Quotient of amalgamated free product by normal subgroup

Question

I was given the following question in my homework:

Given $Γ =G_1∗_HG_2$, suppose that $H$ is a normal subgroup in both $G_1$ and $G_2$.

(a) Show that $H$ is then a normal subgroup in $Γ$.

(b) Show that the quotient group $Γ/H$ is canonically isomorphic to the free product $(G_1/H)∗(G_2/H)$.

I've already submitted my work, but I've noticed that the proof I gave for (b) was wrong. I tried to fix it today (just for the sake of learning) but didn't manage to get a complete solution. My attempt to solve it is as follows (suppose I've solved A correctly):

Using the universal property of $Γ$, and the natural homomorphisms $\phi_1, \phi_2$ from $G_1$ and $G_2$ to $(G_1/H)∗(G_2/H)$, (through the quotients), we get a unique homomorphoism $\theta:Γ\to (G_1/H)∗(G_2/H)$, where all the diagram commutes.

It's easy to see that $H \subseteq Ker(\theta)$ because for each $h \in H \subseteq G_1$ we get $\phi_1(h) = e$, so using the diagram, we get that $\theta(h) = e$ for $h$ as an element of $Γ$.

I'm also pretty sure that $\theta$ is surjective, because $(G_1/H)∗(G_2/H)$ is generated by $Im\phi_1 \cup Im\phi_2$, so using the fact that the diagram commutes, it must be generated by elements that are in the image of $\theta$.

To prove the claim, the only thing I have missing with this approach is to prove that $Ker(\theta)\subseteq H$, and I'll get the desired isomorphism, that $Γ/H \cong (G_1/H)∗(G_2/H)$.

My questions are:

1. Are there any mistakes or nonsense (wrong understanding of definition, etc..) in what I wrote so far?
2. Is that a good approach for this question? Is trying to build an isomorphism by hand an easier approach? How would you try to solve it?
3. If it's indeed a good approach - how do I proof that $Ker(\theta)\subseteq H$?

Answer 1

1. No, it's all correct.
2. Usually, to prove an isomorphism of the form $A/N\cong B$, the first isomorphism theorem seems indeed a good approach.
3. Assume $w=x_1y_1x_2y_2\dots x_ny_n\in \ker\theta\subseteq G_1 *_H G_2,\ x_i\in G_1,\,y_i\in G_2$. We want to prove $w\in H$.
   If any letter $x_i$ or $y_i$ is in $H$, then we can move it to the beginning of $w$, perhaps replacing the letters behind by other elements of the same $G_1$ or $G_2$, then we can ignore it.
   So, we are basically left with the case to consider when $\forall i: x_i,y_i\notin H$.
   But then, $\theta(w)=[x_1][y_1]\dots[x_n][y_n]$, as being an element of the free product, has no chance to have any cancellable substring, so it is definitely not the empty word.

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2. Nevertheless, you might find it easier to prove that $\Gamma/H$ satisfies the universal property of being the free product of $G_1/H$ and $G_2/H$.