Find all groups of order 9, order 10, and order 11

[nhartung]

Homework Statement

Find all groups of order 9, order 10, order 11.

Homework Equations

None

The Attempt at a Solution

We have already done an example in class of groups of order 4 and of order 2,3,5, or 7.
So I'm going to base my proofs on the example of groups of order 4 except for the group of order 11 which I suspect is acting like the groups of order 2,3,5 or 7 since it is also a prime number.

Here is my attempt at groups of order 9, I'm a little unsure about the final part.

Let G be a group of order 9, every element has order 1, 3, or 9. If there is an element g of order 9, then <g> = G. G is cyclic and isomorphic to (Z/9, +).
If there is no element of order 9, the (non-identity) elements must all have order 3.
G = {e, a, a², b, b², c, c², d, d²}
G is isomorphic to Z/3 x Z/3
a³ = e
b³ = e
c³ = e
d³ = e

Now i'll show the mappings of G onto Z/3 x Z/3:
e -> (0,0)
a -> (1,0)
a² -> (2,0)
b -> (0,1)
b² -> (0,2)
c -> (1,1)
c² -> (2,2)
d -> (1,2)
d² -> (2,1)

Did I do everything correctly here, and is this sufficient to find all groups of order 9 as the problem is asking?

 

[micromass]

Uh, yes, this is sufficient.

but you constructed a map from G to Z3 x Z3. I don't find it very obvious that it is an homomorfism. For example, is ab being sent to (1,1). (i.e. do you know for certain that ab=c?)

 

[nhartung]

Maybe it would be better to describe the group as G = {e, a, a², b, b², ab, (ab)², ab², a²b} ?

Or am I getting myself into more trouble here?

 

[micromass]

Yes, you could do that, but the problem remains. Like, what does (ab)a equal? (I know it should equal a²b, but you need to prove it)

 

[nhartung]

Yeah I was just thinking about that, I'm a little stuck on it. I can't assume this group is abelian can I? That would make for a pretty simple proof. Otherwise I guess I can try to use some sort of associativity proof.

Suppose (ab)a ≠ a²b
= (ab)a ≠ a(ab)
= a ≠ a (this is actually true a = a) which contradicts our original supposition which means (ab)a = a²b
(for some reason I have a feeling I can't just divide each side by (ab) like I did. hah). I'll keep trying other methods.

 

[micromass]

No, you can't just divide by ab sorry.

Maybe you should look back to the proof of the group of order 4. How did you show there that the map was a homomorphism. Maybe you could copy that...

Honestly, I don't know any elementary methods for finding the groups of order 9. I guess only a bit of trial-and-error could do the job...

 

[nhartung]

The group of order 4 uses the theorem that if the square of every element in the group = e then the group is Abelian which I can't use for groups of order 9. I did find an example in my book for groups of order 6 which includes an element of order 3. I'll take a look at this

 

[Dick]

Do you know p-groups have a nontrivial center? If not, use the class equation. It's a pretty common group theory exercise to show all groups of order p^2 are abelian.

 

[nhartung]

Ok my professor did an example of groups of order 8 in my last lecture which helped a lot so I think I have it figured out now:

Groups of order 9:
Let G be a group of order 9, every element has order 1, 3, or 9. If there is an element g of order 9, then <g> = G. G is isomorphic to Z/9.

If there is no element of order 9, the (non-identity) elements must all have order 3.
G = {e, a, a², b, b², ab, (ab)², ab², a²b}
Now let's assume the following relationships:
a³ = e
b³ = e
ab = ba (so we are abelian)
Lets check that our assumptions hold true:
a^oddb^odd = ab OR a OR b OR e
a^oddb^even = ab² OR b²
a^evenb^odd = a²b OR a²
a^evenb^even = a²b² = (ab)²
Our assumptions hold and I believe this is isomorphic to Z/3 x Z/3. Let's check the mappings:
e --> (0,0)
a --> (1,0)
a² --> (2,0)
b --> (0,1)
b² --> (0,2)
ab --> (1,1)
(ab)² --> (2,2)
ab² --> (1,2)
a²b --> (2,1)

I think this should be enough to prove that these are the 2 groups of order 9. What do you guys think?

 

[nhartung]

When checking if my assumptions hold true I should also check them with the a and b values flipped to check the ab = ba assumption, I won't write them out here but I know they hold, just assume I wrote them up there.

 

[Dick]

When checking if my assumptions hold true I should also check them with the a and b values flipped to check the ab = ba assumption, I won't write them out here but I know they hold, just assume I wrote them up there.

You already assumed ab=ba in deriving the group structure. There's not much use in 'checking' it now. As I said before, you can prove a group of order 9 is abelian before you start.