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Suppose $a$ and $b$ are elements of a group $G$. If $a^{-1}b^{2}a=b^{3}$ and $b^{-1}a^{2}b=a^{3}$, prove $a=e=b$.

I've been trying to prove but still inconclusive. Please prove to me. Thanks very much for proof.

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    http://mathoverflow.net/questions/76060/prove-in-group-theory-closed2011-09-22
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    Questions aren't generally closed for being too localized on SE.2011-09-22
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    "Please prove to me" is not how it works here. What have you tried?2011-09-22
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    Aside from just playing around with cancellation and substitution, it might be helpful to note that $(xy)^{-1} = y^{-1}x^{-1}$.2011-09-22
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    This problem in the book: Schaum's outline of Theory and Problems of Group Theory, page 127, item 4.102.2011-09-22
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    I have tried to show that $a^{2}=a$ and $b^{2}=b$, but also without success.2011-09-22
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    Off topic: We have five commenters ITT so far, all with names starting with 'a'. Fairly improbable.2011-09-22

3 Answers 3

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In this answer I give credit to Jyrki Lahtonen for the answer he posted.

There are holes in his post, so I sensed the need for a step by step answer (firstly to convince myself, but also other people in doubt), and so here it is.

$\bbox[5px,border:2px solid]{\begin{array}{cc}a^3=b^{-1}a^2b&(\alpha)\\b^3=a^{-1}b^2a &(\beta)\end{array}}$

$b^{6}=b^3b^3=(a^{-1}b^2a)(a^{-1}b^2a)=a^{-1}b^4a\quad(\alpha)$

$b^{12}=b^6b^6=(a^{-1}b^4a)(a^{-1}b^4a)=a^{-1}b^8a\quad(\alpha)$

$b^{18}=b^{12}b^6=(a^{-1}b^8a)(a^{-1}b^4a)=a^{-1}b^{12}a\quad(\alpha)$

$b^{27}=b^{18}b^6b^3=(a^{-1}b^{12}a)(a^{-1}b^4a)(a^{-1}b^2a)=a^{-1}b^{18}a\quad(\alpha)$


$b^{27}=a^{-1}b^{18}a=a^{-2}b^{12}a^2=a^{-3}b^8a^3=\ (\beta)\ =(b^{-1}a^2b)^{-1}b^8(b^{-1}a^2b)=(b^{-1}a^{-2}b)b^8(b^{-1}b^{-1}a^2b)=b^{-1}a^{-2}b^8a^2b=b^{-1}(a^{-1}(a^{-1}b^8a)a)b=b^{-1}(b^{18})b=b^{18}$


$b^{27}=b^{18}\Rightarrow b^9=1$

$b^9=b^3b^3b^3=(a^{-1}b^2a)(a^{-1}b^2a)(a^{-1}b^2a)=(a^{-1}b^6a)=1\Rightarrow b^6=1\quad (\beta)$

$b^3=b^9b^{-6}=1=a^{-1}b^2a\Rightarrow b^2=1$

$b=b^3b^{-2}=1$ and then $a^3=b^{-1}a^2b=a^2\Rightarrow a=1\quad (\alpha)$.

$\bbox[5px,border:2px solid]{a=b=1}$

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In fact, it is true that the presentation $\langle a,b\ |\ a^{-1}b^na=b^{n+1}, b^{-1}a^nb=a^{n+1}\rangle$ always defines the trivial group.

Here's a proof: Let $M=n^{n+1}$, $N=(n+1)^{n+1}$, and check that we have $b^{-(n+1)}a^Mb^{(n+1)}=a^N$.

Now we also know from the relations that $ab^{(n+1)}=b^na$ and similarly $b^{-(n+1)}a^{-1}=a^{-1}b^{-n}$. So we also have $$ a^N=b^{-(n+1)}a^Mb^{(n+1)}=(b^{-(n+1)}a^{-1})a^M(ab^{(n+1)})=a^{-1}b^{-n}a^Mb^na.$$

Thus $a^N=b^{-n}a^Mb^n=a^K$, where $K=n\cdot (n+1)^n$. So we have $1=a^{N-K}=a^L$, where $L=(n+1)^n$. But if $P=n^n$ (sorry for all the letters!), we have $b^{-n}a^Pb^n=a^L=1$, so $a^P=1$.

But $GCD(P,L)=GCD(n,n+1)=1$, so $a=1$, and of course this implies $b^n=b^{n+1}$, so also $b=1$.

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    @ Steve D: Thank you for a new proof.2011-09-22
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    +1 Easy for me to follow this argument :-) Probably well known? @Aj i: Again, just to make sure that you know. If you find Steve's answer more helpful than mine, you are welcome to accept his answer instead of mine. Strictly your call! This is very much in the spirit of the site. A common policy seems to be: 1) be generous with upvotes, but 2) it is usually a good idea to wait a bit for you to accept an answer. A better one may be coming up!!2011-09-22
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    @ Jyrki Lahtonen: Thank you for your good advice for a beginner.2011-09-22
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    @JyrkiLahtonen: It is an exercise in Cohen's Combinatorial Group Theory, somewhere in the first couple of chapters. I didn't remember all the details when I started writing, all I had in mind was "use that n+1 and n are rel. prime" :).2011-09-22
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As it happens I have my copy of Baumslag & Chandler on the shelf. This exercise is listed as very hard. I seem to have pencilled in a solution 27-28 years ago. The text is really worn out, so I can only make out the first few steps. Damn, I really need a prescription for new glasses... Anyway here are the first three consequences of those relations: $$ a^{-1}b^8a=b^{12}, a^{-1}b^{12}a=b^{18}, a^{-1}b^{18}a=b^{27}. $$ As consequences of these I seem to have derived (you must rederive these for full credit) the following: $$ a^{-2}b^{12}a^2=b^{27}=a^{-3}b^8a^3=b^{-1}a^{-2}b^8a^2b. $$ The next consequence seems to be $a^{-2}b^8a^2=a^{-2}b^{12}a^2.$ From that point on the text is too blurred, but I think I might be able to redo that even though my brain has lost most of its agility over the years. Just in case this is homework I will stop here with these hints.

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    Thanks very much for your advice. I'll try again from your instructions.2011-09-22
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    @Aj i: Good luck! To derive the maximum benefit from these hints try to figure out, where this approach came from. I feel that I have, to a great extent, ruined this problem for you. Makes me a bit sad actually.2011-09-22
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    @ Jyrki Lahtonen:I'm trying, but it also proved unsuccessful. Do you have any suggestions for me?2011-09-22
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    @Aj i: Which of the above identities you had trouble with? May be you should try easier questions first. This is *very hard*.2011-09-22
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    @ Jyrki Lahtonen:Thank you very much.2011-09-22
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    @Aj i: You do know that conjugation by an element is an automorphism of the group, and maps powers to powers? That is the starting point. But only a starting point!2011-09-22
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    Please let me know about the conjugation and ways to prove. Thank you very much.2011-09-22
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    @Aj i: If $xyx^{-1}=z$, then $$z^k=(xyx^{-1})^k=(xyx^{-1})(xyx^{-1})\cdots(xyx^{-1})=xy^kx^{-1}.$$2011-09-22
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    In this time, we can prove that $b^{27}=a^{-2}b^{12}a^{2}=a^{-2}b^{8}a^{2}=b^{18}$, so $b^{9}=e$ and $b^{5}=e$. Hence $b=e$. Similarly, $a=e$. Thank you Jyrki Lahtonen.2011-09-22
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    @ Jyrki Lahtonen:Thanks for your very valuable for me. I am sorry for my poor English.2011-09-22