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$G$ 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?

Nick Matteo
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  • That makes more sense ;-) – amWhy Mar 12 '15 at 18:32
  • $n_2 = 3$ gives you a non-trivial homomorphism to $S_3$. – j.p. Mar 12 '15 at 18:35
  • please explain @j.p. by giving an answer – Learnmore Mar 12 '15 at 18:41
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    The non-trivial homomorphism to $S_3$ is given by the action of $G$ by conjugation on the $3$ Sylow $2$-subgroups. – Derek Holt Mar 12 '15 at 19:35
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    @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 Mar 12 '15 at 22:15
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    @sf1 Well yes, the point of this argument is to show that $n_2 \ne 3$. – Derek Holt Mar 12 '15 at 22:23
  • can u please explain it in a more detailed manner @sf1 – Learnmore Mar 13 '15 at 03:19

1 Answers1

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This is a detailed answer based on old comments.

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, $A_4$.

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