 2:44
• I never said the only group in the chain would be that of order $p$... Apr 7, 2013 at 2:46
• I think for a problem at this level it is cheating to use the fact that the smallest nonabelian simple group has order 60, since proving that is probably harder than the problem itself! Apr 7, 2013 at 10:40