## 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 $\mathsf{im}\ \varphi \leq N_{S_{G/H}}(\langle (H\ xH\ \ldots\ x^{q-1}H) \rangle)$.

• 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.