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.

  • 3
    $\begingroup$ 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. $\endgroup$ – anon Oct 4 '14 at 21:31

When people say the normal subgroup generated by a set, they mean the smallest normal subgroup containing that set, namely, the intersection of all normal subgroups containing that set. This is well defined, since the intersection of normal subgroups is again a normal subgroup.

Equivalently, you can define the normal subgroup generated by $S$ to be the subgroup generated by all elements of the form $gsg^{-1}$, with $s\in S$, $g\in G$.

  1. A subgroup does not need to be normal.
  2. There is a subgroup containing all $f_1(h)f_2(h)^{-1}$
  3. There is a smallest such one (by intersection)
  4. There is a normal subgroup containing the smallest one, which we want to be minimal again
  5. Normal subgroup is nice, since we can divide it out and obtain for instance the amalgamated product
  • $\begingroup$ I guess I don't see how is $f_1(h)f_2(h)^{-1}$ a subgroup of $G_1 * G_2$. $\endgroup$ – Yuugi Oct 4 '14 at 21:09
  • $\begingroup$ Read again more carefully. There is one, e.g. the free product itself. $\endgroup$ – Daniel Valenzuela Oct 4 '14 at 21:15
  • $\begingroup$ @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 $\{f_1(h)f_2(h)^{-1}:h\in H\rangle$. $\endgroup$ – anon Oct 4 '14 at 21:38

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