Let $C :=\langle c,d : c^n , d^2, (cd)^2\rangle, A := \langle a,b : a^2 , b^2, (ab)^n\rangle.$ Show that the map $\phi: C\mapsto A: \phi(c) = ab,\phi(d) = b$ is an isomorphism, where $\langle S : R\rangle$ is defined to be $F(S)/N(R),$ the quotient group of the free group on $S$ and the smallest normal subgroup of $F(S) $ containing $R$.
I know that each element of $C$ is of the form $c^{i_1}d^{j_1}\cdots c^{i_k}d^{j_k}$ for some $k\geq 0, i_p, j_p\neq 0$ for $1\leq p\leq k$. I also know that if $\psi : S\to G$ is a function, there is a unique group homomorphism $\tilde{\psi} : F(S)\to G$ with $\tilde{\psi}(s) = \psi(s)$ for all $s\in S.$ This function is defined by $\tilde{\psi}(s_1^{a_1}\cdots s_k^{a_k}) = \psi(s_1)^{a_1}\cdots \psi(s_k)^{a_k}.$ Now if we take two words $w_1 = c^{i_1}d^{j_1}\cdots c^{i_k}d^{j_k}$ and $w_2 = c^{e_1}d^{f_1}\cdots c^{e_l}d^{f_l}$ in $C,\phi(w_1 w_2) = \phi(r),$ where $r$ is the reduced word equivalent to $w_1 w_2.$ Since $(cd)^2 = e, cd=d^{-1}c^{-1}= dc^{-1}$ (as $d^2 = e$). Using this result, one can show by induction that $c^l d = dc^{-l}$ for all $l$, so $c^{i_1}d^{j_1} = d^{j_1}c^{-i_1}.$ Inductively, the word $c^{i_1}d^{j_1}\cdots c^{i_k}d^{j_k}$ is equivalent to $d^{j_1}c^{-i_1+i_2} d^{j_2}\cdots c^{i_k}d^{j_k} = d^{j_1+j_2+\cdots + j_k}c^{-i_k+i_{k-1}+\cdots + (-1)^ki_1}.$ Thus we have that $w_1 w_2 = (d^{j_1+\cdots +j_k}c^{-i_k+i_{k-1}+\cdots (-1)^ki_1})(d^{f_1+\cdots + f_l}c^{-f_l + f_{l-1}+\cdots +(-1)^lf_1}).$ Since $b^2 = e,$ we can apply similar reasoning to above to show that $(ab)^l b = b (ab)^{-l}$ ($((ab)b)^2 = a^2 = e$) for all $l$. One can then show that $\psi(w_1 w_2)$ can be reduced to the same word as $\psi(w_1)\psi(w_2)$ using this property. In fact, it seems that every word in $C$ can be written in the form $c^l d^k$ for some $0\leq l<n, 0\leq k< 2.$
However, I'm not sure if any of these claims are right and how to show $\psi$ is injective. To show surjectivity, it seems that I'd need to reduce words and simplify to get that every word $a^{i_1}b^{j_1}\cdots a^{i_k}b^{j_k}$ is equivalent to a word of the form $(ab)^{i_1'}b^{j_1'}\cdots (ab)^{i_k'}b^{j_k'}$.
In the second presentation we can express $a$ in terms of $ab$ and $b$: since $a=a^{-1}$ and $(ab)^n=1$ we have $$ a = a^{-1} = b (ab)^{n-1}. $$ Hence, letting $z=ab$ we may generate the group by $z$ and $b$ and translate the three relations:
Using the third relation, the first is equivalent to $(bz^{-1})^2=1$, which may be rewritten using $b=b^{-1}$ as $(zb)^2=1$.
Hence we obtain the presentation $$ \langle z,b : z^n, b^2, (zb)^2\rangle, $$ which is exactly the first presentation you are given.