A proof that $\langle u,v\mid u^4=v^3=1, uv=v^2u^2\rangle$ defines the trivial group.

Question

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. $\square$

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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?

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$\dagger$ I've just noticed that a $u$ comes from nowhere in this step. Sorry!

Answer 1

\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 $u^4=1=v^3=u^3$, $u=v=1$.

Answer 2

... 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.

Yeah somehow I didn't see this at first either.

But believe the idea is to start with the LHS. Then apply the relation $u v = v^2 u^2$, once in each direction. \begin{align} v^2 u^3 v &= (v^2 u^2) (u v) \\ &= (u v) (u v) \\ &= (u v) (v^2 u^2) \\ &= u v^3 u^2 \\ &= u^3. \end{align}