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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 $|G|=8p$.

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

1 Answer 1

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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.

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  • $\begingroup$ 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? $\endgroup$
    – Phil-ZXX
    Apr 7, 2013 at 2:37
  • $\begingroup$ It isn't. It can't be: it has order $8$! $\endgroup$ Apr 7, 2013 at 2:41
  • $\begingroup$ But for a group to be solvable, don't we need the factors/quotients to be simple? $\endgroup$
    – Phil-ZXX
    Apr 7, 2013 at 2:44
  • $\begingroup$ I never said the only group in the chain would be that of order $p$... $\endgroup$ Apr 7, 2013 at 2:46
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    $\begingroup$ 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! $\endgroup$
    – Derek Holt
    Apr 7, 2013 at 10:40

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