Cyclic subgroup of a group: Proof If G is a group and $a \in G$, then $\langle a \rangle = \{a^n: n \geq 0\}$ is a subgroup of G.
To prove this, one must prove that there exists the inverse of element $a^n$. Now, if we say that $a^0 = 1$ and say that $1$ is the identity, then $a^n * (a^n)^{-1}=1$. But we know from first principles that for elements $g_1, g_2 \in G$, $(g_1 * g_2)^{-1}=g_2^{-1} * g_1^{-1}$. Thus $(1 * a^n)^{-1}=(a^n)^{-1}={a^{-n}}$. But $-n < 0$. Does this mean that the inverse of $a^n$ does not belong in $\langle a \rangle$?
I got confused a little bit :)
 A: Your statement is not true in general: the set of real numbers of the form $2^n$ with $n \in \Bbb N$ is not a subgroup of the multiplicative group of real numbers. Your statement is true if $G$ is assumed to be finite.
A: What your are trying to prove is in general false. It would amount to saying that for example the  nonnegative integers are a subgroup of the integers, which they just are not. 
The problem is precisely that you cannot show that the inverse elements are in that subset.  
The correct statement is the subgroup generated by $a$ is $\{a^n \colon n \in \mathbb{Z}\}$. Note that $n$ can be negative. 
However, if $G$ is finite it is true, more generally it suffices that $a$ has finite order. If you assume this then you can get by with just considering positive powers. 
If $a$ has finite order then (by definition) there exists some $m \ge 1$ such that $a^m = 1$. So $a (a^{m-1})= 1$ and $a^{-1}= a^{m-1}$ and $m-1  $ is non-negative. 
Depending on what you know you might need to show that in a finite group every element has finite order too. But usually you could know this already. 
