# Prove free groups $\left<a,b \mid ababa = babab\right> \simeq \left<x,\mid x^2 = y^5\right>$.

Prove free groups $$A := \left< a,b \mid ababa = babab \right> \simeq \left =: B$$, where $$\left$$ 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?

• There is a typo in the title $x,|$. Also $A$ and $B$ are not free groups. – 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.