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Let

$$B:=\langle x,y\mid x^3=1,y^2=1,xyx^{-1}y^{-1}=1\rangle$$

My question is: how does one show that

$$\mathbb Z_6 \cong B?$$

My solution:

\begin{array}{ccc} y & \mapsto & 3\\ x & \mapsto & 2\\ \{x,y\} & \rightarrow & B\\ \downarrow & \underset{\tilde{f}}{\nearrow}\\ F_{\{x,y\}} \end{array}

$\tilde{f}(x^3)=\tilde{f}(x)*\tilde{f}(x)*\tilde{f}(x) = 2+2+2=0\equiv1$

$\tilde{f}(y^2)=\tilde{f}(y)*\tilde{f}(y) = 3+3=0\equiv1$

$\tilde{f}(xyx^{-1}y^{-1})=\tilde{f}(x)*\tilde{f}(y)*\tilde{f}(x^{-1})*\tilde{f}(y^{-1}) = 2+3+4+3=12\equiv1 \pmod 6$

Prove $\tilde{f}$ is surjective:

$\tilde{f}(xy)=\tilde{f}(x)*\tilde{f}(y)=2+3=5$ and $5$ is generator of $\mathbb Z_6$ so for each $n \in \mathbb N$ $\tilde{f}(xy)^n \in \langle \rangle.$

Prove $\tilde{f}$ is injective:

Suppose exist $x^a*y^b \neq x^c*y^d \space (a,b,c,d \in \mathbb N) $ that$ \space \tilde{f}(x^a*y^b)=\tilde{f}(x^c*y^d)$

since ${x^3=1} $ and $ {y^2=1} \implies a,c<3 $ and $b,d<2.$

$ \space \tilde{f}(x^a)*\tilde{f}(y^b)=\tilde{f}(x^c)*\tilde{f}(y^d)$

i am not sure i can simply say that $a=c$ and $b=d$.

Is my proof right ?

Shaun
  • 44,997
Algo
  • 2,322

2 Answers2

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You're on the right track.

By Tietze transformations, the relator $xyx^{-1}y^{-1}$ becomes the relation $xy=yx$. Then $B$ becomes the well-known presentation for $\Bbb Z_2\times \Bbb Z_3$; but $\Bbb Z_2\times\Bbb Z_3\cong \Bbb Z_6$.

Shaun
  • 44,997
1

You're trying to use much heavier machinery than you need to.

All elements of the group have the following form: $$x^{n_1}y^{m_1}x^{n_2}y^{m_2}\cdots x^{n_k}y^{n_k}\ ,$$ where the $n_i, m_i\in\mathbb{Z}$.

Now, $x$ has order $3$ and $y$ has order $2$, so $x^{-1}=x^2$ and $y^{-1}=y$.

Also, the $xyx^{-1}y^{-1}=1$ condition means that the group is abelian, so in fact an arbitrary element of the group has the form

$$ x^ny^m\ , $$ where $n=0,1,2$ and $m=0,1$. So we have an abelian group of order $6$. It is generated by $xy$, so it is cyclic.

Teddy38
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