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

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

Let $\struct {G_\infty, \cdot}$ be the algebraic structure defined as follows.

Let $\eqclass {\tuple {x_n, n} } {}, \eqclass {\tuple {y_m, m} } {} \in G_\infty$ be arbitrary elements of $G_\infty$.

Let $l := \max \set {m, n}$.

Let the operation $\cdot$ on $G_\infty$ be defined as:


 * $\tuple {\eqclass {\tuple {x_n, n} } {} \cdot \eqclass {\tuple {y_m, m} } {} } := \eqclass {\tuple {\map {g_{n l} } {x_n} \map {g_{m l} } {y_m}, l} } {}$

Then $\struct {G_\infty, \cdot}$ is a group.

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 product element is independent of this choice.

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

Let $l' := \max \set {n', m'}$.

, suppose that $l' \ge l$.

We have that:
 * $\tuple {x_n, n} \sim \tuple {x_{n'}, n'}$

and:
 * $\tuple {y_m, m} \sim \tuple {y_{m'}, m'}$

and so:
 * $\map {g_{n, l'} } {x_n} = \map {g_{n', l'} } {x_{n'} }$

and:
 * $\map {g_{m, l'} } {y_m} = \map {g_{m', l'} } {y_{m'} }$

Then we have, since all our maps are group homomorphisms:

that is:
 * $\map {g_{n, l} } {x_n} \map {g_{m,l} } {y_m} \sim \map {g_{n', l'} } {x_{n'} } \map {g_{m', l'} } {y_{m'} }$

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

Group Axioms
, by the definition of the group operation, we may assume that the representatives are always in the same group $G_l \in \sequence {G_n}_{n \mathop \in \N}$.

To see this we note that we always consider a finite collection of group elements
 * $\set {\eqclass {\tuple {x_{n_1}, {n_1} } } {}, \ldots, \eqclass {\tuple {x_{n_k}, {n_k} } } {} } \subset G_\infty$

Define $l:= \max \set {n_1, \ldots, n_k}$.

Then:
 * $\forall i \in \set {1, \ldots, k}: \map {g_{n_i, l} } {x_{n_i} } \in G_n \land \tuple {x_{n_1}, {n_1} } \sim \tuple {\map{ g_{n_i, l} } {x_{n_i} }, l}$

Let $\eqclass {\tuple{x_n, n} } {}, \eqclass {\tuple {y_m, m} } {}, \eqclass {\tuple {y_n, n} } {}, \eqclass {\tuple {z_n, n} } {} \in G_\infty$.

Then:

Let $\eqclass {\tuple {x_n, n} } {} \in G_\infty$ and let $1_n$ be the identity of $G_n$.

Note that
 * $\forall k, n \in \N : \paren {k \ge n \implies \map {g_{n k} } {1_n} = 1_k}$

because the maps $g_{n k}$ are group homomorphisms.

Then:

Similarly we also find that $\eqclass {\tuple {1_n, n} } {} \cdot \eqclass {\tuple {x_n, n} } {} = \eqclass {\tuple {x_n, n} } {}$.

Thus $\eqclass {\tuple {1_n, n} } {}$ is the identity of $G_\infty$.

Let $\eqclass {\tuple {x_n, n} } {} \in G_\infty$.

Then:

Similarly we also find that $\eqclass {\tuple {x^{-1}_n, n} } {} \cdot \eqclass {\tuple {x_n, n} } {} = \eqclass {\tuple {1_n, n} } {}$.

Thus $\eqclass {\tuple {x_n, n} } {}$ has an inverse, that is:
 * $\eqclass {\tuple {x_n^{-1}, n} } {}$

in $G_\infty$.