# A Particular Two-Variable System in a Group

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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Questions aren't generally closed for being too localized on SE. –  Austin Mohr Sep 22 '11 at 5:52
Aside from just playing around with cancellation and substitution, it might be helpful to note that $(xy)^{-1} = y^{-1}x^{-1}$. –  Austin Mohr Sep 22 '11 at 6:01
This problem in the book: Schaum's outline of Theory and Problems of Group Theory, page 127, item 4.102. –  Aj I Sep 22 '11 at 6:06
Off topic: We have five commenters ITT so far, all with names starting with 'a'. Fairly improbable. –  anon Sep 22 '11 at 6:19

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. –  Aj I Sep 22 '11 at 6:41
@Aj i: Which of the above identities you had trouble with? May be you should try easier questions first. This is very hard. –  Jyrki Lahtonen Sep 22 '11 at 8:05
@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! –  Jyrki Lahtonen Sep 22 '11 at 9:22
@Aj i: If $xyx^{-1}=z$, then $$z^k=(xyx^{-1})^k=(xyx^{-1})(xyx^{-1})\cdots(xyx^{-1})=xy^kx^{-1}.$$ –  Jyrki Lahtonen Sep 22 '11 at 9:46
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. –  Aj I Sep 22 '11 at 9:52

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. –  Aj I Sep 22 '11 at 11:56
+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 accept an answer. A better one may be coming up!! –  Jyrki Lahtonen Sep 22 '11 at 12:12
@ Jyrki Lahtonen: Thank you for your good advice for a beginner. –  Aj I Sep 22 '11 at 12:17
@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" :). –  user641 Sep 22 '11 at 12:21