Abstract Algebra Problem involving the order of groups

[xcr]

Homework Statement

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.

Homework 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

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.

 

[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.

 

[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.

 

[I like Serena]

Perhaps you can try and prove that the order of "a" is equal to the order of "ba"?
Start with checking that (ba)⁵=e.
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:

Suppose G is a group. Two elements a and b of G are called conjugate if there exists an element g in G with

gag⁻¹ = b.​

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

 

[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^-1a^mb=(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

 

[I like Serena]

Yes, that looks good!

Consider that (ba)⁵=b(ab)⁴a...

 

[Deveno]

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^-1a^mb=(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

the idea is, if we can get e = (ba)⁵ = b^ra^s, 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...

 

[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.

 

[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.)

 

[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.

 

[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...?

 

[xcr]

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

 

[I like Serena]

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

 

[xcr]

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

 

[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.

 

[xcr]

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

 

[I like Serena]

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

 

[Deveno]

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

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

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

why is this a contradiction?

 

[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=a^m, then b⁵=(a^m)⁵=a^5m=(a⁵)^m=e^m=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.

 

[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.

 

[Deveno]

what you HAVE established, is that:

ba = b^-1ab.

so what happens if b^-1ab = e?

(conjugates really do have the same order, it is never just the case that the order of gxg^-1 merely divides the order of x or vice versa).

so if (ba)⁵ = e and ba ≠ e, then the order of ba is some positive integer that divides 5 which is not 1, so...

 

[xcr]

Ok, so far I have that aba^-1=b², rewritten as b^-1ab=ba. Raising both sides to the 5th power gives us e=(ba)⁵, therefore the order of ba is 5 or m where m|5. so m=1,2,3,4. m can't be 2,3,4 because they don't divide m. 1|5 but then we have ba=e, where ba=b^-1ab. So b^-1ab=e, giving us ab=b, which shows that a=e. This is a contradiction to the assumption that a≠e, so the order of ba does not equal 1 and therefore must equal 5. Since the order of ba is 5, then e=(ba)⁵. Rewrite this as e=b(ab)⁴a=b(ab)(ab)(ab)(ab)a. Using ab=b²a, we can substitute until we get to the point where e=b³¹a⁵. Since the order of a is 5, e=b³¹. Thus, the order of b is 31 or m, where m|31. Since 31 is prime, only 1 and 31 divide 31. If the order of b is 1, then b=e, which is a contradiction to one of our assumptions. Therefore the order of b is 31.

I think that is it but if there is something wrong I am still accepting help :) just trying to make sure I understand this topic before the exam comes up.

The one part that I am a little worried about is do I need to include the part starting at "Thus, the order of b is 31 or..." I don't know if by proving that the order of ab does not equal 1 means that I do not need to prove this last part.

 

[Deveno]

Ok, so far I have that aba^-1=b², rewritten as b^-1ab=ba. Raising both sides to the 5th power gives us e=(ba)⁵, therefore the order of ba is 5 or m where m|5. so m=1,2,3,4. m can't be 2,3,4 because they don't divide m.

i think this is a typo...it should be that 2,3,4 do not divide 5.

1|5 but then we have ba=e, where ba=b^-1ab. So b^-1ab=e, giving us ab=b, which shows that a=e. This is a contradiction to the assumption that a≠e, so the order of ba does not equal 1 and therefore must equal 5. Since the order of ba is 5, then e=(ba)⁵. Rewrite this as e=b(ab)⁴a=b(ab)(ab)(ab)(ab)a. Using ab=b²a, we can substitute until we get to the point where e=b³¹a⁵. Since the order of a is 5, e=b³¹. Thus, the order of b is 31 or m, where m|31. Since 31 is prime, only 1 and 31 divide 31. If the order of b is 1, then b=e, which is a contradiction to one of our assumptions. Therefore the order of b is 31.

I think that is it but if there is something wrong I am still accepting help :) just trying to make sure I understand this topic before the exam comes up.

The one part that I am a little worried about is do I need to include the part starting at "Thus, the order of b is 31 or..." I don't know if by proving that the order of ab does not equal 1 means that I do not need to prove this last part.

yes you DO need to include that part. the primality of 31 is essential, if we had come up with something like b³⁰ = e, we would have to consider every divisor of 30.

 

[xcr]

Yea that was a typo, it should have been 5. So does that prove that the order of b is 31 or is there more that needs to be done?

 

[Deveno]

Rewrite this as e=b(ab)⁴a=b(ab)(ab)(ab)(ab)a. Using ab=b²a, we can substitute until we get to the point where e=b³¹a⁵.

you should be more explicit here. again, showing that ab^k = b^2ka first, is helpful here, because it allows you to streamline the derivation. so you should have something like:

e = (ba)⁵ = b(ab)⁴a
= b(ab)(ab)(ab)(ab)a = b(b²a)(b²a)(b²a)(b²a)a
= b³(ab²)(ab²)(ab²)a²
= ...?

in other words, you need to justify your "repeated reduction".

 

[xcr]

Gotcha. And I could show that ab^k=b^2ka by doing aba^-1=b² and then raising each side to the k gives ab^ka^-1=b^2k. Move the a over and get ab^k=b^2ka. Thank you so much Deveno and I like Serena. Hope ya'll both have a great holiday season.