N is a normal subgroup in $G*H$ generated by $G$ then $H\cong G*H/N$

Question

Let $N$ be a normal subgroup in $G*H$ generated by G. Then $H\cong G*H/N$

$G*H$ is free product of $G$ and $H$

By universal property , there exists a surjective homomorphism $\epsilon:G*H\rightarrow H$ with respect to trivial map $\phi:G\rightarrow H$ and identity map $\psi:H\rightarrow H$.

Since $\epsilon(G)$={$e$} and $N=<<G>>$ (the normal subgroup generated by G), there is a homomorphism $\bar\epsilon:G*H/N\rightarrow H$

Then how can I get the injection?

Answer 1

You can write an element of $G*H$

$x=g_1x_1...g_nx_n$, $x=g_1x_1..g_nx_ng_{n+1}$, $x_1g_1...x_ng_n$, $x_1g_1..x_ng_nx_{n+1}$.

Suppose that $\phi(x)=1$. We consider the case $x=x_1g_1...g_nx_n$, $\phi(x)=1$ implies $x_1...x_n=1$.

We can write $x=x_1g_1{x_1}^{-1}(x_1x_2)g_2(x_1x_2)^{-1}(x_1x_2x_3)g_3(x_1x_2x_3)^{-1}...(x_1...x_{n-1})g_{n-1}(x_1..x_{n-1})^{-1}(x_1...x_n)g_n=x_1g_1{x_1}^{-1}(x_1x_2)g_2(x_1x_2)^{1}(x_1x_2x_3)g_3...(x_1...x_{n-1})g_{n-1}(x_1..x_{n-1})^{-1}g_n$

which is $N$

Use a similar approach for the other cases.

Answer 2

If one thinks a bit about it, the statement should be trivial to prove: The free product $G\star H$ is the universal group equipped with two morphisms into it from $G$ and from $H$. If one divides out $G\star H$ by the smallest normal subgroup $N$ containing $G=G\star\{1\}$, then this quotient group should be the universal group $K$ equipped with two morphisms into it from $G$ and from $H$, the one from $G$ being the trivial morphism. The group $H$, equipped with the trivial morphism from $G$ and the identity from $H$ shares the same universal property. It follows that $K$ is isomorphic to $H$.

Here are the arguments in more detail: let $K=(G\star H)/N$, and let $g\colon G\rightarrow K$ and $h\colon H\rightarrow K$ be the natural morphisms. More precisely, denoting by $i$ and $j$ the inclusion of $G$ and $H$ into $G\star H$, respectively, and by $\pi$ the quotient morphism from $G\star H$ to $K$, one has $g=\pi\circ i$ and $h=\pi\circ j$. Note that $g$ is the trivial morphism since $N$ contains $G\star\{1\}=\mathrm{im}(i)$. Let us prove that $K$ is universal. Let $L$ be a group and let $g'\colon G\rightarrow L$ and $h'\colon H\rightarrow L$ be morphisms with $g'$ trivial. Then, by the universal property of the free product, there is a morphism $f\colon G\star H\rightarrow L$ such that $f\circ i=g'$ and $f\circ j= h'$. Since $g'$ is trivial, $f(\mathrm{im}(i))=\{1\}$ and $f$ induces a morphism $k\colon K\rightarrow L$ such that $k\circ\pi=f$. It follows that there is a morphism $k\colon K\rightarrow L$ such that $k\circ\pi=f$. Hence $k\circ g=k\circ\pi\circ i=f\circ i=g'$ and similarly $k\circ h=h'$. This morphism $k$ is unique. Indeed, suppose that there is another morphism $k'\colon K\rightarrow L$ such that $k'\circ g=g'$ and $k'\circ h=h'$. Then, $(k'\circ\pi)\circ i=g'$ and $(k'\circ\pi)\circ j=h'$. By unicity of $f$, one has $k'\circ\pi=f$. By unicity of $k$ as morphism induced by $f$, one has $k'=k$. This proves the universal property of the group $K$.

It is clear that the group $H$, endowed with the trivial morphism $t$ from $G$ and the identity morphism $\iota$ from $H$, shares the same universal property: Let $L$ be a group and let $g'\colon G\rightarrow L$ and $h'\colon H\rightarrow L$ be morphisms with $g'$ trivial. Then there is a unique morphism $k$ from $H$ to $L$ such that $k\circ t=g'$ and $k\circ\iota=h'$, namely $k=h'$.

It follows that $K$ is isomorphic to $H$ by the usual argument one evokes if two objects satisfy the same universal property.