# $G$ is a simple group of order $60$.Then $G$ contains a subgroup of order 12

$G$ is a simple group of order $60$. Then show that $G$ contains a subgroup of order 12.

* MY TRY:

Suppose $G$ has no subgroup of order 12. Let $n_5$ denote the number of Sylow 5-subgroups of $G$. Then $n_5$ is either 1 or 6. $n_5$ cannot be 1, for this would contradict that the group is simple. If $n_5=6$, then we have 24 elements of order 5.

Now $n_2=1,3,5$, or $15$. If $n_5=15$, let the Sylow 2-subgroups be $B_1,B_2,\dotsc$. Then either $B_i\cap B_j=\{e\}$ or $|B_i\cap B_j|=2$. If $B_i\cap B_j=\{e\}$, then the group would have at least 70 elements (45 + 24 + 1), a contradiction.

If $|B_i\cap B_j|=2$, then $B_i\cap B_j$ is a normal subgroup of both $B_i$ and $B_j$. Then I considered $N(B_i\cap B_j)=\{g\in G:g(B_i\cap B_j)g^{-1}=B_i\cap B_j\}$, and arrived at a contradiction (not providing details though) on considering $o(N(B_i\cap B_j))$.

Thus $n_2\neq 15$. $n_2\neq 1$ as the group is simple.

How can I eliminate the remaining cases that $n_2$ is 3 or 5?

• That makes more sense ;-) Commented Mar 12, 2015 at 18:32
• $n_2 = 3$ gives you a non-trivial homomorphism to $S_3$.
– j.p.
Commented Mar 12, 2015 at 18:35
• The non-trivial homomorphism to $S_3$ is given by the action of $G$ by conjugation on the $3$ Sylow $2$-subgroups. Commented Mar 12, 2015 at 19:35
• @DerekHolt The homomorphism you mention does not exist. Indeed: Assume $n_2 =3$. By conjugation on the 2-Sylowgroups we would obtain $\phi \colon G \to S_3$. By cardinality arguments ker $\phi \neq 1$ and since $G$ is simple thus ker $\phi = G$. But then any 2-Sylowgroup is normal in $G$ which is a contradiction.
– sf1
Commented Mar 12, 2015 at 22:15
• @sf1 Well yes, the point of this argument is to show that $n_2 \ne 3$. Commented Mar 12, 2015 at 22:23

Let $$\mathfrak{X}_p$$ be the set of all Sylow $$p$$-subgroups of $$G$$. Recall that (1) any two Sylow $$p$$-subgroups are conjugate and (2) Every conjugate of a Sylow $$p$$-subgroup is also a Sylow $$p$$-subgroup. Thus $$G$$ acts on $$\mathfrak{X}_p$$ by conjugation. This action induces a homomorphism $$\phi_p \colon G \rightarrow A(\mathfrak{X}_p)$$, where $$A(\mathfrak{X}_p)$$ is the group of all permutations of $$\mathfrak{X}_p$$. That is, $$\phi_p(g) : X \mapsto g.X=g^{-1}Xg$$, where $$X \in \mathfrak{X}_p$$ and $$g \in G$$.
Suppose $$n_2=3$$. Since $$|\mathfrak{X}_2| = n_2=3$$, we can identify $$\phi_2 \colon G \rightarrow A(\mathfrak{X}_2)=S_3$$. Since $$\ker \phi_2$$ is a normal subgroup of $$G$$ and $$G$$ is simple, $$\ker \phi_2 = \{e \}, G$$. (Here $$e$$ denotes the identity element of $$G$$.) By the first isomorphism theroem, $$G / \ker \phi_2 \cong \text{im}(\phi_2)$$. Calculating cardinalities yields $$\frac{|G|}{|\ker \phi_2 |} = |\text{im}(\phi_2) | \leq 6$$
Thus $$\ker \phi_2 = G$$. That is, for any $$g \in G$$, $$\phi_2 (g) : X \mapsto g^{-1}Xg$$ is the identity map. In other worlds, every $$X \in \mathfrak{X}_2$$ is a normal subgroup of $$G$$; contradicts to the simplicity of $$G$$.
Now I'll give a proof for the remaining part. As you showed that $$n_2 \neq 1, 15$$, we have $$n_2 = 5$$. Now $$\phi_2 \colon G \rightarrow A(\mathfrak{X}_2)=S_5$$. Since any two Sylow $$2$$-subgroups of $$G$$ are conjugate, we have $$|\text{im}(\phi_2)| \geq n_2 = 5$$. (Write $$\mathfrak{X}_2 = \{ X_1, X_2, \cdots, X_5 \}$$. Then there exists $$\sigma_j \in \text{im}(\phi_2)$$ sending $$X_1$$ to $$X_j$$, $$j=1, \cdots, 5$$.)
Observe that $$\frac{|G|}{|\ker \phi_2 |} = |\text{im}(\phi_2) | \geq 5$$ Thus $$\ker \phi_2$$ is trivial. That is, $$G$$ is isomorphic to a subgroup of $$S_5$$. Recall that the only subgroup of $$S_5$$ of index $$2$$ is $$A_5$$, so $$G \cong A_5$$. Note that $$A_5$$ has a subgroup of order $$12$$, namely, $$A_4$$.