# Non-Isomorphic Groups of Order 30

1. Mar 12, 2008

### apalmer3

1. The problem statement, all variables and given/known data
How many different nonisomorphic groups of order 30 are there?

2. Relevant equations
The previous parts of the problem dealt with proving that 3-Sylow and 5-Sylow subgroups of G were normal in G when o(G)=30, though I'm not sure how that relates...

3. The attempt at a solution
I'm not even sure what this is asking. Any help would be greatly appreciated!

2. Mar 12, 2008

### morphism

This problem is asking you to classify the isomorphism classes of groups of order 30. For example, there are two isomorphism classes of groups of order 4: the cyclic group with 4 elements C_4, and the direct product of two cyclic groups consisting of two elements C_2 x C_2.

With this in mind, can you think of how you can attempt the problem now?

3. Mar 13, 2008

### apalmer3

Alright, let's see if I can take a crack at this. :-D

C_30 , C_2xC_15 , C_3xC_10, C_5xC_6

So, there are 4? Is that right?

4. Mar 13, 2008

### morphism

Yes, there are 4. But no, those aren't all of them! In fact, you just gave one isomorphism class: the groups you listed are pairwise isomorphic.

I guess my example for groups of order 4 was a bit misleading. Maybe groups of order 6 serve as a better example: they are (up to isomorphism) S_3 and C_6. This time we get a nonabelian group.

You're going to have to work harder to classify groups of order 30. The fact that such a group has normal subgroups of order 3 and 5 is going to be very useful.

5. Mar 13, 2008

### apalmer3

I'm sorry, but I am intensely confused right now. How does nowing that it has normal subgroups of order 3 and 5 help?

Thanks for using your time to help me. :-D

6. Mar 13, 2008

### JasonRox

I'll do groups of order 12 for you.

You have C_12, and C_6xC_2.

Let's say you wanted to add C_4xC_3. Why may this be redundant?

7. Mar 13, 2008

### apalmer3

It would be redudant because... 4=2^2?

8. Mar 13, 2008

### JasonRox

No, because C_4xC_3 is isomorphic to C_12... why?

9. Mar 13, 2008

### morphism

This will let you conclude that any group of order 30 has a (normal) subgroup of order 15 (why?). (Normality follows from the fact that the index of this subgroup is 2.)

10. Mar 13, 2008

### StatusX

For the abelian case, there is a theorem that:

1. All finite abelian groups are isomorphic to a group of the form $\mathbb{Z}_{n_1} \times \mathbb{Z}_{n_2} \times ... \times \mathbb{Z}_{n_m}$.

2. $\mathbb{Z}_n \times \mathbb{Z}_m$ is isomorphic to $\mathbb{Z}_{nm}$ if and only if n and m are coprime.

Have you seen this theorem in class and/or are you allowed to use it? If so, you can use it to answer jason's question, and then find the abelian groups of order 30. If not, this question seems a little unfair, but I guess you can at least use it to find the correct answer, which you can then prove by another method (eg, Z2 x Z2 is not isomorphic to Z4 since the latter has an element of order 4 while the former does not).

11. Mar 13, 2008

### apalmer3

Hey Guys,

StatusX-- Thanks for telling me that theorem! I didn't know it (my professor is a teeny bit incompetent...) So, I can use that to prove that because 4*3=12 means C_4xC_3 is isomorphic to C_12? It makes sense. :-D

Morphism-- I proved that a group G s.t. o(G)=30 has a normal subgroup of order 15 earlier in the problem. So I got that part down. :-D

Okay, so while C_30 is an abelian group of order 30, the others (C_2xC_15 , C_3xC_10, C_5xC_6) are redundant. I got that now.

Now, going back to the whole "The fact that such a group has normal subgroups of order 3 and 5 is going to be very useful." Are you saying that S_3 and S_5 are also groups of order 30?

And even if that was right (which I'm really not sure it is) that's only 3, and Morphism already said it was 4.

Help?

12. Mar 14, 2008

### morphism

Let's list what we know:

(1) The only abelian group of order 30 is C_30.

(2) G always has a normal subgroup N of order 15. It's not too hard to prove that N must be cyclic.

(3) By Cauchy, G also has a subgroup K of order 2. Obviously K is cyclic.

Now let's use (2) & (3) to get the nonabelian groups. Since $N\cap K=\{1\}$ (by Lagrange), we see that |NK|=30, and G is thus isomorphic to $N \rtimes_\varphi K$, where $\varphi$ is a nontrivial homomorphism from K=~C_2 to Aut(N)=~Aut(C_15). It should be straightforward to get all these homomorphisms, and consequently all the nonabelian groups of order 30.

If you're not familiar with semidirect products (i.e. if you don't know what $N \rtimes_\varphi K$ means), then you can 'avoid' them by doing the computations directly. To this end, let x be a generator for N and let y be a generator for K (so x^15=y^2=1). Since G=NK and $N\cap K=\{1\}$, then every element of G is of the form xiyj (1<=i<=15 and 1<=j<=2). Now you have to play around to see what kind of relations these things will satisfy. For instance, since N=<x> is normal, we have that yxy=yxy-1=xk for some k (1<=k<=15), and hence (after applying this repeatedly) x^(k^2) = x. What does this tell us?

13. Mar 14, 2008

### apalmer3

That either k is 0 or x is 1 or 0. Since k cannot equal 0 (because it is between 1 and 15 inclusive) that means that x has to be either 0 or 1. And because x^15=1, we know that the only choice is for x to equal 1.

So there are 4 nonisomorphic groups of order 30... 1 is abelian, and the other three are found through finding the nontrivial homomorphisms from K=~C_2 to Aut(N)=~Aut(C_15). I think I've got it now.

Thank you.

14. Mar 15, 2008

### morphism

That's not quite right. x can't be "1 or 0", because it isn't an integer, it's a generator of a group of order 15. So x^(k^2) = x implies that x^(k^2 - 1) = 1, which in turn implies that o(x) divides k^2 - 1, i.e. that k^2 = 1 (mod 15). This should give you 4 possible values for k, and each of these will get you a distinct presentation for G.