Let $G$ be a group and $G^i$ a collection of subgroups which form a direct system over a directed set $I$, so $i\leq j \iff \exists\; \varphi^i_j: G^i\to G^j$ where $\varphi^i_j$ is the inclusion map.
How do I show that the direct limit $\varinjlim G^i$ is the same as the union $\bigcup G^i$?
I think I need to use the universal property of the direct limit but I'm not sure how.
Any help will be much appreciated!
Let $\tau_i:G^i \to K$ be homomorphisms to a group $K$ that are compatible with the inclusion maps $\varphi^i_j$. This compatibility implies that if $g$ is in two of the groups $G^i$ and $G^j$, then $\tau_i(g) = \tau_j(g)$.
So, for any $g \in \cup G^i$, by defining $\tau(g) = \tau_i(g)$ if $g_i \in G$, we get a well-defined map $\tau:\cup G^i \to K$ that restricts to $\tau_i$ on $G^i$ for each $i \in I$. Also, for any $g, h \in \cup G^i$, there exists an $i$ with we have $g,h \in G^i$, so $\tau(gh)=\tau(g)\tau(h)$ and $\tau$ is a homomorphism.
We have now verified that $\cup G^i$ satisfies the Universal Property for direct limits, so the direct limit is isomorphic to $\cup G^i$.