# Any solvable group of order sixty has a normal subgroup of order five

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.

• 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. – Tob Ernack Dec 2 '17 at 5:20
• Also $A_5$ is the only non-solvable group of order 60. – Tob Ernack Dec 2 '17 at 5:31

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.

• 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 $G$. – Derek Holt Dec 2 '17 at 11:18
• @SteveD How do you know that $G$ can only have minimal normal subgroups of order $2$, $3$, $4$, or $5$? – A.M. Dec 2 '17 at 14:58
• @A.M.: the minimal normal subgroups of a finite solvable group are always elementary abelian groups. – Steve D Dec 2 '17 at 18:19
• @DerekHolt: yes, that's a better way to do it. I'll edit the answer later today. – Steve D Dec 2 '17 at 18:20