Validity of Using Induction to Show Union of an Infinite Ascending Chain of Subgroups is a Subgroup Can this be done by induction instead of just proving the subgroup criterion?  I can prove using the essentials tools of group theory, but looking at the problem, I was wondering if we can simply use an induction argument.

Given we have a chain of ascending subgroups of a group $G$... 
  $$H_1\le H_2\le....$$
  Is the union
  $$\bigcup_{i=1}^\infty{H_i}\le G$$

For the first case, we have that $H_1\le H_2$ and thus, $H_1\subseteq H_2$ means that in terms of their union, 
$$\bigcup_{i=1}^2{H_i}=H_2\le G$$
Thus, for an arbitrary $n>2$ we can assume that 
$$\bigcup_{i=1}^n{H_i}=H_n\le G$$
Then 
$$\bigcup_{i=1}^{n+1}{H_i}=\left(\bigcup_{i=1}^{n}{H_i}\right)\cup H_{n+1}=H_n\cup H_{n+1}=H_{n+1}\le G$$
Thus, for any whole value of $n$, the claim is true.
My worry is the infinite upper bound.  But I feel like induction takes care of that.
 A: Induction (the conventional way) only takes care of the "for any whole value of $n$" part. It can never let you conclude anything for the entire infinite union. There is such a thing as transfinite induction, but even then the induction step is split into separate steps for the cases which have an immediate predecessor and cases which don't. The infinite union doesn't have an immediate predecessor, and thus the standard induction step will never reach it.
However, clearly the union is a subset, and clearly that subset contains the identity element, so we're a good way of the way there. When showing that it has inverses and that it's closed under products, it pays to know that any finite collection of elements in the union is contained in some finite union as well. And you already know that any of the finite unions is a group.
A: Of course $1\in\bigcup_{i=1}^{\infty}H_i$ and for any $g\in H_i$,$h\in H_j$ say without loss of generality that $i\leq j$ then $g\in H_j$ too and since this is a subgroup You have 
$$gh\in H_j\subset \bigcup_{i=1}^{\infty}H_i$$
and thus $\bigcup_{i=1}^{\infty}H_i$ is a subgroup. That's all. You don't argue via induction!
$\textbf{Edit:}$ what was missing, for any $g\in\bigcup_{i=1}^{\infty}H_i$ there exists an $i$ so that $g\in H_i$, actually this implies $g\in H_j$ for all $i\leq j$, but we don't need this here. Since $H_i$ is a subgroup $g^{-1}\in H_i\subseteq \bigcup_{i=1}^{\infty}H_i$.
A: (I should have read the question more carefully. What follows is a proof by the method the OP excluded.)

Since $e\in H_1$, we have $$e\in H_1\subseteq\bigcup_{i=1}^{\infty}H_i=:\mathcal{U}.$$ Hence $\mathcal{U}$ is nonempty.
Let $g, h\in\mathcal{U}$. Then $g\in H_k$ and $h\in H_\ell$ for some $k, \ell$. Assume w.l.o.g. that $k<\ell$. Then the ascending chain condition gives that $g\in H_\ell$. Since $h\in H_\ell$ and $H_\ell$ is a subgroup of $G$, $h^{-1}\in H_\ell$. Thus $gh^{-1}\in H_\ell$. Hence $gh^{-1}\in \mathcal{U}$.
Hence, by the one-step subgroup lemma, we have 
$$\mathcal{U}\le G.$$
