Abstract Algebra Problem involving the order of groups


by xcr
Tags: abstract algebra, groups, order of groups
xcr
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#1
Dec10-11, 03:59 PM
P: 22
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.
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Deveno
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#2
Dec10-11, 05:36 PM
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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
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#3
Dec11-11, 01:20 PM
P: 22
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.

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#4
Dec11-11, 02:32 PM
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Abstract Algebra Problem involving the order of groups


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:
Quote Quote by 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
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#5
Dec12-11, 12:32 AM
P: 22
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
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#6
Dec12-11, 02:50 AM
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Yes, that looks good!

Consider that (ba)5=b(ab)4a...
Deveno
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#7
Dec12-11, 05:11 AM
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Quote Quote by xcr View Post
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
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...
xcr
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#8
Dec12-11, 09:59 AM
P: 22
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.
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#9
Dec12-11, 10:11 AM
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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
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#10
Dec12-11, 01:07 PM
P: 22
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.
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#11
Dec12-11, 01:15 PM
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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
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#12
Dec12-11, 04:14 PM
P: 22
Well if (ba)=e, then (ba)2=(ba)(ba)=ee=e, which is not true. Therefore ba≠e
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#13
Dec12-11, 04:17 PM
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I'm afraid it's still possible that (ba)2=e at this stage.
xcr
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#14
Dec12-11, 04:24 PM
P: 22
I thought that since 2,3, and 4 did not divide 5 that the could not be the order?
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#15
Dec12-11, 04:28 PM
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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
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#16
Dec12-11, 04:37 PM
P: 22
Ok, so ba≠e because then a must equal e and b must equal e, which cannot be true because of our assumptions
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#17
Dec12-11, 04:41 PM
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Close... but how did you get that a must be equal to e?
Deveno
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#18
Dec12-11, 05:50 PM
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Quote Quote by xcr View Post
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 = am, show that b5 = e.

why is this a contradiction?


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