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

## Homework Statement

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

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?

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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?)

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?

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)

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.

No, you cant 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...

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
Homework Helper
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.

Last edited:
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 lets assume the following relationships:
a3 = e
b3 = e
ab = ba (so we are abelian)
Lets check that our assumptions hold true:
aoddbodd = ab OR a OR b OR e
aoddbeven = ab2 OR b2
aevenbodd = a2b OR a2
aevenbeven = 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?

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