4
$\begingroup$

I want to show that any solvable group of order sixty has a normal subgroup of order five.

Recall that a solvable group is a group that has a normal series in which every factor is abelian.

I think that I can prove this by looking at the Hall subgroups of the group. Let $G$ be a solvable group of order sixty. We know that $60=2^2\cdot 3\cdot 5$. A Hall divisor $d$ of $G$ is a divisor of $60$ such that $\gcd(d, 60/d)=1$, and $H\leq G$ is a Hall subgroup if the order of $H$ is a Hall divisor. In particular, I know that $5$ is a Hall divisor of $G$.

Now since $5$ is a prime divisor of $G$, I know that $G$ has a $5$-Sylow subgroup $S$, and I also know that its order has to be $5$, by Lagrange's Theorem. If I can show that $S$ is the only $5$-Sylow subgroup, then I know it is normal, and I am done. I don't know if this is the case, and I don't know how the fact that $G$ is solvable helps.

Any help is appreciated, thank you.

$\endgroup$
  • $\begingroup$ From Sylow's third theorem we have the number of Sylow $5$ subgroups $n_5$ divides $m = 12$ and $n_5 \equiv 1 \pmod 5$. The only possibilities are $1$ and $6$, so have to eliminate the case of 6. Not sure how to proceed from here yet. $\endgroup$ – Tob Ernack Dec 2 '17 at 5:20
  • $\begingroup$ Also $A_5$ is the only non-solvable group of order 60. $\endgroup$ – Tob Ernack Dec 2 '17 at 5:31
2
$\begingroup$

Let $K$ be a minimal normal subgroup of $G$; $|K|$ is either $2$, $3$, $4$, or $5$.

  • If $K$ has order $3$, then $G/K$ has order $20$, and has a normal Sylow 5-subgroup (by Sylow's theorems). The preimage is a normal subgroup of order $15$ in $G$. Again, by Sylow's theorems, the Sylow 5-subgroup is normal (in the subgroup, and then the whole group).
  • If $K$ has order $4$, $G/K$ has order $15$, and the arguments above are basically reversed. Namely, $G/K$ has a normal Sylow 5-subgroup, we take the preimage, and get a normal subgroup of order $5$ again.
  • If $K$ has order $5$, we're done.

What remains is the case $K$ has order $2$, and thus $G/K$ has order $30$. If the Sylow 5-subgroup of $G/K$ is normal, then just like above, $G$ has a normal subgroup of order $5$ (since the Sylow 5-subgroup of a group of order $10$ is always normal).

That means all that is left is to prove a group $H$ of order $30$ always has a normal Sylow 5-subgroup. We can proceed the same way we did above: find a minimal normal subgroup, look at the quotient, etc. The same arguments show $H$ has a normal Sylow 5-subgroup. [Thanks to @DerekHolt for mentioning this approach, which is pedagogically cleaner.]

[Original Version: This can be done by considering the action of $H$ on itself, by right multiplication. An element of order $2$ will act as an odd permutation, which shows $H$ has a normal subgroup of order $15$. Once more, we finish by noting that a group of order $15$ always has a normal Sylow 5-subgroup.]

Here's a second approach, which is more case-by-case, but also more elementary:

Assume $G$ has no normal Sylow 5-subgroup. Then it has $6$ such subgroups, and a case-by-case analysis shows it has $10$ Sylow 3-subgroups. Counting elements then shows $G$ has $5$ Sylow 2-subgroups. The conjugation action on these Sylow 2-subgroups embeds $G$ in $S_5$, which means $G\cong A_5$, contradicting solvability.

$\endgroup$
  • 1
    $\begingroup$ I think it is easier to use the same approach on the group of order $30$ as on the group of order $60$. In this case it has a normal subgroup $L/K$ of order $2$, $3$ aor $5$, and in each case $G/L$ has a normal Sylow $5$-subgroup, hence so does $G/K$, hence so does $G$. $\endgroup$ – Derek Holt Dec 2 '17 at 11:18
  • $\begingroup$ @SteveD How do you know that $G$ can only have minimal normal subgroups of order $2$, $3$, $4$, or $5$? $\endgroup$ – A.M. Dec 2 '17 at 14:58
  • $\begingroup$ @A.M.: the minimal normal subgroups of a finite solvable group are always elementary abelian groups. $\endgroup$ – Steve D Dec 2 '17 at 18:19
  • $\begingroup$ @DerekHolt: yes, that's a better way to do it. I'll edit the answer later today. $\endgroup$ – Steve D Dec 2 '17 at 18:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.