Exercise 1.2.18 (Collapsing of a group)

Proof.

(a)

From $v^3=1$, multiplying each member by $v^{-1}$, we obtain $v^2=v^{-1}$.

(b)

Note that $$\begin{array}{llll}\hfill v^2{u}^{3}v& =(v^2{u}^2)(uv)& \hfill & \\ \hfill & =(uv)(uv)(\text{since }{v}^2{u}^2=uv)& \hfill & \\ \hfill & =(uv)(v^2{u}^2)(\text{since }uv=v^2{u}^2)& \hfill & \\ \hfill & =u{v}^3{u}^2& \hfill & \\ \hfill & =u^3(\text{since }{v}^3=1)& \hfill & \end{array}$$

using part (a) (or multiplying left by $v$), we obtain

$$u^{3}v=v{u}^3,$$

so $v$ commutes with $u^3$.

(c)

Since $u^4=1$, $u^9=u{(u^4)}^2=u$. Then, using $u^{3}v=u{v}^3$ three times, $$uv=u^{9}v=u^3{u}^3{u}^{3}v=u^3{u}^{3}v{u}^3=u^{3}v{u}^3{u}^3=v{u}^3{u}^3{u}^3=v{u}^9=vu,$$

so $v$ commutes with $u$.

(d)

Since $v$ commutes with $u$, the last relation $uv=v^2{u}^2$ gives $vu=v^2{u}^2$. Multiplying left by $v^{-1}$ and right by $u^{-1}$, we obtain $$1=vu=uv.$$

(e)

Since $v$ commutes with $u$, ${(uv)}^3=u^3{v}^3$, therefore, using $uv=1$, $$u=u{(uv)}^3=u{u}^3{v}^3=u^4{v}^3=1.$$

The third relation gives $uv=v^2{u}^2$, with $u=1$, so $v=v^2$, and $v=1$.

We can conclude $Y=⟨u,v⟩=\{1\}$.