Existence and Uniqueness of Direct Limit of Sequence of Groups

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Theorem

Let $\sequence {G_n}_{n \mathop \in \N}$ be a sequence of groups.

Let $\sequence {g_n}_{n \mathop \in \N}: g_n: G_n \to G_{n + 1}$ be a sequence of group homomorphisms.


Then their direct limit $G_\infty$ exists and is unique up to unique isomorphism.


Proof


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.

$\Box$


Uniqueness

A direct proof of uniqueness can be formulated as follows:


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 $u_\infty$.


$\blacksquare$



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.

$\blacksquare$


Sources