Semi-Direct Product & Non-Abelian Groups

[OMM!]

Homework Statement

Let p, q be distinct primes s.t. p $\equiv$ 1 (mod q). Prove that there exists a non-Abelian group of order pq and calculate the character table.

Homework Equations

Semi-Direct Product: Let H = < Y | S > and N = < X | R > be groups and let $\phi$ : H $\rightarrow$ Aut (N) be a homomorphism. Then the SDP has the presentation:

N x H = < X, Y | R, S, $y^{-1}xyw_{x,y}^{-1}$ >

With x in X, y in Y, $w_{x,y}$ = ($\phi$(y(x))) (i.e. $\phi$ of y and then $\phi$ of y of x etc.) in < Y > = N

The Attempt at a Solution

I've done a specific example where p=7 and q=3 and found a group of order 21, using the semi-direct product of 2 cyclic groups $C_{7}$ and $C_{3}$.

But can't see how to prove that there exists a group more generally. So far I have.

If N = $C_{p}$ = < x > and H = $C_{q}$ = < y >, then $\phi$ : $C_{q}$ $\rightarrow$ Aut ($C_{p}$) = $C_{p-1}$ = < $\alpha$ >, then we have y $\mapsto$ $\phi$(y) too.

When p=7 and q=3

I had that $\phi$(y) = {1, $\alpha^{2}$, $\alpha^{4}$}

If $\phi$(y) = 1, then we'd have that the SDP of $C_{7}$ and $C_{3}$ would be ismorphic to $C_{7}$ x $C_{3}$.

If $\phi$(y) = $\alpha^{2}$, then the SDP of $C_{7}$ and $C_{3}$ would be equal to < x, y >

Then by the relation in the SDP, we'd have: yx$y^{-1}$ = $\alpha^{2}$(x) = x^{t}

And we have that: t = {1, 2, 3, 4, 5, 6}, with p=7 and q=3, then: t^{3} $\equiv$ 1 (mod 7), and so: t = 1 and t = 2 and t = 4, thus we have groups:

G = < X, Y | $x^{7} = y^{3} = 1$, $yxy^{-1} = x$ > = $C_{7}$ x $C_{3}$

G = < X, Y | $x^{7} = y^{3} = 1$, $yxy^{-1} = x^{2}$ >

G = < X, Y | $x^{7} = y^{3} = 1$, $yxy^{-1} = x^{4}$ >

But how would I get to this stage more generally?! Thanks for any help!

 

[micromass]

Hi OMM!

I think it's clear that you will want to form the group $C_p\rtimes C_q$. The first thing you will want to do is check if there exists a suitable homomorphism

$$\phi:C_q\rightarrow Aut(C_p)$$

 

[OMM!]

Hi micromass, thanks for your help once again!

So if we say, < y > = $C_{q}$, then we need to show there is a homomorphism $\phi$ that sends y to $\phi$(y) which is an element of Aut $C_{p}$ = $C_{p-1}$.

And since p $\equiv$ 1 (mod q), then $C_{p-1}$ isomorphic to $C_{q}$?

And clearly there exists a homomorphism from one group to itself (endomorphism?).

Am I on the right track?!

 

[micromass]

You're certainly on the right track, but there is a small mistake. You say that $C_q$ is isomorphic to $C_{p-1}$, but this is not true.

All you know is that p-1=0 (mod q), but this does not imply that q=p-1. It merely implies that q divides p-1.

 

[OMM!]

So for y in $C_{q}$, we have $y^{q}$ = 1

And if q divides p-1, then $y^{p-1}$ = 1?

Does the order of the element $\phi$(y) divide p-1? Or does it divide q?

(Sorry, probably obvious, but having a mental moment!)

Then to find the possible element for $\phi$(y) for $C_{p-1}$ = < $\alpha$ >

$\phi$(y) in {1, $\alpha$, $\alpha^{2}$, ... , $\alpha^{p-1}$}

Show which powers of $\alpha$, when put to the power of q, give a power 1 (mod p)? (i.e. yx$y^{-1}$ = $x^{t}$, we need $t^{q} \equiv$ 1 (mod p))

Am I missing something out here or is it just a case now of showing that basically t = 1 is one of these values of t, thus getting the non-abelian cyclic group $C_{p}$ x $C_{q}$ of order pq?

 

[micromass]

Hmm, I think you're making it a bit too hard on yourself.

We know that q divides p-1, does that mean that we can always find an element of order q in $C_{p-1}$?

 

[OMM!]

I'd say yes (With very little conviction! Haha)

As I said in the last post, if we have an element y in $C_{q}$ which is obviously of order q, then y is also in $C_{p-1}$?

 

[micromass]

I'd say yes (With very little conviction! Haha)

As I said in the last post, if we have an element y in $C_{q}$ which is obviously of order q, then y is also in $C_{p-1}$?

Aah, yes. I seem to be a bit confused with the notation. You are identifying $C_q$ with elements on the unit circle of $\mathbb{C}$ yes??

 

[OMM!]

I'm not entirely sure!

I am basically saying that C_{q} is the cyclic group of order q, thus is generated by < y >.

In other words: C_{q} = < y | y^{q} = 1 >

The elements of which are: {1, y, y^{2}, ... , y^{q-1}}

 

[micromass]

I'm not entirely sure!

I am basically saying that C_{q} is the cyclic group of order q, thus is generated by < y >.

In other words: C_{q} = < y | y^{q} = 1 >

The elements of which are: {1, y, y^{2}, ... , y^{q-1}}

Yes, that's what I thought.

There are several methods to prove the existence of an element of order q in $C_{p-1}$, the easiest makes use of the fundamental theorem of abelian groups.

However, you seem to want to do another method (which is also fine): you want to say that if y is an element of $C_q$, then it is an element of $C_{p-1}$. Unfortunately, this does not make much sense: the elements in the cyclic group might look quite different from each other!
However, every cyclic group is isomorphic to a part of the unit circle by

$$\Phi:C^q\rightarrow S^1:y^k\rightarrow e^{2\pi i k/q}$$

this is an group monomorphism. So now you can indeed say that an element of $C_q$ is also an element of $C_{p-1}$!

 

[OMM!]

And the order of that element divides q?

It can't be 1, else the group would be Abelian i.e. yxy^{-1} = x ===> yx = xy

So the order must be q, as q is prime?

 

[micromass]

Well, you can choose that element such that the order is q!

 

[OMM!]

Thanks for your help, I'll give it another shot in the morning. Found a bit on Frobenius Groups which seems quite related, which I think may be my bed-time reading!

Thanks again, you're a great help!