$Ka=K$ iff $a\in K$ Let $K$ be a subgroup of of a group $G$. Let $a \in G$ 
Prove $Ka=K$ iff $a\in K$

need to show $Ka=K \Rightarrow a \in K $ and $a\in K \Rightarrow Ka=K$
$\Rightarrow]$($Ka=K \Rightarrow a \in K $)  (no clue appreciate a hint )
$\Leftarrow]$  ($a\in K \Rightarrow Ka=K$ (same dont know how to approciate)

Def $$ Ka =\{ka: k\in K \}$$
 A: $K$ is a subgroup and $e\in K, Ka=K$ implies $ea=a\in K$.
On the other hand if $a\in K, a^{-1}\in K, x\in K$ implies $x=(xa^{-1})a, xa^{-1}\in K$ since $x\in K, a^{-1}\in K$ so $x=(xa^{-1})a\in K$. This implies that $K\subset Ka, Ka\subset K$ since a group is stable by multiplication.
A: Maybe this can help.
$\Rightarrow$ Assume that $Ka=K$. You know that $ea=a$ where $e$ is the identity in $G$ and hence, an identity in $K$ so that $e\in K$. Since $ea\in Ka$ and $Ka=K$, we get $ea\in K$, that is, $a\in K$.
$\Leftarrow$ Assume that $a\in K$. We need to consider the following.
$i.$ Let $x\in Ka$. Then there exists $k\in K$ such that $x=ka$. So, we have $a,k\in K$ and because $K$ is a subgroup of $G$, we get $ka\in K$. Because $ka=x$, we get $x\in K$. Hence, $Ka\subset K$.
$ii.$ Let $x\in K$. Because $K$ is a subgroup of $G$, we get $a^{-1}\in K$ and hence, $xa^{-1}\in K$. This shows that $(xa^{-1})a\in Ka$. But $(xa^{-1})a=x$. Thus, $x\in Ka$ and so, $K\subset Ka$.
Combining $(i)$ and $(ii)$, we get $Ka=K$.
