Union of an infinite number of subgroups Let $H_{1}, H_{2}$ be subgroups of $G$ satisfying, $H_{1}\leq H_{2}\leq G$. It is easy to prove that $\bigcup_{i=1}^{2}H_{i}$ is a subgroup of G. Consider an infinite ascending chain of subgroups of G, i.e. $H_{1}\leq H_{2}\leq \cdots$. 
I know that by fixing an $n\in \mathbb{N}$ we can show that the union of the first $n$ subgroups is again a subgroup of $G$.
Is it meaningful to talk about the union of such groups? And is it possible to extend this to the infinite case?
 A: Yes. The union of an infinite chain of subgroups is again a subgroup. This also works for subrings, subspaces, in fact, most every sub-thing.
Suppose $H_i$ is a subgroup of $G$ for each $i=1,2,\dots$ and in addition suppose that $H_i \subseteq H_{i+1}$ for each $i$. 
Let $H=\cup_{i=1}^\infty H_i$. This is certainly a nonempty subset of $G$. Next, let $a,b \in H$. This means $a \in H_i$ for some $i$ and $b \in H_j$ for some $j$. Either $i<j$ and so $H_i \subseteq H_j$ and thus both $a$ and $b$ are in $H_j$ or vice-versa: $j<i$ and both elements are in $H_i$. Suppose both are in $j$. Then so is $ab$ and $a^{-1}$ since $H_j$ is a subgroup. Therefore, $ab,a^{-1} \in H_j \subseteq \cup_{k=1}^\infty H_k=H$. Therefore, $H$ is a subgroup of $G$.
It's easy to see why this type of argument should work in many different contexts. Also, note that the index set "$1,2,\dots$" can be replaced with any totally ordered set or partially ordered sets such that for all $i,j$ there is $k$ such that $i,j \leq k$.
Here's an example: We can stick $S_n$ inside $S_m$ for any $n<m$. All of these symmetric groups can be viewed as subgroups of $S_\infty$ (permutations of $\mathbb{Z}_{>0}={n \in \mathbb{Z} \,|\, n \geq 0 }$). The subgroup we get $S = \cup_{i=1}^\infty S_i$ is the set of all permutations fixing all but finitely many positive integers.
A: This is an example of a rather general phenomenon. Let $G$ be any group, and let $\mathscr{H}$ be a chain of subgroups of $G$, meaning that for any $H_0,H_1\in\mathscr{H}$, either $H_0\subseteq H_1$ or $H_1\subseteq H_0$. Then $H=\bigcup\mathscr{H}$ is a subgroup of $G$.
Certainly $H\subseteq G$. Now suppose that $h_0,h_1\in H$; to show that $H \le G$, we must show that $h_0h_1,h_0^{-1}\in H$. Since $h_0\in H$, there must be some $H_0\in\mathscr{H}$ such that $h_0\in H_0$. $H_0$ is a subgroup of $G$, so $h_0^{-1}\in H_0 \subseteq H$, as desired. Similarly, there must be some $H_1\in\mathscr{H}$ such that $h_1\in H_1$. Since $\mathscr{H}$ is a chain, either $H_0 \subseteq H_1$ or $H_1 \subseteq H_0$. Without loss of generality $H_0 \subseteq H_1$, in which case $h_0,h_1\in H_1$. But $H_1$ is a subgroup of $G$, so $h_0h_1\in H_1 \subseteq H$, and the proof that $H\le G$ is complete.
