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Let $X$ and $Y$ be disjoint sets and $\langle X|R\rangle$ be the group given by the generators $X$ and relations $R$. Similarly for $\langle Y|S\rangle$. Is there a simple relation between $\langle X\cup Y|R\cup S\rangle$ and $\langle X|R\rangle,\langle Y|S\rangle$? (Here we identify $R$ and $S$ in the free group $F(X\cup Y)$ via the map $F(X)\to F(X\cup Y)$ induced from the inclusion $X\to X\cup Y$).

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    $\begingroup$ This is the coproduct of $\langle X\mid R\rangle$ and $\langle Y \mid S\rangle$ in the category of groups, but I've forgotten the proof. $\endgroup$ – neth Oct 21 '15 at 23:37
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    $\begingroup$ en.wikipedia.org/wiki/Free_product#Presentation $\endgroup$ – janmarqz Oct 21 '15 at 23:39
  • $\begingroup$ Neat, a proof in the answer would be good though. $\endgroup$ – Zero Oct 21 '15 at 23:40
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    $\begingroup$ what do you what to prove? $\endgroup$ – janmarqz Oct 21 '15 at 23:42
  • $\begingroup$ The proof is straightforward: group presentations describe a group as a certain colimit of free groups, and colimits commute with colimits. $\endgroup$ – Qiaochu Yuan Oct 21 '15 at 23:45
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Let $\phi \colon F(X) \to A$ and $\psi \colon F(Y) \to B$ be the canonical morphisms with kernel $\langle \langle R \rangle \rangle$ and $\langle \langle S \rangle \rangle$ respectively (i.e. $A = \langle X \mid R \rangle$ and $B = \langle Y \mid S \rangle$).

Let $\theta \colon F(X \cup Y) \to A * B$ be the morphism coinciding with $\phi$ on $X$ and with $\psi$ on $Y$. Obviously $\langle \langle R \cup S \rangle \rangle \subseteq \ker \theta$.

Let now $g = g_1 g_2 \cdots g_n \in \ker \theta$, where $g_i \in (F(X) \cup F(Y)) \backslash \{1\}$ and adjacent factors $g_i, g_{i+1}$ do not lie in the same group $F(X)$ or $F(Y)$. Use induction on the length $n$. Since $\theta(g_1) \theta(g_2) \cdots \theta(g_n) = 1$ in $A * B$ there exists $i$ such that $\theta(g_i) = 1$, i.e. $g_i \in \langle \langle R \rangle \rangle$ or $g_i \in \langle \langle S \rangle \rangle$. Thus $\theta(g_1 g_2 \cdots g_{i-1} g_{i+1} \cdots g_n) = 1$. Induction!

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