Existence and Uniqueness of Direct Limit of Sequence of Groups/Lemma 2

Lemma
The following defines a group structure on $G_\infty$:

Let $\eqclass{\tuple{x_n,n}}{}, \eqclass{\tuple{y_m,m}}{} \in G_\infty$ and let $l:= \max\{m,n\}$.

Define the group operation on $G_\infty$ via:
 * $\tuple{ \eqclass{\tuple{x_n,n}}{}, \eqclass{\tuple{y_m,m}}{} } \longmapsto \eqclass{\tuple{g_{nl}(x_n)g_{ml}(y_m),l}}{}$.

Well-Definedness
The definition depends on the choice $\tuple{x_n,n}$ and $\tuple{y_m,m}$ of representatives of $\eqclass{\tuple{x_n,n}}{}$ and $\eqclass{\tuple{y_m,m}}{} $.

We have to show that the multiplication is independent of this choice.

Let $\tuple{x_{n'},n'}$ and $\tuple{y_{m'},m'}$ be different representatives of the chosen equivalence classes.

Let $l' := \max\{n',m'\}$ and suppose w.l.o.g. that $l' \geqslant l$. First note that $g_{n,l'}(x_n) = g_{n',l'}(x_{n'})$, since $(x_{n},n) \sim (x_{n'},n')$ and analogously for $y$. Then we have, since all our maps are group homomorphisms:

i.e. $\map{ g_{n,l} } { x_n } \map{ g_{m,l} } { y_m } \sim \map{ g_{ n',l' } } { x_{n'} } \map{ g_{ m',l' } } { y_{m'} }$.

This proves that our definition is independent of the choice of representative.

Group Axioms
Since multiplication reduces to multiplication in one of the groups $G_l \in (G_n)_{n \in\mathbb{N}}$, the group axioms follow directly from those on $G_l$.