Is it true that groups $\langle a,b \mid a^n b^k=b^ka^{n+1}, b^la^s=a^sb^{l+1}\rangle$ are nontrivial for almost all (in any sense:))) $n,k,l,s\in\mathbb N$?
The group is always trivial. Indeed, $a^{ns}b^{l^n}a^{ns}=b^{(l+1)^n}$. On the other hand, $a^{ns}=b^ka^{(n+1)s}b^{k}$. Substitute $a^{ns}$ from the second equality to the first. You will get that $a^{(n+1)s}$ conjugates $b^{l^n}$ to $b^{(l+1)^n}$. Since $a^{ns}$ does the same, you get that $a^s$ commutes with $b^{l^n}$. Hence $b^{l^n}=b^{(l+1)^n}$ and $b$ is torsion, $b^p=1$. Similarly, $a$ is torsion, $a^q=1$. Notice also that $p=(l+1)^nl^n$ is coprime with $l$ and $q$ is coprime with $n$. Now we have $b^{kp}a^{n^p}b^{kp}=a^{(n+1)^p}$. Hence $a^{n^p}=a^{(n+1)^p}$. We already know that $a^{n^l}=a^{(n+1)^l}$. Suppose that the order of $a$ is $t$. Then $t$ divides both $(n+1)^ln^l$ and $(n+1)^pn^p$ and is coprime with $n$. Then $t$ divides $$(n+1)^pn^p(n+1)^p+n^l*(n+1)^{pl}=n^p+n^l*(n+1)^{pl}.$$ Since $t$ and $n$ are coprime, $t$ divides $(n+1)^{pl}n^{pl}$. Proceed by using the Euclidean algorithm. Since $p$ and $l$ are coprime, we get that $t$ divides $(n+1)n=1$, so $a=1$. Similarly $b=1$ and the group is trivial.

$\begingroup$ I'm not following this. Why does $a^s$ commute with $b^{l^n}$? $\endgroup$ – Neil Strickland Sep 27 '11 at 13:45

$\begingroup$ Elementary, Neil: Substitute $a^{ns}$ from the second equality to the first, then cancel extra $b$'s. $\endgroup$ – Mark Sapir Sep 27 '11 at 14:35


$\begingroup$ how did you obtain the first identity equation? $\endgroup$ – Nikita Kalinin Sep 27 '11 at 19:34

$\begingroup$ @Nikita: Suppose for simplicity that $a^2b^3a^{2}=b^7$ (the same works for every other choice of exponents). Then $$a^2b^9a^{2}=(a^2b^3a^{2})^3=b^{21}.$$ Now $$a^4b^9a^{4}=a^2b^{21}a^{2}=(a^2b^3a^{2})^7=a^{49}.$$ Notice that $9=3^2, 49=7^2$. I guess you can do it in general now. $\endgroup$ – Mark Sapir Sep 27 '11 at 20:30
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