Is it true that $hH$ is subset of $H$ Let $G$ be a Group and let $H <G$ and if we choose a fixed element $h \in H$ then is it always TRUE that $hH \subset H$ .If So how can we prove it.
I came to know after seeing this property being true for the following example:
$1.$ If $G=Z_6=\left\{0,1,2,3,4,5\right\}$ and $H=\left\{0,3\right\}<G$
we have $0+H \subset H$ and $3+H \subset H$
EDIT:
According to the claim of Geoffrey:
we have $hH=H$
but while proving theorem $19.2$ in this link http://people.virginia.edu/~mve2x/3354_Spring2015/lecture19.pdf
the author has taken the following steps:
If $g^{-1}k=h$
Then $k=gh$
$\implies$
$kH=(gh)H$
$\implies$
$kH=g(hH)$
Then $kH \subseteq gH$
But since $hH=H$ why can't we write directly conclude $kH=gH$
 A: Hint:  $H$ is a subgroup, so $H$ is closed under the operation.
A: The authors have not proven that $hH=H$ for all $h\in H$ when they are proving Theorem 19.2; as such, they cannot invoke that result. Instead, they simply note that for every $x\in H$, since $h,x$ are both in $H$, then $hx\in H$, hence $hH\subseteq H$. This suffices for their purposes, which is to show that if $g^{-1}k\in H$ then $gH=kH$, since they can use that inclusion to show that $kH\subseteq gH$ (and then symmetrically, using the fact that $k^{-1}g=(g^{-1}k)^{-1}$ is also in $H$) that $gH\subseteq kH$.  
In fact, this result can be used to prove that $hH=H$ for all $h\in H$, since we will have $he^{-1}=h\in H$, hence $hH=eH=H$. 
It can also be proven directly; as noted above, we have that $hH\subseteq H$. To prove the converse inclusion, let $x\in H$. Then $h^{-1}x\in H$ as well, since both $h^{-1}$ and $x$ are in $H$; then $x=h(h^{-1}x)\in hH$. Thus, $H\subseteq hH$, proving the equality. 
A: If $H$ is a subgroup of $G$ and $h \in H$, then $hH = H$. This is because $hH \subseteq H$ since $H$ being a group is closed under the group operation, and given an arbitrary arbitrary $k \in H$, since $H$ is a group and $h \in H$ we know $h^{-1} \in H$ and so if $k \in H$ then $h^{-1}k \in H$ so $k = hh^{-1}k \in hH$, and since $k \in H$ was arbitrary $hH \subseteq H$. Hence $hH = H$.
That being said I have seen people use "$<$" to denote subset instead of subgroup, so if $H$ is a subset but not necessarily a subgroup then $hH$ might not be a subset of $H$. For example if $\textbf{Z}_6$ and $H = \{1, 2, 3\}$ and $h = 2$ then $hH = \{3, 4, 5\}$
