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 $l' := \max \set {n', m'}$. , 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: {{begin-eqn}} {{eqn | l = \map {g_{ l, l'} } {\map {g_{n, l} } {x_n} \map {g_{m, l} } {y_m} }       | r = \map {g_{ l, l'} } {\map {g_{n, l} } {x_n} } \map{ g_{ l,l' } } { \map{ g_{ m,l } } {y_m} }       | c = }} {{eqn | r = \map {g_{n, l'} } {x_n} \map {g_{m, l'} } {y_m}        | c =  }} {{eqn | r = \map{ g_{ n',l' } } { x_{ n' } } \map{ g_{m',l'} } { y_{ m' } }       | c = }} 		 {{end-eqn}}	 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. {{qed|lemma}} === 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:  {{begin-eqn}} {{eqn | l = \paren{\eqclass{\tuple{x_n, n} }{} \cdot \eqclass{ \tuple{y_n, n} }{} } \cdot \eqclass{ \tuple{z_n, n} }{}       | r = \eqclass{ \tuple{ x_n y_n, n} }{} \cdot \eqclass{ \tuple{z_n, n} }{}          | c =  }} {{eqn | r = \eqclass{ \tuple{ \paren{x_n y_n}z_n, n} }{}       | c =  }} {{eqn | r = \eqclass{ \tuple{ x_n \paren{y_n z_n}, n} }{}       | c = $G_n$ is a group }} {{eqn | r =\eqclass{ \tuple{ x_n, n} }{} \cdot \eqclass{ \tuple{ y_n z_n, n} }{}        | c =  }} {{eqn | r = \eqclass{ \tuple{ x_n, n} }{} \cdot \paren{\eqclass{ \tuple{ y_n, n} }{} \cdot \eqclass{ \tuple{ z_n, n} }{} }       | c =  }} {{end-eqn}} ==== $\text G 2$: 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:  {{begin-eqn}} {{eqn | l = \eqclass{\tuple{x_n, n} }{} \cdot \eqclass{ \tuple{1_n, n} }{}        | r = \eqclass{\tuple{x_n 1_n, n} }{}        | c =  }} {{eqn | r = \eqclass{\tuple{x_n, n} }{}       | c =  }} {{end-eqn}} 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 : {{begin-eqn}} {{eqn | l = \eqclass{\tuple{x_n, n} }{} \cdot \eqclass{ \tuple{x_n^{-1}, n} }{}        | r = \eqclass{\tuple{x_n x^{-1}_n, n} }{}        | c =  }} {{eqn | r = \eqclass{\tuple{1_n, n} }{}       | c =  }} {{end-eqn}} 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 $G_\infty$.