Normal group of order 60 isomorphic to A_5

[fishturtle1]

Homework Statement

Prove that any simple group $G$ of order $60$ is isomorphic to $A_5$. (Hint. If $P$ and $Q$ are distinct Sylow $2$ subgroups having a nontrivial element $x$ in their intersection, then $C_G(x)$ has index $5$; otherwise, every two Sylow $2$ subgroups intersect trivially and $N_G(P)$ has index $5$.

Relevant Equations

Theorem: Let $H$ be a subgroup of $G$ with index $n$. Then there exists homomorphism $\psi : G \rightarrow S_n$.

Sylow's Theorem: Let $p$ be a prime divisor of $G$ and $n_p$ be the number of Sylow p subgroups of $G$. Then $n_p \vert \vert G \vert$ and $n_p = 1 (mod p)$. Also, the Sylow $p$ subgroups are conjugate to one another.

Proof: We note $60 = 2^2\cdot3\cdot5$. By Sylow's theorem, $n_5 = 1$ or $6$. Since $G$ is simple, we have $n_5 = 6$. By Sylow's theorem, $n_3 = 1, 4,$ or $10$. Since $G$ is simple, $n_3 \neq 1$. Let $H$ be a Sylow $3$ subgroup and suppose $n_3 = 4$. Then $[N_G(H) : G] = 4$ which implies there is homomorphism $\psi : G \rightarrow S_4$. But $\ker \psi = 1$. By 1st iso, $S_4$ contains a subgroup isomorphic to $G$. But $S_4$ contains strictly less elements than $G$, contradiction. Hence, $n_3 \neq 4$. We may conclude $n_3 = 10$. Hence $G$ contains $1$ element of order $1$, $24$ elements of order $5$, and $20$ elements of order $3$.

By Sylow's theorem, $n_2 = 1, 3, 5,$ or $15$. By a similar argument as above, we see $n_2 \neq 1, 3$. I think we need to show $n_2 = 5$ but I'm not sure how.

I'm also having trouble with the first part of the hint: Suppose there are distinct Sylow $2$ subgroups $P \cap Q$ that don't interact trivially. Let $x \in P \cap Q$ be a nonidentity element. Then $x$ has order $2$. I see that if I can prove the hint, then there is an homorphism $\psi : G\rightarrow S_5$ with $ker \psi = 1$. And therefore $G$ iso. to $A_5$(i think...) But I'm having trouble with the first part of the hint. How can I proceed, please?

 

[member 587159]

Let me start with proving the hint.

Note first that it is possible that $x$ has order 4. You don't necessarily know that $x$ has order $2$. What you do know is that by Cayley's theorem that $P\cap Q$ contains some element of order $2$. Alright, since $P,Q$ are groups of order $4$ , they are abelian so you know that $x$ commutes with every element in $P$ and $Q$ hence, since P
$P$ and $Q$ are distinct you know that $x$ commutes with at least 5 distinct elements´. Hence $|C_G(x)|\geq 5$ and thus since this order also divides $60$ we have $|C_G(x)|\geq 12$ (order $6$ is not possible since then this centraliser contains a subgroup of order 3 and then we have $5+2$ distinct elements already), so the index is at most $5$.

If $n=[G:C_G(x)]$, then $G$ acts on the cosets of $C_G(x)$ and we have an induced group morphism $G\to \operatorname{Sym}(G/C_G(x))\cong S_n$ and since $G$ is simple the kernel is trivial so this is an injection´. Hence, $60=|G|\leq n!$, from which it follows that $n\geq 5$.

 

[fishturtle1]

Thank you! My only question is, how did we rule out $C_G(x) \neq 10$?

Proof: Let $P, Q$ be distinct Sylow $2$ subgroups of $G$. Suppose $P, Q$ do not intersect trivially. Then there exists non identity element $x \in P \cap Q$. Since $P, Q$ have order $4$, they are abelian. Hence, $x$ commutes with at least $5$ elements. We note these $5$ elements have order $1, 2$ or $4$. So, $\vert C_G(x) \vert \ge 5$. We know $\vert C_G(x) \vert \vert 60$. If $C_G(x) \ge 15$, then we can find an injective homomorphism $\psi : G \rightarrow S_k$ for some $k \le 4$. Since $G$ has more elements than $S_k$ for $k \le 4$, we can conclude $C_G(x) \le 12$.

From a previous note, $\vert C_G(x) \vert \neq 5$ since it contains an element of order $2$ or $4$. So, $\vert C_G(x) \vert = 10$ or $12$. So, the index of $C_G(x)$ is $5$ or $6$.

If $n = [G : C_G(x)]$, then there is an homomorphism $\psi : G \rightarrow S_n$ with $\ker \psi = 1$. Hence, $G$ is isomorphic to some subgroup of $S_n$. This implies $\vert G \vert \le n!$. So, $n \ge 5$.

On the other hand, suppose any two distinct Sylow $2$ subgroups of $G$ intersect trivially. In the OP, we see $n_2 = 5$ or $15$. If $n_2 = 15$, then we get $45$ elements in addition to the $45$ we've already found, a contradiction. Hence, $n_2 = 5$. Let $P$ be a Sylow $2$ subgroup. Then $\vert \mathcal{O}(P) \vert = 5$ where $\mathcal{O}(P)$ is the orbit of $G$ acting on $P$ by conjugation.

In either case, we've found a subgroup of $G$ with index $5$. So there is an homomorphism $\psi: G \rightarrow S_5$ with $\ker\psi = 1$. So, $G$ is isomorphic to some subgroup of $S_5$ of order $60$. The only such subgroup of $S_5$ is $A_5$. Hence, $G \cong A_5$.[]

 

[mathwonk]

In my notes, it seems also to be left as an exercise, #36, on p.60, but maybe the hint and previous discussion are helpful.
https://www.math.uga.edu/sites/default/files/inline-files/843-1.pdf

 

[fishturtle1]

Response to post #4

I think I can do the element of order $2$ part, but showing $N_G(x)$ has index $5$ has me stuck, unless we use post #2.

Proof: Continuing from the OP, we have $n_2 = 5$ or $15$. Suppose $n_2 = 5$ and let $P$ be a Sylow 2 subgroup of $G$. Then $N_G(P)$ has index $5$.

On the other hand, suppose $n_2 = 15$. Then there's exactly $15$ distinct Sylow 2 subgroups. If all pairs of these intersect trivially, then we get $45$ new elements in addition to the 45 we already found (in the OP), a contradiction. So there must be distinct Sylow 2 subgroups $P$ and $Q$ such that $P \cap Q \neq 1$. Let $x \in P \cap Q$ be a nonidentity element. If the order of $x = 4$, then it generates both $P$ and $Q$, a contradiction. Hence, the order of $x$ is $2$.

 

[member 587159]

Thank you! My only question is, how did we rule out $C_G(x) \neq 10$?

$C_G(x)$ contains $P$, a group of order $4$. Thus $4$ divides the order of $C_G(x)$ and thus this cannot be of order $6$ or $10$.

Let me know if you need any further help!

 

[fishturtle1]

$C_G(x)$ contains $P$, a group of order $4$. Thus $4$ divides the order of $C_G(x)$ and thus this cannot be of order $6$ or $10$.

Let me know if you need any further help!

That makes sense, thank you again!