Non-Isomorphic Groups of Order 30

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?

 

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.

 

Why not start with the abelian cases?

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?

 

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

 

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.)

 

For the abelian case, there is a theorem that:

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

2. [itex] \mathbb{Z}_n \times \mathbb{Z}_m[/itex] is isomorphic to [itex] \mathbb{Z}_{nm}[/itex] 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).

 

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 [itex]N\cap K=\{1\}[/itex] (by Lagrange), we see that |NK|=30, and G is thus isomorphic to [itex]N \rtimes_\varphi K[/itex], where [itex]\varphi[/itex] 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 [itex]N \rtimes_\varphi K[/itex] 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 [itex]N\cap K=\{1\}[/itex], then every element of G is of the form xⁱy^j (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=x^k for some k (1<=k<=15), and hence (after applying this repeatedly) x^(k^2) = x. What does this tell us?

 

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.