Let and be primes with . Prove that a nonabelian group of order has a nonnormal subgroup of index , so that there exists an injective homomorphism . Deduce that is isomorphic to a subgroup of the normalizer in of the cyclic group generated by .

If , then is abelian, a contradiction. If is or , then is cyclic, so that is abelian. Hence .

Let be a nonidentity element; note that is either or , since if we have cyclic, a contradiction. Moreover is not all of , since otherwise . Since and is maximal by Lagrange, the -conjugacy class containing has order . That is, every nonidentity element is in a conjugacy class containing either elements (if ) or elements (If ). Thus the class equation for is , where is the number of order conjugacy classes and the number of order conjugacy classes.

Now has a subgroup of order by Cauchy’s Theorem. Suppose is normal; then is a union of conjugacy classes. However, the smallest possible union of conjugacy classes which contains 1 and some nonidentity element has order . , a contradiction. Thus no subgroup of order (in particular ) is normal.

Let denote the permutation representation of induced by the action of on the left cosets of by left multiplication, and let denote the kernel of this action. Note that and that is normal in . Since has prime order, . Thus is injective.

Let be an element of order by Cauchy’s Theorem. Note that is a maximal subgroup of , and that all conjugates of either coincide or intersect trivially, since has prime order. If is a proper subgroup of , we have . By a previous result, the number of nonidentity elements in which are in some conjugate of is at most ; in fact this number is equal to since two distinct conjugates intersect trivially. Thus has at least elements of order .

Similarly, since is not normal, exactly nonidentity elements of are in conjugates of , so that has at least elements of order . Combining these results, we see that , a contradiction. Thus is in fact all of .

Suppose now that for some and , we have . Then ; because and are distinct primes, we have . Thus mod . Thus the cosets are distinct; because is finite, these are precisely the elements of . We can see that . Since , we have ; hence .

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You have: “However, the smallest possible union of conjugacy classes which contains 1 and some nonidentity element has order p+1, a contradiction.”

If p=2 and q=3, would it be a contradiction?

No; what we have in is a subgroup of order which is a union of conjugacy classes. So for some nonnegative integers and . Since , we have . Then ( prime) clearly has no solutions in the integers.

“the smallest possible union of conjugacy classes which contains 1 and some nonidentity element has order p+1”

Do you mean q+1?

We assumed at the beginning that . is a union of conjugacy classes, each of which has order 1, , or . The smallest possible union of conjugacy classes which contains 1 and a nonidentity element thus has elements.