{{refactor|Transclude the lemmata}} === Existence === Define $\widehat G_\infty$ by: :$\widehat G_\infty := \displaystyle \bigsqcup_{n \mathop \in \N} G_n$ where $\displaystyle \bigsqcup$ signifies a disjoint union. For $m \ge n$, denote with $g_{n, m}$ the composition $g_{m - 1} \circ \cdots \circ g_n: G_n \to G_m$. In particular, $g_{n, n}: G_n \to G_n$ is taken to be the identity mapping on $G_n$. Now define an equivalence relation $\sim$ on $\hat G_\infty$ by putting: :$\tuple {x_n, n} \sim \tuple {y_m, m} \iff \exists k \ge n, m: \map {g_{n, k} } {x_n} = \map {g_{m, k} } {y_m}$ which is established to be indeed an equivalence relation in Lemma 1. Now define $G_\infty$ as the quotient set: :$G_\infty := \widehat G_\infty / \sim$ We can equip $G_\infty$ with a group structure as follows: Let $\eqclass {\tuple {x_n, n} } {}, \eqclass {\tuple {y_m, m} } {} \in G_\infty$, where $\eqclass {\tuple {x_n, n} } {}$ denotes the equivalence class of $\tuple{x_n,n}$. Let $l:= \max \set {m, n}$. We define a group operation on $G_\infty$ via  :$\tuple {\eqclass {\tuple {x_n, n} } {}, \eqclass {\tuple {y_m, m} } {} } \longmapsto \eqclass {\tuple {\map {g_{n l} } {x_n} \map {g_{m l} } {y_m}, l} } {}$. We shall also refer to this operation as multiplication. A proof that this is indeed a well-defined group can be found in Lemma 2. Now define $u_n: G_n \to G_\infty$ by: :$\map {u_n} {x_n} := \eqclass {\tuple {x_n, n} } {}$ That this is a group homomorphism follows immediately from the definition of the group structure. Now let $H$ be a group and $h_n: G_n \to H$ be group homomorphisms such that $h_{n + 1} \circ g_n = h_n$. Define $h_\infty: G_\infty \to H$ by: :$\map {h_\infty} {\eqclass {\tuple {x_n, n} } {} } = \map {h_n} {x_n}$ Under this definition: :$\forall x_n \in G_n: \map {h_\infty \circ u_n} {x_n} = \map {h_n} {x_n}$ By Equality of Mappings, it follows that: :$h_\infty \circ u_n = h_n$ It remains to be shown that $h_\infty$ is well-defined and a group homomorphism. This is done in Lemma 3. {{qed|lemma}} === Uniqueness === A direct proof of uniqueness can be formulated as follows: {{tidy}} Let $G_\infty'$ be a second direct limit group  with homomorphisms $u_n': G_n \rightarrow G_\infty'$.  By the above universal property we can construct homomorphisms $u'_{\infty} : G_{\infty} \rightarrow G_{\infty}'$ (taking $H = G_{\infty}'$ and $h'_n = u'_n$) and $u_{\infty} : G'_{\infty} \rightarrow G_{\infty}$ (taking $H = G_{\infty}$ and $h_n = u_n$).  We shall prove that $u'_\infty$ is a left inverse of $u_{\infty}$. Showing that it is also a right inverse follows analogously by exchanging $u_{\infty}$ and $u'_{\infty}$.  From  :$u'_{\infty} \circ u_n = u_n' = \mathop{Id}_{G'_\infty} \circ u'_n$ we find that the map $h_\infty : G'_\infty \to G'_\infty$ corresponding to $H = G'_\infty$ and $h_n = u'_{\infty} \circ u_n$ must be $\mathop{Id}_{G'_\infty}$ by uniqueness. We also have :$u'_{\infty} \circ u_n  = u'_{\infty} \circ \mathop{Id}_{G_\infty}  \circ u_n $ thus by the same argument $h_\infty$ must also equal $u_\infty'\circ u_\infty$.  We conclude that $u_\infty'$ is a right inverse of $u_\infty$. {{qed}} {{refactor|Extract the below and implement it as a second proof on its own page.}} For a more categorical point of view: From Equivalence of Definitions of Direct Limit of Sequence of Groups, $G_\infty$ can be described as a colimit in the category of groups $\mathbf{Grp}$. By Colimit is Unique, it therefore is unique up to unique $\mathbf{Grp}$-isomorphism. By Isomorphism in Category of Groups iff Group Isomorphism, the result follows. {{qed}}