Classify nonabelian groups of order pq, p and q distinct primes

Let p and q be primes with p < q. Prove that a nonabelian group G of order pq has a nonnormal subgroup of index q, so that there exists an injective homomorphism G \rightarrow S_q. Deduce that G is isomorphic to a subgroup of the normalizer in S_q of the cyclic group generated by (1\ 2\ \ldots\ q).


If Z(G) = G, then G is abelian, a contradiction. If |Z(G)| is p or q, then G/Z(G) is cyclic, so that G is abelian. Hence Z(G) = 1.

Let x \in G be a nonidentity element; note that |x| is either p or q, since if |x| = pq we have G cyclic, a contradiction. Moreover C_G(x) is not all of G, since otherwise x \in Z(G). Since \langle x \rangle \leq C_G(x) \leq G and \langle x \rangle is maximal by Lagrange, the G-conjugacy class containing x has order |G|/|x|. That is, every nonidentity element is in a conjugacy class containing either p elements (if |x| = q) or q elements (If |x| = p). Thus the class equation for G is |G| = 1 + mp + nq, where m is the number of order p conjugacy classes and n the number of order q conjugacy classes.

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

Let \varphi : G \rightarrow S_{G/H} denote the permutation representation of G induced by the action of G on the left cosets of H by left multiplication, and let K denote the kernel of this action. Note that K \leq H and that K is normal in G. Since H has prime order, K = 1. Thus \varphi is injective.

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

Similarly, since H is not normal, exactly pq-q nonidentity elements of G are in conjugates of H, so that G has at least pq-q elements of order p. Combining these results, we see that |G| \geq 1 + pq-q + pq-p = 1 + 2pq - (p+q) \geq pq+1 > |G|, a contradiction. Thus N_G(x) is in fact all of G.

Suppose now that for some a and b, we have x^aH = x^bH. Then x^{a-b} \in H; because |x| = q and |H| = p are distinct primes, we have x^{a-b} = 1. Thus a = b mod q. Thus the cosets H, xH, x^2H, \ldots, x^{q-1}H are distinct; because G/H is finite, these are precisely the elements of G/H. We can see that \varphi(x) = (H\ xH\ \ldots\ x^{q-1}H). Since \langle x \rangle \vartriangleleft G, we have \langle \varphi(x) \rangle \vartriangleleft \mathsf{im}\ \varphi; hence \mathsf{im}\ \varphi \leq N_{S_{G/H}}(\langle (H\ xH\ \ldots\ x^{q-1}H) \rangle).

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Comments

  • Peter  On October 13, 2011 at 1:30 pm

    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?

    • nbloomf  On October 13, 2011 at 3:32 pm

      No; what we have in H is a subgroup of order p which is a union of conjugacy classes. So p = 1 + ap + bq for some nonnegative integers a and b. Since p < q, we have b = 0. Then p = 1+ap (p prime) clearly has no solutions in the integers.

  • Peter  On October 13, 2011 at 6:10 pm

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

    Do you mean q+1?

    • nbloomf  On October 14, 2011 at 12:33 am

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

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