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 :)