# Class equation of G is 60=1+15+20+12+12. Then G is simple.

Hi: I think I have a proof, but I'm not quite 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?