# Group of order pq && nonnormal subgroup

#### hermanni

I saw the following problem on my abstract algebra book (dummit && foote) , I tried to solve it but I couldn't :
Let p, q be primes with p < q . Prove that a nonabelian group G of order pq
has a nonnormal subgroup of index q , so there exists an injective
homomorphism into Sq. Deduce that G is isomorphic to a subgroup of the normalizer in S(q) of the cyclic group
generated by the q-cycle.
I think I need to construct the group and see it's nonabelian.I thought of using
conjugacy and group actions , but I could't get anywhere.Can someone help?

#### lavinia

Gold Member
I think you have the right idea. Here is a start. Suppose there is an element of the group that does not commute with all of the others and whose order is p. Let Zp be the cyclic group that it generates.

Then I think that this group can not be normal.

Since conjugation is a homomorphism, if the group were normal then the action would determine an isomorphism of order q of Zp.
But a cyclic group of prime order can not have an isomorphism of order q > p, I think. So Zp can not be normal.

Is this wrong?

#### Tinyboss

A more straightforward way to show the first part:

By Cauchy's theorem, there are elements of order p and q, which generate subgroups of index q and p, respectively. The subgroup of index p is normal since its index is the smallest prime dividing |G|, so the other one can't be normal, else G would be abelian (the direct product Zp x Zq).

#### hermanni

Ok , tinyboss I totally agree with this simple and compact solution. What about the second part? Let H be that subgroup of q and we see G acts on H by conjugation. So there exists a homomorphism from G into Sq , how do we see it's injective ? (is it relevant to the H's nonabelianness? ) Even if we see it , I couldn't get the relationship with this and the third part of the question.

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