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:
\(\ds \map {g_{ l, l'} } {\map {g_{n, l} } {x_n} \map {g_{m, l} } {y_m} }\) | \(=\) | \(\ds \map {g_{ l, l'} } {\map {g_{n, l} } {x_n} } \map{ g_{ l,l' } } { \map{ g_{ m,l } } {y_m} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \map {g_{n, l'} } {x_n} \map {g_{m, l'} } {y_m}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \map{ g_{ n',l' } } { x_{ n' } } \map{ g_{m',l'} } { y_{ m' } }\) |
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:
\(\ds \paren{\eqclass{\tuple{x_n, n} }{} \cdot \eqclass{ \tuple{y_n, n} }{} } \cdot \eqclass{ \tuple{z_n, n} }{}\) | \(=\) | \(\ds \eqclass{ \tuple{ x_n y_n, n} }{} \cdot \eqclass{ \tuple{z_n, n} }{}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \eqclass{ \tuple{ \paren{x_n y_n}z_n, n} }{}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \eqclass{ \tuple{ x_n \paren{y_n z_n}, n} }{}\) | $G_n$ is a group | |||||||||||
\(\ds \) | \(=\) | \(\ds \eqclass{ \tuple{ x_n, n} }{} \cdot \eqclass{ \tuple{ y_n z_n, n} }{}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \eqclass{ \tuple{ x_n, n} }{} \cdot \paren{\eqclass{ \tuple{ y_n, n} }{} \cdot \eqclass{ \tuple{ z_n, n} }{} }\) |
$\text G 2$: 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:
\(\ds \eqclass{\tuple{x_n, n} }{} \cdot \eqclass{ \tuple{1_n, n} }{}\) | \(=\) | \(\ds \eqclass{\tuple{x_n 1_n, n} }{}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \eqclass{\tuple{x_n, n} }{}\) |
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 :
\(\ds \eqclass{\tuple{x_n, n} }{} \cdot \eqclass{ \tuple{x_n^{-1}, n} }{}\) | \(=\) | \(\ds \eqclass{\tuple{x_n x^{-1}_n, n} }{}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \eqclass{\tuple{1_n, n} }{}\) |
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 $G_\infty$.
$\blacksquare$