The class equation of G is 60=1+15+20+12+12. Then G is simple.

I think I found a proof but not sure. Let A_i denote the classes.

Class Number of elements in the class

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A1 1

A2 15

A3 20

A4 12

A5 12

Let N be normal in G, g in N, g not 1. Then g belongs to, say, A2. But every conjugate of g is in N because N normal. Hence N contains A2. But 1 does not belong to A2. Hence |N| = 20 or |N| = 30. In any case there exists h in N such that h belongs to, say A3. But then A2 union A3 contained in N. So |N| > 35. Then N = G. Repeating this procedure for the different choices, I get N = G in every case. But I distrust this proof. It seems too simple and is based on a lot of manual computations. Is it correct?

Re: The class equation of G is 60=1+15+20+12+12. Then G is simple.

If such a G of order 60 has that class equation, it indeed must be simple. The main point of your proof (although it is hard to understand) is this:

A normal subgroup must be a union of conjugacy classes (since a normal subgroup contains conjugates of all its elements). But the following sums are not divisors of 60:

1+15

1+20

1+12

1+12+12

1+12+15

any other sums will total more than 30, and if a subgroup contains more than half the elements of a given group, it must contain the entire group.

thus the only PROPER subgroup that is normal is the trivial subgroup, hence G is simple.

Re: The class equation of G is 60=1+15+20+12+12. Then G is simple.

Thank you very much Deveno.