I know that every element in $G$ is also in $gG$ for all $g\in G$, but does this really mean that $G=gG$? I mean, If we set $f_a(g)=a*g$ for some $a\in G$, the only thing $f_a$ really does is rearranging the elements in $G$. So, is $gG$ equal to $G$ or is $gG$ isomorphic to $G$? In other words, does order (not the size of the group, but the $a$ before $b$ or $a$ after $b$ type of order) matter?
I realize that if $a\neq b$, that then $f_a\neq f_b$, but does this also mean that $$a\neq b\implies \{f_a(g):g\in G\}\neq \{ f_b(g):g\in G \}$$
By the way, I thought of this when I was trying to understand Caley's theorem.
Back to the basics: two sets are equal precisely if each one is contained in the other.
$G \subset gG$: any $h \in G$ can be written as $h = gg^{-1}h \in gG$.
$G \supset gG$: any product of $g$ with another group element belongs to $G$, by definition of a group.