This appears to be new to MSE.
I'm reading "Abstract Algebra (Third Edition)," by Dummit & Foote.
This is based on Exercise 1.2.18.
Question: Show that $$Y=\langle u,v\mid u^4=v^3=1, uv=v^2u^2\rangle$$ is defines the trivial group.
My Attempt:
Tietze transformations are not yet introduced in the text and so they do not apply here.
From $v^3=1$ and $u^4=1$, we have $v^2=v^{-1}$ and $u^3=u^{-1}$, respectively. Also $u^{-2}=u^2$.
Also, we have
\begin{align}
\color{red}{v^2u^3v}
&=(v^2u^2)(uv) \\
&=v^2u^{-1}v \\
&=v^{-1}u^{-1}v \\
&=(uv)^{-1}v \\
&=(v^2u^2)^{-1}v \\
&=u^{-2}v^{-2}v \\
&=\color{red}{u^3v},^\dagger
\end{align}
so that $v^2=1$. Now $v^2=v^{-1}$ implies $v=1$.
Thus we can kill $v$:
$$Y=\langle u\mid u^4=1, u=u^2\rangle,$$
but now $u=u^2$ implies $u=1$.
Hence the group given by $Y$ is trivial as all its generators are trivial. $\square$
My problem is that the proof Exercise 1.2.18 ibid. guides the reader through a different path.
Thus I would like to know if my proof holds.
The only thing that I stumble on in the proof outlined in the exercise is part (b), which gets the reader to show, in effect, that
$$v^2u^3v=u^3.$$
This ought to be simple for me but I just don't see it. The rest of the exercise seems easy.
Why do I get an extra $v$ in the RHS?
$\dagger$ I've just noticed that a $u$ comes from nowhere in this step. Sorry!
Best Answer
\begin{align*}uv^2 &= v^2u^2v\\ &= v^2u(uv)=v^2uv^2u^2\\ &= v^2(uv)vu^2=vu^2vu^2\\ &= vu(uv)u^2=vuv^2\\ &= v(uv)v=u^2v\end{align*} Hence $u=v$; since $u^4=1=v^3=u^3$, $u=v=1$.