Register to reply 
Abstract Algebra Problem involving the order of groups 
Share this thread: 
#1
Dec1011, 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 nm. 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
Dec1011, 05:36 PM

Sci Advisor
P: 906

one idea to try:
see if you can prove that ab^{n} = b^{2n}a. 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
Dec1111, 01:20 PM

P: 22

I understand how you got to this point: ab^{n} = b^{2n}a. The steps to get there go like this:
aba^{1}=b^{2} so raising both sides to the nth power would look like (aba^{1})^{n}=(b^{2})^{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 ab^{n}a^{1}=b^{2n}. Multiplying by a on the right gives ab^{n}=b^{2n}a. 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
Dec1111, 02:32 PM

HW Helper
P: 6,187

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


#5
Dec1211, 12:32 AM

P: 22

ab=b^{2}a
b^{1}ab=ba (b^{1}ab)^{5}=(ba)^{5} (b^{1}a^{5}b)=(ba)^{5} b^{1}eb=(ba)^{5} b^{1}b=(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^{1}a^{m}b=(ba)^{m}, which ≠e? Then do I just show that (aba^{1})^{n}=(b^{2})^{n}, which can be rearranged as b^{n}ab^{n}=b^{n}a. Not sure what to do from here. If I raise both sides to the 5th power, I would get e=(b^{n}a)^{5} 


#7
Dec1211, 05:11 AM

Sci Advisor
P: 906




#8
Dec1211, 09:59 AM

P: 22

ok so I took e=(ba)^{5}=b(ab)^{4}a=b(ab)(ab)(ab)(ab)a. I then took ab=b^{2}a 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=b^{31}a^{5}, but a^{5}=e, so e=b^{31}e, or e=b^{31}. Therefore the order of b is 31. Felt like I was doing the right steps but not very confident in that answer.



#9
Dec1211, 10:11 AM

HW Helper
P: 6,187

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
Dec1211, 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)^{2}b≠e, (ab)^{4}=a(ba)^{3}b≠e.



#11
Dec1211, 01:15 PM

HW Helper
P: 6,187

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
Dec1211, 04:14 PM

P: 22

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



#13
Dec1211, 04:17 PM

HW Helper
P: 6,187

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



#14
Dec1211, 04:24 PM

P: 22

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



#15
Dec1211, 04:28 PM

HW Helper
P: 6,187

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
Dec1211, 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



#18
Dec1211, 05:50 PM

Sci Advisor
P: 906

if b is any power of a, say b = a^{m}, show that b^{5} = e. why is this a contradiction? 


Register to reply 
Related Discussions  
Abstract Algebra Question: order, stabilizer, and general linear groups  Calculus & Beyond Homework  0  
Abstract Algebra  Cyclic groups  Calculus & Beyond Homework  0  
Abstract Algebra Groups  Calculus & Beyond Homework  4  
Groups, Normalizer, Abstract Algebra, Dihedral Groups...help?  Calculus & Beyond Homework  12  
Abstract Algebra: Groups of order 21  Introductory Physics Homework  0 