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]

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.

 

[blue_raver22]

I would approach this problem by first understanding the definitions and concepts involved. A group is a mathematical structure consisting of a set of elements and an operation that satisfies certain properties such as closure, associativity, identity, and inverse. The order of a group is the number of elements in the group. A subgroup is a subset of a group that also satisfies the properties of a group. A normal subgroup is a subgroup that is invariant under conjugation by elements of the group.

In this problem, we are given a group of order pq, where p and q are primes and p < q. We are asked to prove that this group has a nonnormal subgroup of index q, meaning there is a subgroup of size p in this group that is not normal.

To prove this, we can start by considering the group G itself. Since G is nonabelian, we know that the elements do not commute with each other. This means that there must exist at least one pair of elements in G that do not commute. Let's call these elements a and b.

Now, let's consider the subgroup generated by a and b, denoted <a, b>. This subgroup will have size p since it is generated by two elements. We can see that this subgroup is nonnormal because if we conjugate it by the element b, we get a different subgroup, namely <b, ab>. This shows that <a, b> is not invariant under conjugation by elements of G.

Next, we need to show that the index of <a, b> in G is q. This means that there are q distinct cosets of <a, b> in G. We can see this by considering the elements of G and how they are related to <a, b>. Since a and b do not commute, we can see that the elements of <a, b> and their conjugates will fill up q distinct cosets in G.

Now, let's consider the symmetric group Sq. This group has order q! and contains all possible permutations of q elements. We can define an injective homomorphism from G to Sq by sending each element of G to a permutation of the cosets of <a, b> in G. This shows that G is isomorphic to a subgroup of Sq.

Finally, we can deduce that G is isomorphic to a subgroup of the normalizer in Sq of the cyclic group generated by the q-cycle. This is