Showing a Mapping Between $\left\langle a,b \mid abab^{-1}\right\rangle$ and $\left\langle c,d \mid c^2 d^2 \right\rangle$ is Surjective

Question

Hypothesis:

1. Let

   $$ G \cong \left\langle a,b \mid abab^{-1}\right\rangle $$

   $$ H \cong \left\langle c,d \mid c^2 d^2 \right\rangle $$

2. Let the function $f$ be defined as follows. First let $f(a) = cd$ and $f(b) = d^{-1}$. For all other elements $g$ of $G$, define $f(g)$ as follows:

   $$ f(g) = f(a^{\alpha_1} b^{\beta_1} \cdot \ldots \cdot a^{\alpha_k}b^{\beta_k}) = f(a)^{\alpha_1} f(b)^{\beta_1} \cdot \ldots \cdot f(a)^{\alpha_k}f(b)^{\beta_k} $$

   such that $a^{\alpha_1} b^{\beta_1} \cdot \ldots \cdot a^{\alpha_k}b^{\beta_k}$ is the fully reduced and unique word representation of $g$ in $G$.

   Then $f$ is a well-defined mapping from $G$ to $H$.

Goal: Show that $f$ is an isomorphism. As my attempt below will reflect, I know how to show that $f$ is a surjective homomorphism, however I don't know how to show that it is an injection.

Attempt:

1. We need only check that $f(abab^{-1}) = e_H = c^2d^2$ in order for $f$ to be a homomorphism. To do this we have

   $$ f(abab^{-1}) = f(a)f(b)f(b)f(b)^{-1} = (cd)(d^{-1})(cd)(d^{-1})^{-1} = c^2d^2 = e_H $$

   as desired.

2. To show that $f$ is surjective, we note that

   $$ f(ab) = f(a)f(b) = (cd)(d^{-1}) = c $$

   $$ f(b^{-1}) = f(b)^{-1} = (d^{-1})^{-1} = d $$

   so that if $h = c^{\alpha_1}d^{\beta_1} \cdot \ldots \cdot c^{\alpha_k}d^{\beta_k}$ we have that

   $$ f\left((ab)^{\alpha_1}(b^{-1})^{\beta_1} \cdot \ldots \cdot (ab)^{\alpha_k}(b^{-1})^{\beta_k}\right) = c^{\alpha_1}d^{\beta_1} \cdot \ldots \cdot c^{\alpha_k}d^{\beta_k} = h $$

   as desired.

Question: Why is $f$ injective?

Answer 1

Let's observe that an element in $G$ can be written as $a^nb^m$ for integers $n,m$. The relation $abab^{-1}=e\Rightarrow ba^{-1}=ab$ or $a^{-1}b=ba$. So given some word $$a^{k_1}b^{k_2}...a^{k_{r-1}}b^{k_r}$$ we can always move the $a^{k_j}$ to the left of the $b^{k_{j-1}}$. Then an arbitrary element in $G$ given by $a^nb^m$ mapped to $e$ by $f$ means $$f(a^nb^m) = f(a)^nf(b)^m = (cd)^n(d)^{-m} = e \Rightarrow n=0, m=0$$

Basically, $(cd)^n(d)^{-m}$ cannot be made to be the identity using $c^2d^2=e$ without $m=n=0$.

Therefore the kernel of $f$ is trivial.

Answer 2

An alternative way to do this is to show that the map $g:c \mapsto ab,\,d \mapsto b^{-1}$ extends to a homomorphism $g:H \to G$ (to do that, you just need to check that $g(c^2d^2)=1$), and then show that $fg:H \to H$ and $gf:G \to G$ are the identity maps on $H$ and $G$, which is easy: just check that they map the group generators to themselves. So $f$ and $g$ are $2$-sided inverse maps, and must be isomorphisms.

Answer 3

A different way to do this problem is to use Tietze transformations. These are specific transformations you can do to group presentations. The key result is that two presentations $\mathcal{P}$ and $\mathcal{Q}$ define isomorphic groups if and only if there exists a sequence of Tietze transformations which takes $\mathcal{P}$ to $\mathcal{Q}$. In this example we can do the following. $$\begin{align*} \langle a, b; abab^{-1}\rangle &\cong \langle a, b, c; abab^{-1}, c=ab\rangle&\text{add in new generator }c\\ &\cong \langle a, b, c; ababb^{-2}, c=ab\rangle\\ &\cong \langle a, b, c; c^2b^{-2}, c=ab\rangle&\text{replace }ab\text{ with }c\text{ throughout}\\ &\cong \langle a, b, c; c^2b^{-2}, cb^{-1}=a\rangle\\ &\cong \langle b, c; c^2b^{-2}\rangle&\text{remove generator }a\\ &\cong \langle b, c, d; c^2b^{-2}, d=b^{-1}\rangle&\text{add in new generator }d\\ &\cong \langle b, c, d; c^2d^{2}, d=b^{-1}\rangle&\text{replace }b\text{ with }d^{-1}\text{ throughout}\\ &\cong \langle b, c, d; c^2d^{2}, d^{-1}=b\rangle\\ &\cong \langle c, d; c^2d^{2}\rangle&\text{remove generator }b\\ \end{align*}$$ et voila! The groups are isomorphic. Note that in practice you would just write the following. $$\begin{align*} \langle a, b; abab^{-1}\rangle &\cong \langle a, b, c; abab^{-1}, c=ab\rangle\\ &\cong \langle a, b, c; c^2b^{-2}, cb^{-1}=a\rangle\\ &\cong \langle b, c; c^2b^{-2}\rangle\\ &\cong \langle c, d; c^2d^{2}\rangle\\ \end{align*}$$

I realise that this doesn't answer your specific problem, but thought you might be interested anyway :-)