# How does $f_1(h)f_2(h)^{-1}$ generate a normal subgroup of $G_1 * G_2$?

This question is about a normal subgroup of a free product.

Suppose we have $$G_1$$ and $$G_2$$ groups and let $$f_1: H \to G_1$$ $$f_2: H \to G_2$$ be group homomorphisms.

The amalgamated product $$G_1 *_{H} G_2$$ is defined as follows: Let $$N$$ be the normal subgroup of $$G_1 * G_2$$, the free product, that is generated by $$f_1(h)f_2(h)^{-1}$$ for all $$h \in H$$.

Please would someone explain how $$f_1(h)f_2(h)^{-1}$$ generates a normal subgroup of $$G_1 * G_2$$?

This fact is just stated without any explanation or motivation.

• It's not saying the subgroup they generate is normal. Terminology: The subgroup generated by a subset is the unique subgroup minimal among those containing that subset; similarly, the normal subgroup generated by a subset is the unique subgroup which is minimal among normal subgroups containing that subset. – anon Oct 4 '14 at 21:31

Equivalently, you can define the normal subgroup generated by $S$ to be the subgroup generated by all elements of the form $gsg^{-1}$, with $s\in S$, $g\in G$.
2. There is a subgroup containing all $f_1(h)f_2(h)^{-1}$
• I guess I don't see how is $f_1(h)f_2(h)^{-1}$ a subgroup of $G_1 * G_2$. – Yuugi Oct 4 '14 at 21:09
• @ggfgfg First off, $f_1(h)f_2(h)^{-1}$ is an element (depending on $h$). What you really need to be talking about is the subset $\{f_1(h)f_2(h)^{-1}:h\in H\}$. But that is not a subgroup, and nobody claims it is. It does generate a subgroup, $\langle f_1(h)f_2(h)^{-1}:h\in H\rangle$, but that is not guaranteed to be normal, and again nobody claims it is. Rather, we are constructing "the normal subgroup generated by" the subset $\{f_1(h)f_2(h)^{-1}:h\in H\rangle$. – anon Oct 4 '14 at 21:38