Does $Kg=K$ implies that $g=e$? $K$ is a subgroup of the group $G$ and for a element $g\in{G}$,does  $Kg=K$ implies that $g=e$? where $e$ is the identity element of $G$.
 A: Assume $Kg=K$. Since $e\in K$, we have $g=eg\in Kg=K$, i.e. $g\in K$.
On the other hand,  if $g\in K$ and $K$ is a subgroup, then for all $k\in K$ also $kg\in K$, i.e.$ Kg\subseteq K$, and from $k'=kg^{-1}\in K$ we have $k=k'g\in Kg$, i.e. also $K\subseteq Kg$.
We conclude that 
$$ Kg=K\iff g\in K.$$
So unless $K$ is the trivial subgroup $\{e\}$, we cannot infer that $g=e$.
A: If $$Kg=K$$, then there exists $k_1$ and $k_2$ $\in K$ such that $$k_1g=k_2$$
$$\implies g= k_1^{-1}k_2$$. They cann't be $e$ all the time. 
A: Maybe we can also interpret the question as whether it is possible for a subgroup $K$ of a group $G$ to have an identity $g$ different from that of $G$ , i.e., can we have a $g$ so that, $kg=k$ for all $k$ in $K$? The answer is then no: consider the identity $e$ of $G$; we then have $ge=g$ for all $g \in G$; in particular, for $k \in K$: $$ke=k=kg \implies k^{-1}ke=e=k^{-1}kg=g $$.
So the only element that can act like the identity in a subgroup $K$ of a group $G$ is the identity of the group $G$.
