For any subgroup H of the group $G$, let $H^2$ denote the product $H^2=HH$. Prove that $H^2=H$. 
For any subgroup of the group $G$, let $H^2$ denote the product $H^2=HH$. Prove that $H^2=H$.

This question seems simple but I do not know how can I prove.
 A: A start: To prove that two sets $A$ and $B$ are equal, we often use the following strategy: (i) We show that every element of $A$ is an element of $B$ and  (ii) We show that  every element of $B$ is an element of $A$.  It is usually best to tackle these two items separately. 
(i) Can you show that everything in $H^2$ is in $H$? 
(ii) Can you show that everything in $H$ is in $H^2$?  
To do either half, recall the meaning of $H^2$. It is the set of all objects of the form $h_1h_2$, where $h_1\in H$ and $h_2\in H$. The objects $h_1$ and $h_2$ need not be equal.  For (ii), the identity element of the group will be useful.
A: Its obvious that all elements of $H$ are in $H^2$, right?  Think for any $h$ in $H$, 1*$h$ is in $H^2$.
On the other hand, for any $h,k$ in $H$, we have $hk$ as an element in $H^2$ (by definition) and as a subgroup is closed under multiplication $hk$ is in $H$.  Therefore, $H=H^n$ for all natural numbers.
A: First, prove that for all $h\in H$, $Hh=hH=H$.
Now, we are ready to prove that,  $HH=H$ :
$$HH=\{h_1h_2|h_1,h_2\in H\}= \underset{h_1\in H}{\cup}\{h_1h_2|h_2\in H\}=\underset{h_1\in H}{\cup}h_1H=\underset{h_1\in H}{\cup}H=H$$