Notice that what made this work is that each of the conditions that had to be checked involved only finitely many elements of $H$. Specifically, they were closure conditions of the following form:
$$\text{If }h_0,\dots,h_n \in H,\text{ then }\phi(h_0,\dots,h_n)\in H.\tag{1}$$ 
Suppose that $S$ is any structure with the property that a subset $H$ of $S$ is a substructure iff it satisfies a list of condition of the form $(1)$. Then if $\mathscr{H}$ is a chain of substructures of $S$, and $H = \bigcup\mathscr{H}$, the same kind of argument that I used for the group $G$ will show that $H$ is a substructure of $S$:

If $h_0,\dots,h_n\in H$, there are substructures $H_0,\dots,H_n \in \mathscr{H}$ such that $h_k\in H_k$ for $k=0,\dots,n$. The $H_k$ with $k=0,\dots,n$ are linearly ordered by $\subseteq$, so one of them contains all the rest. Let that one be $H_i$. Then $h_0,\dots,h_n\in H_i$, and since $H_i$ is a substructure of $S$, it’s closed under the function $\phi$, and hence $\phi(h_0,\dots,h_n)\in H_i \subseteq H$.

A: This is possible, and the way to do it is made precise using the notion of a colimit (more specifically, a direct limit).  This is a general categorical construction, and gives you a way to take a diagram of groups, and spit out another group in a meaningful way.
In your case, the diagram we are looking at is $H_1\to H_2\to H_3\to\cdots$, where the arrows are the inclusion maps.  The colimit of this diagram will be a group $G$ such that we have maps $H_i\to G$ for all $i$ such that the big diagram containing our original diagram of $H_i$'s together with all the maps into $G$ commutes.  It is universal in the sense that if we have any other set of maps $H_i\to K$ such that the same big diagram commutes, then there is a unique map $G\to K$ making everything commute.
In the category of sets, if we have inclusions $X_1\subseteq X_2\subseteq X_3\subseteq\cdots$, then the colimit of this diagram is $\bigcup_{i=1}^\infty X_i$.  It is worth noting that if each of your groups $H_i$ is a subgroup of some group $G$, then as a set, this colimit will just be the union.
If you want an example of such a colimit where all the groups don't live inside some larger group, consider the inclusions 
$$\mathrm{GL}_1(F)\to\mathrm{GL}_2(F)\to\mathrm{GL}_3(F)\to\cdots$$
where the arrows are inclusions are into the upper left hand corner.  The colimit of this diagram is $\mathrm{GL}_\infty(F)$, and appears in algebraic topology.
A: More generally,  let $I$ be an index set of arbitrary non-zero cardinality, and suppose that for every $\alpha\in I$, $H_\alpha$ is a subgroup of $G$. 
Suppose also that for any $\alpha, \beta \in I$, we have $H_\alpha \subseteq H_\beta\,$ or $\,H_\beta \subseteq H_\alpha$.  Then $\bigcup_\alpha H_\alpha$ is a subgroup of $G$.  
The question that was asked has $I=\mathbb{N}$, and $H_m \subseteq H_n$ if $m \le n$.
The proof is straightforward. We first show that if $u$ and $v$ are in $\bigcup_\alpha H_\alpha$, then $uv\in \bigcup_\alpha H_\alpha$.
Since $u \in \bigcup_\alpha H_\alpha$, there is an index $\beta$ such that $u \in H_\beta$. Similarly, $v \in H_\gamma$ for some $\gamma\in I$.  By our condition on the $H_\alpha$, we have $H_\beta \subseteq H_\gamma$, or $H_\gamma \subseteq H_\beta$.  Without loss of generality we may assume that $H_\beta \subseteq H_\gamma$.  Then $u$ and $v$ are both in $H_\gamma$, and therefore so is their product $uv$. It follows that $uv \in \bigcup_\alpha H_\alpha$.
The other things we need to check are easier. The unit element (of $G$) is in all the $H_\alpha$, so it is in their union. Also, if $x \in \bigcup_\alpha H_\alpha$, then $x \in H_\beta$ for some $\beta$, so $x^{-1} \in H_\beta$, and therefore $x^{-1} \in  \bigcup_\alpha H_\alpha$. 
Comment: The condition that for all $\alpha, \beta \in I$, we have $H_\alpha \subseteq H_\beta\,$ or $\,H_\beta \subseteq H_\alpha$ can be weakened. Everything goes through if for all  $\alpha, \beta \in I$, there is a $\gamma \in I$ such that  $H_\alpha \subseteq H_\gamma\,$ and $\,H_\beta \subseteq H_\gamma$. 
