# Frattini subgroup

1. Jul 3, 2007

### happyg1

1. The problem statement, all variables and given/known data
Find
$$Frat(D_{2n}),Frat(D_{\infty})$$

2. Relevant equations

Frat is the set of all nongenerators of a group.

3. The attempt at a solution
I know that D_2n is generated by a rotation of 360/n with order n, and a reflection, f of order 2.

So D_2n=<r,f> and any element that can be added to that would be a nongenerator. So the Frattini would be something like $$\{r^2,r^3,...,r^{n-1},rf,r^2f,r^3f,.....r^{n-1}f\}$$

I'm not sure if my reasoning here is correct. I'm still working on the generators for $$D_{\infty}$$

Please tell me if I'm on the right track here.
CC

Last edited: Jul 3, 2007
2. Jul 3, 2007

### Kreizhn

It's been a while since I took Group theory, but is there a restriction on your term for generator? I mean, your reasoning is certainly correct for $$D_{2n}$$, but there is another way to reason about this.

We see that every rotation can be generated by r, but every rotation can also be generated by $$r^m$$ where $$gcd(m,n)=1$$. I might be wrong, I can't quite remember. But in this case we could take the generator subset to be $$< r^m, f >$$ which would generatate an isomorphic group to the one you presented above.

3. Jul 3, 2007

### Kreizhn

For your $$Frat(D_\infty)$$, note that $$D_\infty$$ is isomorphic to the semidirect product of $$\mathbb{Z}$$ and $$\mathbb{Z}_2$$. Furthermore, generators are mapped to generators (I should be able to scrape up a proof if you want). Thus if you can find the generators of the semidirect product, then you can get them for $$D_\infty$$

4. Jul 3, 2007

### NateTG

Let's consider a simpler example like:
$$Frat(\mathbb{Z}_7)=\{e\}$$
Since any non-identity element will generate the group.

Similarly, if you look at:
$$D_{2 \times 7}$$
Any one of the rotations will generate all seven rotations, so all of them are generators in combination with any one of the reflections.

The 'non-generator' condition is that for any list of elements that
generates the group will still generate the group of all of the non-generators are removed from it.

5. Jul 3, 2007

### Kreizhn

Very good point. I apparently forgot what the problem was when doing my rationale. Though I believe I'm still right about the whole coprime argument (not that you're saying that I'm wrong). $$D_{2\times7}$$ has every rotation as a rotational generator since every power would be coprime to 7. Clearly this doesn't hold in $$D_{2\times 3}$$ Where $$\{ r^2, r^3, r^4 \}$$ are not generators. Thus I would dare say that

$$Frat(D_{2n}) = < r^m | gcd(m,n)\neq1>$$

Any objections?

Last edited: Jul 3, 2007
6. Jul 3, 2007

### happyg1

I was just coming back on to asl that exact question. After I really started tearing it apart I saw the$$D_{2\times7}$$ counter example.

I agree with your result. It's the same thing that I came up with. I hope it's correct.

I'm still hjaving trouble on the $$D_{\infty}$$ one. I'm stuck on the semidirect product thing. I see that Z2 has generator 1 and Z has generators everything except 0. So does that mean that the semidirect product in generated by everything except 0 and 1? The semidirect product is still a very fresh idea to me.

Next is:
I tried to find the generators of $$D_{\infty}$$ by saying that
$$D_{\infty}$$ is generated by a rotation of $$\frac{360}{k}$$ where k does not divide 360, and a reflection f of order 2. Now there are infinitely many k's (but does k have to be an integer? I think it does.....) that work as far as I can tell. So the only nongenerators of the r would be divisors of 360?

Tell me what you think.
CC

Last edited: Jul 3, 2007
7. Jul 3, 2007

### Kreizhn

The semidirect product result might be the wrong way to go about it, but you can continue trying it if you like. I realized this after thinking about it for a bit.

I'm curious though how you might think that every non identity element of $$\mathbb{Z}$$ is a generator. Consider 3: 3 will never generate the number 2, and so 3 isn't a generator. As a matter of fact, for any positive $$n \neq 1$$, n will not generate any integer less than itself. Thus the only generator of $$\mathbb{Z}$$ is 1, which makes sense since $$\mathbb{Z}$$ is isomorphic to the infinite cyclic group.

Thus if I had to guess about $$Frat(D_\infty)$$ I would say

$$Frat(D_\infty)=<r^n | n \neq 1 >$$

8. Jul 3, 2007

### happyg1

It would have to be all rational multiples of a full rotation....I think.
$$D_{\infty}$$ happens when the r is a nonrational multiple of a full rotation, according to wikipedia.
So how to write that nicely....If it's right...
CC
EDIT:
So would that be $$Frat(D_{\infty})=\{r^n|n\in \mathbb{R}\backslash \mathbb{Q}\}$$?

