H and K are subgroups of G then H=K Assume $H \leq G$ and $K \leq G$ and assume $aH=bK$ for some $a,b \in G$. Prove that $H=K$
would someone give me a hint. I know that definition of being a subgroup. I know that the element look like this $ah_1=bk_1$ for all $h_1 \in H$ and for all $k_1 \in K$. I do not have an idea from where should I get started. 
 A: Multiply $aH=bK$ by $b^{-1}$ to get
$$b^{-1}aH=K$$
Since there is a $g\in G$ such that $g=b^{-1}a$ we get $gH=K$. This means that for some $h\in H$ and some $k\in K$ we get the equality $gh=k$. $H$ is a subgroup which means that $e\in H$. Set $h=e$ to get $g=k$ and therefore $g\in K$. $K$ is a subgroup which means that also $g^{-1}\in K$. Multiply $gH=K$ by $g^{-1}$ to get 
$$H=g^{-1}K=K$$
A: Assume $H \subseteq G$ and $K \subseteq G$ and assume $aH=bK$ for some $a,b \in G$. 
Then $\forall h \in H \exists k \in k :ah = bk
$ and of course that $k$ may depend on $h$, so let us label one of the (possibly multiple) values of $k$ that satisfies $ah = bk$ as $k(h)$.
Consider the  $e$ (the group identity) which must be an particular element  $h$:
$$
a e = bk(e) \implies a = bk(e) 
$$ 
Then for any arbitrary $h\in H$, 
$$ ah = bk(h) \implies (bk(e) )h = bk(h) \implies b^{-1}(bk(e) )h =b^{-1}bk(h)
\implies k(e) h =k(h)
$$
But $k(e) \in k$ so $[k(e)]^{-1}\in k$.  
Left-multiply both sides by $[k(e)]^{-1}$:
$$h =[k(e)]^{-1}k(h)$$
Can you see now that $h\in k$?
So for all $h$, $h\in H \implies h \in K$.  Similarly you show that  for all $k$, $k\in K \implies k \in H$.  You can take it from there... 
