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I have seen here that given two groups $G=\langle X|R \rangle := F(X)/N(R)$ and $H=\langle Y|S \rangle := F(Y)/N(S)$, then their semidirect product can be written as: $$ G\rtimes_\phi H \;=\; \langle X, Y \mid R,\,S,\,yxy^{-1}=\phi(y)(x)\text{ for all }x\in X\text{ and }y\in Y\rangle \tag{1} $$

I'm trying to prove this result with an epimorphism $\theta:F(G\times H)\rightarrow G\rtimes_\phi H$, where $\theta ((g,h))=(g,h)$ for all the "letters" $(g,h) \in G\times H$, and then considering the kernel: $$\ker(\theta)=\{w\in F(G\times H): \theta (w)=(1,1)\},$$ where $(1,1)$ is the identity element of $G\rtimes_\phi H$, and then $F(G\times H)/\ker(\theta) \cong G \rtimes_{\phi}H$. In $F(G\times H)$ we have a generic element: $$w=\prod_{i}(g_{i},h_{i})^{r_{i}}=\prod_{i}(g_{i}^{r_{i}},h_{i}^{r_{i}}),$$ which is transformed by $\theta$: $$\theta (w)=\prod_{i}{}^{'}(g_{i}^{r_{i}},h_{i}^{r_{i}})=\left(\prod_{i}\phi\left(\prod_{j=1}^{i-1}g_j{}^{r_{j}} \right)(h_{i}^{r_{i}}) , \prod_{i}h_{i}^{r_{i}} \right),$$ where the primed product means that it is the product of $G\rtimes_{\phi}H$. Here I'm stuck.

I don't know if the relations "$R,\,S,\,yxy^{-1}=\phi(y)(x)\text{ for all }x\in X\text{ and }y\in Y$" have something to do with $\ker(\theta)$. Also I don't know if this approach to proving (1) is correct, nor if there is a more "natural" way of viewing this. Unfortunately, I can't get the book "Presentations of Groups" by D.L. Johnson where it's said that this is proved.

Any help will be greatly appreciated!

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    $\begingroup$ If $F(G \times H)$ is really the free group on the elements of $G \times H$ then you do not have $(g_i,h_i)^{r_i} = (g_i^{r_i},h_i^{r_i})$. $\endgroup$ – Derek Holt Sep 22 '16 at 9:33

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