# Can we apply relations in a group presentation one by one?

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.

• I think there is one one too many in the title. Commented May 1, 2021 at 7:32

## 1 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 $$N(a,b)$$.

• Thank you, for some reason I thought it would be difficult to change the order of the product in general Commented Apr 30, 2021 at 13:50