# 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.

## Proof

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

Without loss of generality, 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:

 $\displaystyle \map {g_{ l, l'} } {\map {g_{n, l} } {x_n} \map {g_{m, l} } {y_m} }$ $=$ $\displaystyle \map {g_{ l, l'} } {\map {g_{n, l} } {x_n} } \map{ g_{ l,l' } } { \map{ g_{ m,l } } {y_m} }$ $\displaystyle$ $=$ $\displaystyle \map {g_{n, l'} } {x_n} \map {g_{m, l'} } {y_m}$ $\displaystyle$ $=$ $\displaystyle \map{ g_{ n',l' } } { x_{ n' } } \map{ g_{m',l'} } { y_{ m' } }$

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.

$\Box$

### Group Axioms

By the definition of the group operation, we may assume, without loss of generality, 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

$\{ \eqclass{ \tuple{ x_{n_1}, {n_1} } }{}, \dots, \eqclass{ \tuple{ x_{n_k}, {n_k} } }{} \} \subset G_\infty$.

Define $l:= \max\{n_1,\dots, n_k\}$. Then

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

#### $\text G 1$: Associativity

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:

 $\displaystyle \paren{\eqclass{\tuple{x_n, n} }{} \cdot \eqclass{ \tuple{y_n, n} }{} } \cdot \eqclass{ \tuple{z_n, n} }{}$ $=$ $\displaystyle \eqclass{ \tuple{ x_n y_n, n} }{} \cdot \eqclass{ \tuple{z_n, n} }{}$ $\displaystyle$ $=$ $\displaystyle \eqclass{ \tuple{ \paren{x_n y_n}z_n, n} }{}$ $\displaystyle$ $=$ $\displaystyle \eqclass{ \tuple{ x_n \paren{y_n z_n}, n} }{}$ $G_n$ is a group $\displaystyle$ $=$ $\displaystyle \eqclass{ \tuple{ x_n, n} }{} \cdot \eqclass{ \tuple{ y_n z_n, n} }{}$ $\displaystyle$ $=$ $\displaystyle \eqclass{ \tuple{ x_n, n} }{} \cdot \paren{\eqclass{ \tuple{ y_n, n} }{} \cdot \eqclass{ \tuple{ z_n, n} }{} }$

#### $\text G 2$: Identity

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

Note that

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

because the maps $g_{nk}$ are group homomorphisms.

Then:

 $\displaystyle \eqclass{\tuple{x_n, n} }{} \cdot \eqclass{ \tuple{1_n, n} }{}$ $=$ $\displaystyle \eqclass{\tuple{x_n 1_n, n} }{}$ $\displaystyle$ $=$ $\displaystyle \eqclass{\tuple{x_n, n} }{}$

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 $G_\infty$.

#### $\text G 3$: Inverses

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

Then :

 $\displaystyle \eqclass{\tuple{x_n, n} }{} \cdot \eqclass{ \tuple{x_n^{-1}, n} }{}$ $=$ $\displaystyle \eqclass{\tuple{x_n x^{-1}_n, n} }{}$ $\displaystyle$ $=$ $\displaystyle \eqclass{\tuple{1_n, n} }{}$

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 $G_\infty$.

$\blacksquare$