# Abstract Algebra Problem involving the order of groups

1. Dec 10, 2011

### xcr

1. The problem statement, all variables and given/known data

Let G be a group with identity e. Let a and b be elements of G with a≠e, b≠e, (a^5)=e, and (aba^-1)=b^2. If b≠e, find the order of b.

2. Relevant equations

Maybe the statement if |a|=n and (a^m)=e, then n|m.

Other ways of writing (aba^-1)=b^2:
ab=(b^2)a
b=(a^-1)(b^2)a
a=(b^2)a(b^-1)

Also, if the order of a=5, then |a|=|(b^2)a(b^-1)|=5

3. The attempt at a solution

My work is kinda in the relevant equations. I have manipulated the given formula and looked at the statement listed above but can't see if these will get me anywhere or started in the right direction.

2. Dec 10, 2011

### Deveno

one idea to try:

see if you can prove that abn = b2na.

since a is a conjugate of ba, a = b(ba)b-1, |a| = |ba|.

now write e = (ba)5 by getting all the b's first.

3. Dec 11, 2011

### xcr

I understand how you got to this point: abn = b2na. The steps to get there go like this:
aba-1=b2 so raising both sides to the nth power would look like (aba-1)n=(b2)n. On the left side we would have n terms of aba-1, but regrouping them would cause then the aa-1 terms to cancel. yielding abna-1=b2n. Multiplying by a on the right gives abn=b2na.

We haven't covered conjugates in my course so I don't understand the second hint and I think your third hint follows from that.

4. Dec 11, 2011

### I like Serena

Perhaps you can try and prove that the order of "a" is equal to the order of "ba"?
Then prove that (ba)m≠e for m<5.

For background, conjugate elements are rather important in group theory.
Basically it means that 2 elements "behave the same".
In this case Deveno uses the fact that 2 conjugate elements have the same order.

From wiki:

For the last step that Deveno suggested, can you rewrite (ba)5?

5. Dec 12, 2011

### xcr

ab=b2a
b-1ab=ba
(b-1ab)5=(ba)5
(b-1a5b)=(ba)5
b-1eb=(ba)5
b-1b=(ba)5
e=(ba)5

Therefore |a|=|ba|

As for proving that (ba)m≠e for m<5, do I just show that when m=1,2,3,4, you get b-1amb=(ba)m, which ≠e?

Then do I just show that (aba-1)n=(b2)n, which can be rearranged as b-nabn=bna.

Not sure what to do from here. If I raise both sides to the 5th power, I would get e=(bna)5

6. Dec 12, 2011

### I like Serena

Yes, that looks good!

Consider that (ba)5=b(ab)4a...

7. Dec 12, 2011

### Deveno

the idea is, if we can get e = (ba)5 = bras, with hopefully (if things go well) s = 5, well then we'll have an equation that tells us something about the order of b, which is what we were after in the first place. I like Serena has given you a VERY good place to start...

8. Dec 12, 2011

### xcr

ok so I took e=(ba)5=b(ab)4a=b(ab)(ab)(ab)(ab)a. I then took ab=b2a and substituted it in. After doing this REPEATEDLY I moved all the b's to the left and all the a's to the right. I ended up with e=b31a5, but a5=e, so e=b31e, or e=b31. Therefore the order of b is 31. Felt like I was doing the right steps but not very confident in that answer.

9. Dec 12, 2011

### I like Serena

Right! :)

Btw, taken literally you have not found the order of b yet.
What you found is a power of b that brings b to the identity.
So the actual order of b is at most 31.
To finish things neatly, you should proof that it has to be exactly 31.
Can you?
(Same thing for the order of "ba", where you actually concluded an order of 5 prematurely, but I didn't want to nitpick at the time.)

10. Dec 12, 2011

### xcr

I would just need to show that (ab)m, where m=1,2,3,4 does not equal e. So just show that (ab)1≠e, (ab)2=a(ba)b≠e, (ab)3=a(ba)2b≠e, (ab)4=a(ba)3b≠e.

11. Dec 12, 2011

### I like Serena

Yes, but for (ba)m instead of (ab)m.

To make it easier, consider that the order of (ba) has to divide 5, so it can't be 2,3, or 4.
That leaves the question whether it's possible that ba=e...?

12. Dec 12, 2011

### xcr

Well if (ba)=e, then (ba)2=(ba)(ba)=ee=e, which is not true. Therefore ba≠e

13. Dec 12, 2011

### I like Serena

I'm afraid it's still possible that (ba)2=e at this stage.

14. Dec 12, 2011

### xcr

I thought that since 2,3, and 4 did not divide 5 that the could not be the order?

15. Dec 12, 2011

### I like Serena

Indeed, but you should realize why.
The order is the lowest power to yield identity.
If that were for instance 2, then the power 5 could not be identity.
But the order could still be 1, in which case all powers 2,3,4,5 are identity.

16. Dec 12, 2011

### xcr

Ok, so ba≠e because then a must equal e and b must equal e, which cannot be true because of our assumptions

17. Dec 12, 2011

### I like Serena

Close... but how did you get that a must be equal to e?

Last edited: Dec 12, 2011
18. Dec 12, 2011

### Deveno

if ba = e, what happens when you multiply both sides of this equation by a4?

if b is any power of a, say b = am, show that b5 = e.

why is this a contradiction?

19. Dec 12, 2011

### xcr

If you multiply both sides by a4 on the right, you get b=a4, so I see now that there is a possibility that ba=e where a and b do not equal e. But if you do b=am, then b5=(am)5=a5m=(a5)m=em=e. Not sure if this is correct but I'd say that this is a contradiction because 5 does not divide 31 and we proved earlier that b31=e.

20. Dec 13, 2011

### I like Serena

I'm afraid you are doing things out of order.

The proof that (ba) has order 5 is earlier in the proof, and indeed is used to prove that b31=e.
So you can't use it the other way around.

You also haven't proven yet that b5≠e.