Consider a group presentation $\langle x,y \mid xy=yx, x^7=y^3 \rangle$. By definition, this is $F(\{x,y \})/N(xyx^{-1}y^{-1},x^7y^{-3})$ where $F(S)$ denotes the free group on the set $S$ and $N(R)$ denotes the normal subgroup generated by $R$. Intuitively, when identifying this group from its presentation, I first apply the relation $xy=yx$, which gives the group $\mathbb{Z}^2$, and then i further apply the relation $x^7=y^3$ to the groups $\mathbb{Z}^2$, which gives $\mathbb{Z}$ as a final answer. As a sanity check, I'm trying to prove that this approach is valid in general, but I can't quite seem
Claim: Let $G$ be a group and $a,b\in G$ and $\pi: G \to G/N(a)$ be the quotient map. Then $G/N(a,b) \cong \big (G/N(a) \big )/N(\pi(b))$
Attempted proof:
We know that $N(\pi(b))= \pi^{-1}(N(\pi(b)))/N(a)$ and then by the third isomorphism theorem we have that $\big (G/N(a)\big )/N(\pi(b)) \cong G/\pi^{-1}(N(\pi(b)))$ so all we need to prove is that $N(a,b) = \pi^{-1}(N(\pi(b)))$.
We have $z\in\pi^{-1}(N(\pi(b))) \Leftrightarrow \pi(z) \in N(\pi(b)) \Leftrightarrow zN(a) =N(bN(a)) \Leftrightarrow zN(a) = \big( \prod_{g_i \in G} g_i^{-1}b^{\epsilon_i}g_i \big)N(a)$ for some $\epsilon_i$ all $\pm 1$. This is equivalent to $z = \big( \prod_{g_i \in G} g_i^{-1}b^{\epsilon_i}g_i \big)\big( \prod_{h_i \in G} h_i^{-1}a^{f_i}h_i \big)$.
Now, if $G$ were Abelian I could claim that this was equivalent with $z \in N(a,b)$, but without that assumption I'm not sure how to proceed. Is the claim incorrect? Or do I just need to do a little more work?
Not a duplicate of Why is $\langle S\mid R\cup R'\rangle $ a presentation for $G/N(R')$, where $G$ is a group with presentation $\langle S\mid R\rangle?$ because I'm asking about specifically the part of the proof which the only answer there omits.
Best Answer
Every element of $N(a,b)$ can be written as a product of terms of the form $g^{-1}a^\epsilon g$ and $g^{-1} b^\epsilon g$ for elements $g \in G$.
To complete the proof, you just need to show that you can rewrite this product with all conjugates of $b^\epsilon$ coming before all conjugates of $a^\epsilon$.
To do that just note that, for $g,h \in G$, $$(g^{-1}a^\epsilon g)(h^{-1} b^{\epsilon'}h) = (h^{-1} b^{\epsilon'}h) (w^{-1} a^\epsilon w),$$ where $w = gh^{-1} b^{\epsilon'}h$.
Alternatively, this follows from $N(a,b) =N(a)N(b)$. Clearly $N(a)N(b) \le N(a,b)$, and since $N(a)N(b)$ is a normal subgroup of $G$ containing $a$ and $b$, it must contain $N(a,b)$.