For a finite multiplicative group G, show that the subset $H={g, g^2, ...} $ that contains all powers of g in G is a subgroup of G. It should be enough to show that the product of any two elements in this subset is also in the subset, but how can we use the finite order of G?  What if G had infinite order?
 A: It’s not enough to show that $H$ is closed under multiplication: you must also show that $H$ is closed under taking inverses. If $G$ is infinite, this need not be the case. Consider, for instance, the group $\Bbb Q^+$ of positive rational numbers under multiplication, and let $g=\frac12$. Then
$$H=\left\{\frac1{2^n}:n\in\Bbb Z^+\right\}=\left\{\frac12,\frac14,\frac18,\ldots\right\}$$
is closed under multiplication, but it does not contain the multiplicative inverse of any of its elements: those inverses are the integers $2^n$ for $n\in\Bbb Z^+$. It also doesn’t contain the identity.
When $G$ is finite, you know that every element has finite order, so some $g^n\in H$ will be the identity. Once you have that, it’s not hard to show that $H$ is closed under taking inverses.
A: When a group $G$ is finite, a nonempty subset $H$ is a subgroup if and only if, for $a,b\in H$, we have $ab\in H$.
In other words, just closure under products is sufficient (provided the subset is not empty, of course).
Here's the proof. First, let $c\in H$, which exists because $H$ is not empty. Then, by the assumption, the map
$$
f\colon H\to H,\qquad f(a)=ca
$$
is well defined and injective; indeed, $f(a)=f(b)$ implies $ca=cb$, so $c^{-1}ca=c^{-1}cb$ (we know $c$ has an inverse in $G$).
Finiteness of $G$ ensures the map is also surjective, so there is $e\in H$ such that $f(e)=c$. Therefore $ce=c$ and so $e=1\in H$.
Now $1\in H$, so, by surjectivity, there is $a\in H$ with $f(a)=1$, which means $ca=1$ and so $a=c^{-1}\in H$. Since $c$ was an arbitrary element of $H$, we have proved that the inverse of every element of $H$ belongs to $H$.
Final note: the above proof uses just the finiteness of $H$.
In your case, from $g^{m}g^{n}=g^{m+n}$, we deduce that $H$ is (not empty) and closed under products, so it is a subgroup. Note, however, that finiteness of $G$ is needed here, otherwise we could not state that $H$ is finite (it could not be, just think to the integers with respect to addition and $g=2$, or any other nonzero element).
A: In any group $G$ (not just a finite one) we have that if $g \in G$ and $k < m$ are two positive integers with $g^k = g^m$, then $g$ is of finite order.
The reason being that one can show $g^{m-k} = e$ (you can figure out how).
Now this doesn't show that the order of $g$ is $m-k$, but it does show that the set:
$S = \{g^n: g^n = e, n \in \Bbb N^+\}$
is non-empty, and thus has a smallest element (this conclusion is a consequence of the well-ordered property of the natural numbers).
That smallest element of the set $S$ listed above is indeed (by definition) the order of $g$.
By your definition, it is easy to see that $S \subseteq H$, and so to show that $e \in H$, what is needed is to show there are two elements of:
$\{g^k: k \in \Bbb N^+\}$ that coincide for distinct $k$.
(Note that saying $S \neq \emptyset$ is the same as saying $e \in H$).
If $G$ is infinite, this may not be the case-for example, we have have $G = \Bbb Z$ (under addition), with $g = 1$, and every "power" (multiple) of $1$ is distinct (this is how we build the natural numbers, by assuming $n+1 \neq n$, which then naturally leads to an infinite set).
If, however, $G$ is finite, Then the map $k \mapsto g^k$ cannot be one-to-one (why)? It follows a fortiori (a fancy phrase meaning "all the more") that $H \subseteq G$ is likewise finite- note that $H$ is the image of our map $k \mapsto g^k$.
I note in passing that the mapping $k \mapsto g^k$ is, for a $g$ in a group $G$ group, the basic example of a homomorphism; in this case, from $(\Bbb Z, +) \to (G, \ast)$. The image of this homomorphism is the subgroup $\langle g\rangle$. This (class of) homomorphism(s) turns out to be extremely important, especially in the study of abelian groups, allowing us to "transfer" things we learned long ago about integers to abelian groups, which proves very handy.
A: Consider the sequence $H = \{ g^i, i \ge 1 \}$, of positive powers of any element $g \in G$; it is clearly closed under the group operation:  for any positive integers $m, n$, $g^m g^n = g^{n + n} \in H$.  Since $G$ is finite, $H \subset G$ is also finite, and so the sequence $g^i$ must begin to repeat itself at some point; i.e., there exist $k, l \in \Bbb N$, the natural numbers, with $l > k$ and
$g^l = g^k; \tag{1}$
then
$g^{l - k} = g^l g^{-k} = g^k g^{-k} = e, \tag{2}$
$e$ being the identity element of $G$.  We may take $k$ to be the smallest positive integer such that there exists $l > k$ with (1) binding, and $l$ as well to be the least such that (1) holds.  
Since $l > k$,
$l - k \ge 1, \tag{3}$
and (2) shows that $e$ is in fact a positive power if $g$; thus
$e \in H. \tag{4}$
Furthermore, from (2) we have
$gg^{l - k - 1} = g^{l - k} = e; \tag {5}$
thus
$g^{-1} = g^{l - k - 1} \in H; \tag{6}$
since
$g^{-j} = (g^{-1})^j = (g^{l - k - 1})^j$
$= g^{(l - k - 1)j}, \tag{7}$
we infer, if in fact $l - k > 1$ so that $l - k -1$ is positive, that
$g^{-j}$ may be expressed as a positive power of $g$; thus
$g^{-j} \in H, \tag{8}$
and $H$ contains the inverse of each of its elements; being closed under the group operation, containing $e$ as well, it is a subgroup of $G$.  In the event that $l - k = 1$, we have via (2) that $g = e$; in this case $e = e^{-1} \in H = \{ e \}$, and there is not a lot to discuss: $\{ e \}$ is manifestly a subgroup (of $G$).
We thus see that for finite $G$, $H = \{ g^i, i \ge 1 \}$ is in fact a subgroup.  QED.
Finally, this result no longer holds if $G$ is infinite; as a counterexample, consider $G$ to be the additive group of integers $\Bbb Z$.  Taking $g = 2 \in \Bbb Z$, we have
$H = \{ 2, 4, 6, 8, . . . \}, \tag{9}$
i.e. $H$ consists of the positive even whole numbers; $0 \notin H$, and $-m \notin H$ for any $m \in H$; in this case, $H$ is not a subgroup.  Similar examples abound: set $g = m$ for any integer and analogous remarks apply.
