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,, 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 $G_\infty$.