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 $G=H$.

  • $\begingroup$ 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! $\endgroup$
    – Limsup
    Oct 15 '20 at 12:08

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