I asked this question on Stackexchange more than 24 hours ago and I got no answser, so I take leave to ask it here.

It is well known that all simple groups of ordre 360 are isomorphic with the alternating group $A_{6}$. Cole's original proof is here on Stackexchange :

In these forum discussions : http://mathforum.org/kb/message.jspa?messageID=3699550 and http://fr.sci.maths.narkive.com/7uqHmNIX/groupe-simple-d-ordre-360#post8

there is an allusion to a proof using a presentation, but this proof is not given explicitly.

I was interested in such a proof and I think I found one. It uses the following facts :

Fact 1. Let $G$ be a simple group of order 360. For every Sylow 3-subgroup $P$ of $G$, $P \setminus \{1\}$ consists in 8 elements of order 3.

Fact 2. Let $G$ be a simple group of order 360. The Sylow 3-subgroups of $G$ intersect pairwise trivially. This implies that every element of $G$ normalizing a subgroup $Q$ of order 3 of $G$ normalizes the (unique) Sylow 3-subgroup of $G$ that contains $Q$.

Fact 3. Let $G$ be a simple group of order 360. $G$ has exactly 144 elements of order 5.

Fact 4. Let $G$ be a simple group of order 360. $G$ has exactly 45 elements of order 2. For every Sylow 3-subgroup $P$ of $G$, there are exactly 9 elements of order 2 of $G$ normalizing $P$. Thus, for every Sylow 3-subgroup $P$ of $G$, there are exactly 36 elements of order 2 of $G$ that don't normalize $P$.

I found proofs of facts 1 to 4 on Internet some years ago, perhaps in the two forum discussions mentioned above. In any case, I can give complete proofs of these facts if someone asks for it.

Now, let $G$ be a simple group of order 360. Choose a Sylow 3-subgroup $P$ of $G$. Let $E$ denote the set of elements of order 2 of $G$ that don't normalize $P$. In view of Fact 4, $E$ has exactly 36 elements.

Let $P_{1}$ denote $P \setminus \{1\}$. Thus $P_{1}$ has cardinality 8.

Let us prove that the set $P_{1} E$ has exactly $8 \cdot 36 = 288$ elements. It suffices to prove that the surjective function

$f : P_{1} \times E \rightarrow P_{1} E : (x, y) \mapsto xy$ is injective.

Let $(x, y)$ and $(x', y')$ be elements of $P_{1} \times E$ such that

(1) $xy = x'y'$.

Then $y = x^{-1} x' y'$. Since the left member is of order 2, the second member is of order 2, thus

$(x^{-1} x') y' (x^{-1} x') y' = 1$

$y' (x^{-1} x') y' = (x^{-1} x')^{-1}$.

Since $y'$ is of order 2, this can be written

$ y' (x^{-1} x') y'^{-1}= (x^{-1} x')^{-1}$.

Thus, if $x$ and $x'$ were distinct, $y'$ should normalize a subgroup of order 3 of $P$; in view of fact 2, $y'$ should normalize $P$; this is false, since $y'$ is in $E$.

Thus $x = x'$, so (1) gives $y = y'$, thus $f$ is injective. As noted, this proves that the set $P_{1} E$ has exactly 288 elements. If none of these elements was of order 5, then , in view of fact 3, $G$ should have at least 288 + 144 = 432 elements, which is false since $G$ is assumed to have order 360.

Thus

(2) $G$ has an element $a$ of order 2 and an element $b$ of order 3 such that $ab$ is of order 5.

But the presentation group $<a, b \vert a^{2} = b^{3} = (ab)^{5} = 1>$ is isomorphic to $A_{5}$, as proved here :

https://math.stackexchange.com/questions/1999312/group-presentation-of-a-5-with-two-generators

Thus every group generated by elements $a, b$ such that $a^{2} = b^{3} = (ab)^{5} = 1$ is a homomorphic image of $A_{5}$. Since $A_{5}$ is simple, every nontrivial group generated by elements $a, b$ such that $a^{2} = b^{3} = (ab)^{5} = 1$ is thus isomorphic with $A_{5}$. In view of our result (2), $G$ has thus a subgroup isomorphic to $A_{5}$. Such a subgroup is of order 60 and thus of index 6 in $G$. Classically, a simple group that has a subgroup of index 6 is isomorphic with a subgroup of $A_{6}$, thus $G$ is isomorphic with a subgroup of $A_{6}$. Since both $G$ and $A_{6}$ have order 360, $G$ is isomorphic with $A_{6}$.

I presume that this proof is already in the literature. My question is : do you know a reference to the literature for this proof ? Thanks in advance.