Show a group of order 60 is simple. I am given that $G$ is a group of order $60$, with $20$ elements of order $3$, $24$ elements of order $5$ and $15$ elements of order $2$. I have to show that $G$ is isomorphic to $A_5$.
I think that the best way to go about this is to prove that $G$ is itself simple, rather than list out the elements of $A_5$, and then it just follows that it is isomorphic to $A_5$, but this where I am stuck. I have tried using Sylow's Theorem to show there are no normal subgroups but to no avail.
I would appreciate if someone could help me or even point me in the right direction. Thanks.

(1) There are more than one Sylow-$2$ subgroups (order $4$). [Hint: see no. of elements of order $2$].
(2) Let $H_1,H_2$ be two (abelian) Sylow-$2$ subgroups. Suppose they intersect non-trivially. Then $|H_1\cap H_2|=2$. 


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*(2.1) Let $x\in H_1\cap H_2$ with $x\neq 1$. Show: $|C_G(x)|\geq 8$. [Hint $C_G(x)\supseteq H_i$]

*(2.2) Show $|C_G(x)|$ can not be $8$ (Hint: what is order of $G$?)

*(2.3) Note that $C_G(x)$ contains $H_1$ so $|H_1|$ divides $|C_G(x)|$. By (2.1) and (2.2), show $3$ or $5$ divides $|C_G(x)|$. 

*(2.4) $C_G(x)$ contains an element of order $3$ or $5$, say $z$. Then order of $xz$ is $2.3$ or $2.5$, contradiction; why? [see hypothesis].
(3) Show that any two Sylow-$2$ subgroups intersect trivially; each contains $3$ elements of order $2$. Deduce that there are exactly $5$ Sylow-$2$ subgroups. 
(4) Let $H_1,H_2,H_3,H_4,H_5$ be the five Sylow-$2$ subgroups. $G$ acts on them by conjugation, and transitively (why?) [Hint: Recall Sylow's theorems]
(5) Thus we get a homomorphism from $G$ to $S_5$. The order of image is at least $5$ (there is $g_i$ taking $H_1$ to $H_i$, $i=1,2,3,4,5$). Hence order of kernel is at most $12$. 
(6) By arguments of Derek Holt in comment, deduce that kernel should be trivial.
(7) Deduce that $G$ is isomorphic to subgroup of $S_5$. 
(8) Take a long breath!!!!
