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

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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.



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:

$\tuple {x_n, n} \sim \tuple {x_{n'}, n'}$


$\tuple {y_m, m} \sim \tuple {y_{m'}, m'}$

and so:

$\map {g_{n, l'} } {x_n} = \map {g_{n', l'} } {x_{n'} }$


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

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.


Group Axioms

Without loss of generality, 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 $l:= \max \set {n_1, \ldots, n_k}$.


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

Group Axiom $\text G 1$: Associativity

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


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

Group Axiom $\text G 2$: Existence of Identity Element

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.


\(\ds \eqclass {\tuple {x_n, n} } {} \cdot \eqclass {\tuple {1_n, n} } {}\) \(=\) \(\ds \eqclass {\tuple {x_n 1_n, n} } {}\)
\(\ds \) \(=\) \(\ds \eqclass {\tuple {x_n, n} } {}\)

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 $G_\infty$.

Group Axiom $\text G 3$: Existence of Inverse Element

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


\(\ds \eqclass {\tuple {x_n, n} } {} \cdot \eqclass {\tuple {x_n^{-1}, n} } {}\) \(=\) \(\ds \eqclass {\tuple {x_n x^{-1}_n, n} } {}\)
\(\ds \) \(=\) \(\ds \eqclass {\tuple {1_n, n} } {}\)

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 $G_\infty$.