# How to find group types for a particular order?

1. Sep 22, 2015

### jackmell

Hi,

How do we determine the group types for a particular order? I know the example of using the Cayley table to show there are only two types of groups of order 4 but do not know how to determine this for other groups. For example, suppose I wanted to show there is a group of order 10 in the alternating group $A_5$. Now, according to Wikipedia, this group must be cyclic or dihedral. Wikipedia lists a table of small groups:

https://en.wikipedia.org/wiki/List_of_small_groups

How do we determine this?

Here's what I have so far:

(1) If the group order is prime then it must be a cyclic group. Therefore, there are no non-abelian groups with prime order.

(2) Let G be a ﬁnite Abelian group. Then G is isomorphic to a product of groups of the form $H_p = Z/p^{e1}Z×···×Z/p^{en}Z$, in which $p$ is a prime number and $1\leq e1 \leq···\leq en$ are positive integers

not real clear on #2 but let's assume I got that. So that leaves non-abelian groups. Take for example a group of size 6. How do I conclude that the only non-abelian group of order 6 is $D_6$ and all other non-abelian groups with that size are isomorphic to $D_6$?

Then take for example a non-abelian group of size $8$. Well $D_8$ and $Q$ (quaternions) are of that size. So then any other are isomorphic to those two. How would I prove this? That no matter what kind of non-abelian group of size 8 I could invent, it must be isomorphic to these two.

(3) Let G by a group of order 2p with p a prime number, then G is either cyclic or isomorphic to $D_p$. (need to review this proof I think). Ok, then that answers my question for the case of $|G|=10$.

Guess I should know these already.

What about when it's not 2p though, say for 12? Well, one of them is dihedral, $D_{12}$. Is that the only non-abelian group of size 12?. I think there is a dicyclic also however. Certainly exists a cyclic group $C_{12}$ also. But there is also a product one $Z_6\times Z_2$. But then why isn't there a $Z_3\times Z_4$ then? Maybe they're isomorphic. Need to check.

There is a $Z_3\times Z_4$. Also not hard to show $Z_{10}\cong Z_5\times Z_2$. Same dif for $Z_{12}\cong Z_3\times Z_4$. However, $Z_6\times Z_2 \ncong Z_3\times Z_4$ (try coming up with an isomorphism) and so therefore $Z_{12}\ncong Z_6\times Z_2$.

I think I'm getting it guys.

Thanks,
Jack

Last edited: Sep 22, 2015
2. Sep 25, 2015

### mathwonk

3. Sep 25, 2015

### jackmell

Thanks mathwonk. That's very informative and I'm sure I'll use it while I'm studying abstract algebra. For example, just briefly running though it I noticed it had a very good summary of the various generators for the symmetric group including the one I had worked on recently here and elsewhere about when $n$ is a prime $p$ than any transposition and any $p$-cycle generates the group. I'm also finding GAP very useful in that regards (thanks to micromass).

Jack