Cfrgtkky

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

$A_6 simeq mathrm{PSL}_2(mathbb{F}_9) $ only simple group of order 360

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, $G$ shoul 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 $;langle a, b vert a^{2} = b^{3} = (ab)^{5} = 1rangle;$ is isomorphic to $A_{5}$, as proved here :

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 elemnts $a, b$ 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.

Edit : I now asked the question on Mathoverflow :
https://mathoverflow.net/questions/293043/isomorphy-of-simple-groups-of-order-360-a-proof-with-a-presentation

But there, they closed the question as off-topic, because it does not appear to be about research level mathematics within the scope defined in the help center. Well, it seems impossible to know if this proof is already in the literature...