Union of elements in quotient group For a quotient group G/H, is is true that union of all elements in G/H equals G?
I think it is true because union of all cosets relative to H covers G. But when I asked this question to two of my professors, both of them said it is not so true because we should view each thing in G/H as a single element of G/H, not a set. But I think being a set does not prevent them from being an element of G/H? Or is there some convention behind?
 A: Don't get confused. If $N\trianglelefteq G$ then $G/N$ is a group of the left cosets of $N$ with the operation being multiplication of cosets. It means that the elements of $G/N$ are the cosets themselves (as sets), not the elements which are inside the cosets. So it's not true to say that the union of the elements of $G/N$ give the group $G$ because the elements of $G/N$ are not elements of $G$ at all. For example, $\langle 2 \rangle\trianglelefteq\mathbb{Z_4}$. Then: 
$\mathbb{Z_4}/\langle 2\rangle=\{\langle 2\rangle, 1+\langle 2\rangle\}$. 
But are $\langle 2\rangle$ and $1+\langle 2\rangle$ elements of $\mathbb{Z_4}$? No, they aren't. The elements which are inside these cosets are elements of $\mathbb{Z_4}$, not the cosets themselves. 
A: You're right. By definition, $G/H$ is the set of equivalent classes of the equivalence relation $\sim$ on $g$ defined by: $$\forall g,g'\in G,g\sim g'\Leftrightarrow gg'^{-1}\in H.$$
Given any $g\in G$, the element $gH$ of $G/H$ is, still, the equivalence class of $g$ under $\sim$, hence a subset of $G$.
Since any element of $g$ is in $gH$, it does follow that $G=\bigcup G/H$, or in other words, $G=\bigcup_{g\in G}gH$.
