**Problem**. Suppose that are three prime numbers which are not necessarily distinct. Prove that if is a group of order then is not simple.

*Proof*. I proved in this post that a group of order where are (not necessarily distinct) primes is not simple. So I may assume that are distinct and, say, . For any prime number let be the number of Sylow -subgroups of Now suppose that is simple. So and, by the Sylow theorem, But we cannot have or because and hence Thus and so, since every Sylow -subgroup of has exactly elements of order the number of elements of order in is We also have and Thus there are more than elements of order Hence the number of elements of order is less than which is impossible because

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Quick question: Why does assuming that G is simple give n_s > 1 ? I know that for a group to be simple, it has no proper and non-identity normal subgroups, but how does that lead us to say n_s >1 ?

In general, If a (finite) group G has only one subgroup H of a given order, then H is normal in G because for every g in G we have |H|=|gHg^{-1}| and so H=gHg^{-1}.

Why is it that you can assume that the intersection of all the conjugate Sylow-r subgroups is only the identity element. (You make that assumption when you say that the number of elements of order r in G is pq(r-1))

If H and K are two distinct r-Sylow subgroups of G, then |H| = |K| = r. Let L be the intersection of H and K. Then L is a subgroup of both H and K. So |L| divides r = |H| = |K|. But r is a prime number and so either |L| = 1 or |L| = r = |H| = |K|. If |L| = 1, we are done. Otherwise, since L is a subgroup of H and K and |L|= |H| = |K|, we get L = H = K, which is false because H and K were distinct.