Group of order $60$ with $20$ elements of order $3$

Question

I am trying to show that a group $G$ of order $60$ with $20$ elements of order $3$ is isomorphic to $A_5$.

I know that $A_5$ is the unique simple group of order $60$. I can conclude if I show that $n_5 = 6$.

I know $n_3 = 10$. I have made multiple attempts, counting arguments don't seem to be working. Most successful so far I showed that if $H$ is a non trivial normal subgroup of $G$, if $3$ divides $|H|$ then $H$ must have order $30$.

I am out of ideas.

Answer 1

I think the following argument works.

Suppose that $n_5 \neq 6$. The only other possible value is $1$, so $n_5=1$. Let $P$ be the unique (i.e. normal) Sylow $5$-subgroup of $G$, and let $Q$ be a Sylow $3$-subgroup of $G$. Since there are exactly $20$ elements of order $3$ in $G$, it follows that $|N_G(Q)|=6$.

But $P$ is normal in $G$, thus it permutes with every other subgroup. In particular, $PQ$ is a subgroup of $G$ and it has order $15$. Now every group of order $15$ is abelian, so $P$ centralises $Q$ and therefore $P \leq N_G(Q)$. This implies that $|P| = 5 \mid 6 = |N_G(Q)|$, a contradiction.

So we see that $n_5=6$, and from this it follows that $G \cong A_5$. (I'm not providing a proof of this last claim because in your question you mention that you know this.)

Answer 2

Another approach is to use the counting portion of Philip Hall's generalization of Sylow's theorem to Hall subgroups of solvable groups.* That implies in particular that in any solvable group the number of $p$-Sylows is not only congruent to $1\pmod p$ but also a product of prime powers each of which is congruent to $1 \pmod p$. This shows immediately that no solvable group can have $n_3=10$. But since $A_5$ is the smallest non-abelian simple group, it is the only non-solvable group of order $60$.

• The full statement is that for any solvable finite group $G$ and set of primes $\pi$, then

(1) There exist Hall-$\pi$ subgroups (of order divisible only by primes in $\pi$ and index relatively prime to all primes in $\pi$)

(2) All Hall-$\pi$ subgroups are conjugate

(3) The number of Hall-$\pi$ subgroups is a product of prime powers that are orders of chief factors of $G$, each of which is congruent to $1$ modulo some prime in $\pi$.

Here a chief factor is a factor group in chief series, which is just like a composition series except that all the groups in the series are normal in $G$, not just in the next group up. Chief factors are always direct products of one or more copies of the same simple group. For solvable groups, this means they are elementary abelian and therefore have prime power order.