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Let $H,G,G'$ be groups such that $H$ injects into both $G$ and $G'$. Then we may form the amalgamated product $G\ast_H G'$. Is the canonical map $G\rightarrow G\ast_H G'$ always injective?

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up vote 3 down vote accepted

Yes. The free product with amalgamation contains isomorphic copies of $G$ and $G'$ which intersect at $H$, via the canonical embeddings (which are induced by the canonical embeddings into the free product $G*G'$, followed by the quotient map modulo the normal subgroup generated by the elements that identify the images of $H$ in both $G$ and $G'$).

You can also show this by determining the normal form and multiplication rules for $G*_HG'$: let $\{g_{\alpha}\}_{\alpha\in A}$ and $\{g'_{\beta}\}_{\beta\in B}$ be sets of left coset representatives for $H$ in $G$ and $G'$, respectively. Then every element of $G*_HG'$ can be written uniquely as a word that alternates $g_{\alpha}$s and $g_{\beta}$s, followed by an element of $H$. Multiplication is by juxtaposition followed by rewriting, "pushing" the $h$ that is now in the middle of the word all the way to the right by rewriting products $xg_{\alpha}$ or $xg'_{\beta}$, with $x\in H$, into products of the form $g_{\alpha'}y$ or $g'_{\beta'}y$ with $y\in H$. The canonical embeddings send $g\in G$ to $g_{\alpha}h$, where $g_{\alpha}$ is the unique coset representative such that $g\equiv_H g_{\alpha}$ and $x=g_{\alpha}h$; and similarly for $g'\in G'$.

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