Prove free groups $A := \left< a,b \mid ababa = babab \right> \simeq \left<x,y \mid x^2 = y^5\right> =: B$, where $\left<x_i \mid r_j = s_j\right>$ stands for the quotient group generated by $\{x_i, i \in I\}$ by the normal subgroup generated by the elements $r_js_j^{-1}$.

This has been addressed in here, but I am still not sure how to prove it.

My attempts: Define $\phi: B \to A$ by $\phi(x) = ababa, \phi(y) = ab$ and $\phi(xy) = \phi(x)\phi(y)$, we could see that $\phi$ is a homomorphism and $\phi(x^2) = (ababa)(ababa) = (ababa)(babab) = (ab)^5 = \phi(y^5)$. We could solve $a, b$ in terms of $x, y$: $\phi(x(y^{-1})^2)= b$ and $\phi(y^{-1}x^{-1}) = a$, so $\phi$ is surjective.

My question: How to show $\phi$ is injective? Why the answer in the above link says that we "implicitly apply universal property of group presentations" and how the inverse map in the above link works?

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    $\begingroup$ There is a typo in the title $x,|$. Also $A$ and $B$ are not free groups. $\endgroup$ – Lozenges Jul 19 at 8:43

Let $\langle a,b\rangle$ denote the free group generated by $\{a,b\}$ and similarly for $\{x,y\}$, so that in my notation $\langle a,b\rangle\not=\langle x,y\rangle $ but these groups are isomorphic. Let $N_1$ denote the smallest normal subgroup of $\langle a,b\rangle$ containing the element $ababab^{-1}a^{-1}b^{-1}a^{-1}b^{-1}$ and let $N_2$ denote the smallest normal subgroup of $\langle x,y\rangle$ containing the element $ x^2y^{-5}$. There is a unique homomorphism $f\colon \langle x,y\rangle\to\langle a,b\rangle$ satisfying $f(x)=ababa$ and $f(y)=ab$. We wish to show that $f$ descends to a homomorphism on the quotients $q\colon \langle x,y\rangle/N_2\to \langle a,b\rangle/N_1$ and that this map $q$ is in fact an isomorphism.

Given a coset $wN_2$ with $w\in\langle x,y\rangle$, we define $q$ of this coset to be the coset $f(w)N_1$ in $\langle a,b\rangle$. For this to be well-defined, we must show that the coset $f(w)N_1$ does not depend on which representative $w$ was chosen. It amounts to showing that $f(N_2)\subseteq N_1$, which in turn amounts to showing that $f(x^2y^{-5})\in N_1$, i.e. $$ (ababa)^2(ab)^{-5}\in N_1. $$ To see this, we compute that $$ (ababa)^2(ab)^{-5}=ababa\cdot \bigl(ababab^{-1}a^{-1}b^{-1}a^{-1}b^{-1}\bigr)\cdot (ababa)^{-1}, $$ and this is an element of $N_1$ (since it is conjugate to the middle parenthesized term, which belongs to $N_1$ by definition). Thus we have a well-defined mapping $q$ on the quotient group, and it inherits the homomorphism property from $f$.

Finally, we show that $q$ is an isomorphism by exhibiting an inverse. Define $g\colon \langle a,b\rangle \to\langle x,y\rangle$ to be the unique homomorphism satisfying $g(a)=y^3x^{-1}$ and $g(b)=xy^{-2}$. Similarly to the above, to show that $g$ descends to a quotient function $r$ amounts to checking that $$ g(ababab^{-1}a^{-1}b^{-1}a^{-1}b^{-1})\in N_2, $$ which is straightforward to see since $ g(ab)=y $ implies that $$ g(ababab^{-1}a^{-1}b^{-1}a^{-1}b^{-1})=y^2 g(a) y^{-2} g(b)^{-1}=y^2 \cdot y^3 x^{-1} \cdot y^{-2} \cdot y^2 x^{-1}=y^5 x^{-2}\in N_2. $$ Finally, since $f\circ g$ satisfies $$ f\circ g(a)=f(y^3x^{-1})=(ab)^3(ababa)^{-1}=a, $$ and $$ f\circ g(b)=f(xy^{-2})=ababab^{-1}a^{-1}b^{-1}a^{-1}\in bN_1, $$ it follows that $q\circ r$ is the identity, as desired.


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