Generating set for factor group Let $G$ be a group and let $H$ be a normal subgroup.
Prove that if $S\subseteq G$ generates $G$,
then the set $\{sH\mid s∈S\} ⊆ G/H$  generates  $G/H$.
I have no idea how to deal with the question above.
Can somebody please give me some help?
 A: You can prove in general that if $\psi:G_1\to G_2$ is a surjective group homomorphism, then if $S\subseteq G_1$ generates $G_1$, then $\psi(S)=\{\psi(s)\mid s\in S\}$ generates $G_2$. The proof is quite straightforward, just follows the meaning of being a generating set. 
Now, to conclude what you need to show, just remember that for any quotient construction, there is an associated natural surjection: $G\to G/H$, given by $g\mapsto gH$.  
A: I think, the best to consider any quotient set $A/\sim$ as that it contains elements of the set $A$ itself but the equality sign is replaced by $\sim$ (we consider $a$ and $b$ equal in $A/\sim$ if $a\sim b$).
In your case the set is the group $G$ and $x\sim y$ iff $x^{-1}y\in H$, and replacing $=$ by this in $G$ gives exactly the quotient group $G/H$. (Observe that $y\in xH \iff x^{-1}y\in H$.)
Now take an element of $G/H$, i.e. a $g\in G$ is given (formally you would write $gH\in G/H$). As $S$ generates $G$, we have in particular $g=s_1s_2s_3\dots s_k$ for some elements $s_i\in S\cup S^{-1}$ in $G$, but the same continues to hold in $G/H$.
(Formally, you would write $s_1H\cdot s_2H\cdot \ldots\cdot s_kH=gH$.)
