Permutation Representations of Non-Abelian Groups of Semiprime Orders (4.3.28)

Dummit and Foote Abstract Algebra, section 4.3, exercise 28:

Let G be a non-abelian group of order pq, where p and q are prime and p < q. Prove that G has a nonnormal subgroup of index q, so that there exists an injective homomorphism into S_q. Deduce that G is isomorphic to a subgroup of the normalizer of < ( 1 2 ... q ) > in S_q.

Proof: Take x and y from G such that | x | = p and | y | = q by Cauchy's Theorem.

We have that < x, y > = G since | < x > | and | < y > | both divide | < x, y > | by Lagrange's Theorem, therefore | < x, y > | = pq.

We have that < y > is normal since it is of index p in G, and since p is the smallest prime dividing | G |, by a previous result we have our justification.

We have that < x > is nonnormal. If it weren't, then it would be the union of conjugacy classes, and since for every g∈G we have that the order of the conjugacy class of g is either 1, p, q, or pq (since the order of a conjugacy class divides the order of the group), and since 1∈< x > has a conjugacy class of order 1, it must be that every element in < x > has a conjugacy class order of 1, ie every element in < x > is in Z(G). This includes x, which implies xy = yx, which would imply G abelian by the fact that these two elements generate G.

Now, let < x > = H and let π_H be the left regular permutation representation of G into S_{| G : H |} = S_q. Since the kernel of the action is the largest normal subgroup of G contained in H, and H is nonnormal and without nontrivial proper subgroups by the primality of its order, we have that the kernel is the identity element; thus the mapping is injective. Thus we know that π_H(y) is of order q in S_q, so π_H(y) is a q-cycle as q is prime and therefore not the result of a combination of smaller cycles whose least common multiple of orders is q (and this is the desired q-cycle if the cosets of H that y permutes are numbered appropriately). Now, since g< y >g^-1 = < y > for all g∈G by the demonstrated normality of < y >, and since π_H is an isomorphism, we have that π_H(g)< π_H(y) >π_H(g)^-1 = < π_H(y) >, so that ∀g∈G, π_H(g)∈N_G(< π_H(y) >).

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