# Prove that the given simple group can be generated by two elements.

Given a simple group $$G$$ of order $$60$$. I have to prove that $$G = \langle a, b \rangle$$, where $$a, b \in G$$, $$a$$ has order $$5$$ and order of $$b$$ is $$3$$.
I don't quite know how to act in this situation (in general how to prove that some group is generated by two elements, especially when the group is not abelian). For example, I don't see how do we get an element of order two.
We can use the fact that $$G \simeq A_6$$, but I would like to understand how to solve these kind of exercises in general.

First of all, by Cauchy's theorem we know there are some elements $$a,b$$ with orders $$5$$ and $$3$$ respectively. By Lagrange's theorem it follows that $$|\langle a,b\rangle|$$ is divisible by both $$3$$ and $$5$$, hence it is divisible by $$15$$. Also, $$\langle a,b\rangle\leq G$$, and so $$|\langle a,b\rangle|$$ must divide $$60$$. So it follows that the order of $$H:=\langle a,b\rangle$$ must be either $$15, 30$$ or $$60$$.
If $$|H|=30$$ then it means $$H$$ is a subgroup of $$G$$ with index $$2$$, and hence it is normal in $$G$$. This is a contradiction to $$G$$ being simple.
Now assume $$|H|=15$$. Then it is a subgroup of $$G$$ of index $$4$$. We can define an action of $$G$$ on the left cosets $$G/H$$ by $$g.xH=gxH$$. As always, an action induces a homomorphism $$\varphi: G\to S_{G/H}$$ by $$\varphi(g)(xH)=gxH$$. Since $$G$$ is simple the homomorphism must be either trivial or injective. It obviously can't be injective because $$|G|=60$$ and $$|S_{G/H}|=24$$. Also, take any element $$g\notin H$$. Then $$\varphi(g)(H)=gH\ne H$$, i.e $$\varphi(g)$$ is not the identity permutation. So $$\varphi$$ also isn't the trivial homomorphism. Again, a contradiction.
So we have no choice, he must have $$|H|=60$$, and so $$G=H$$.
• Ok, I see now. I was going the hard way, considering all the elements of $G$ and trying to prove they can be expressed as a "combination" of $x$ and $y$, which obviously is not a useful method. Thanks! Oct 15 '20 at 12:08