Last edited: Jul 3, 2007
9. Jul 3, 2007

### NateTG

In the context of $D_\infty$ what rotations will:
$$r^n,r^{n-1}$$
generate?

Last edited: Jul 3, 2007
10. Jul 3, 2007

### Kreizhn

In all honesty, I'm not 100% sure. I don't see a problem in letting $$n \in \mathbb{Z}$$ since we still get an infinite set. Plus if we let $$r\in\mathbb{R}\setminus\mathbb{Q}$$ couldn't we get things that seemed to conflict with the multiplicative representation (since not all irrationals are transcedental)

Edit: Did you change $$r\in\mathbb{R}\setminus\mathbb{Q}$$ to $$n\in\mathbb{R}\setminus\mathbb{Q}$$? Then this clearly can't be correct since we don't have the notion of an irrational power in our definition of the binary operator

11. Jul 3, 2007

### happyg1

I see what you're saying about Z. I just get so easily turned around.
In your solution, wouldn't n=2 generate a noninfinite dihedral group? Or am I missing something?

up above I edited my Frat. I meant for n, not r to be not rational.

Last edited: Jul 3, 2007
12. Jul 3, 2007

### Kreizhn

Instead of considering the abstract infinite cyclic group, let's instead think of $$\mathbb{Z}$$ under addition. The moving from multiplicative notation to additive, letting n = 2 is equivalent to saying the element 2n generates all integers, where $$n\in \mathbb{Z}$$

13. Jul 3, 2007

### happyg1

I understand about the Z generators now. The only possible generators for that semirect product are 1. So the infinite dihedral group only has 1 generator. I get confused because I'm used to thinking about the dihedral group itself, not the semidirect product.

So we're saying that the infinite dihedral group has infinitely many generators? There's ONLY one generator for it?

I have to wrap my mind around that from the dihedral group perspective.

14. Jul 3, 2007

### Kreizhn

I'd say there's two. Don't forget f. Not to mention if there was only 1 generator, then $$D_\infty$$ would be isomorphic to $$C_\infty$$ (infinite cyclic group) which is clearly not true. Thus I'd say the only generators of $$D_\infty$$ are r and f.

Last edited: Jul 3, 2007
15. Jul 3, 2007

### happyg1

OK, so it has 2. So the generators for the semidirect product are 1 and 1, right? So then when we write down the Frat, since it's the same generator, we only have to name it once?

16. Jul 3, 2007

### Kreizhn

No, since we have to take the disjoint union of the two generators, which means that although in each of their groups they represent the same number, they are not actually the same generator.

17. Jul 3, 2007

### happyg1

I understand that. My confusion lies in that there is one 1 in the Frat. I think I need to walk away for a minute and let this settle in my head.

18. Jul 3, 2007

### Kreizhn

Think about it this way. The isomorphism between the semidirect product and $$D_\infty$$ is

$$\alpha: D_\infty \to \mathbb{Z}\times \mathbb{Z}_2$$
$$(r^n f^m) \to (n, m$$ mod 2)

Since (1,0) and (0,1) are the generators in the semi-direct product, then

$$\alpha^{-1} (1,0) = r$$
$$\alpha^{-1}(0,1) = f$$

Edit : This could probably be better represented by limiting $$m \in \{0,1\}$$. Otherwise I just know someone is going to rag on me about defining the inverse the way I did.

Last edited: Jul 3, 2007
19. Jul 3, 2007

### Kreizhn

Why didn't I just do that in the beginning? It probably would've made more sense.

20. Jul 3, 2007

### NateTG

It might help if you think of the Frattini subgroup as the intersection of maximal proper subgroups rather than as the set of non-generators.

Now, if you apply that to the dihedral groups, it's clear that the set of rotations is a proper maximal subgroup, so the Frattini subgroup cannot contain any of the reflections.

Moreover, for the finite dihedral groups, $D_{2 \times n}$, $$|r^p,f|[/itex] is a proper maximal subgroup for any prime factor $p$ of $n$. Clearly, substituting a non-prime power of the generating rotation will not generate a maximal proper subgroup. Thus $$Frat(D_{2xn})=\mathbb{Z}_k$$ where $$k=n \prod_{p | n} \frac{1}{p}$$ Now, for the infinite case, for any $r^k \neq 1$ there is some prime $p>k$$ and [itex]|r^p,f|$ is a maximal proper subgroup that does not contain it, so
$$Frat(D_\infty)=\{1\}$$

Last edited: Jul 3, 2007