Find the Quotient group of normal subgroup of order 4 of quaternion group. $H=(\pm 1 , \pm i , \pm j, \pm k)$.
I know
$$c(i)=( \pm 1, \pm i) \quad  c(j)=( \pm 1, \pm j)\quad  c(k)=( \pm 1, \pm k) $$
and thus the class equation of $H$ is $8= 2+2+2+2$ because $\lvert H\rvert =\lvert Z| +\sum \mathrm{conj}_H$ and $\lvert \mathrm{conj}_H(a)|= \lvert H|/\lvert c(a)|$. 
Now normal subgroups are unions of conjugacy classes. So a normal subgroup of order $4$ of $H$ is $N=( \pm 1, \pm i)$. Take $\phi\colon H \to H/N$ whose kernel is $N$. I'm basically trying to find the $H/N$ which should be of order $2$, because $\lvert H/N|=\lvert H|/\lvert N|$.
Now I find images of each element of $H$ by the fact that elements of $H/N$ look like $hN$ i.e.  


*

*$\pm 1 \mapsto N $ (as $\pm 1 (\pm 1, \pm i) = (\pm i, \pm 1))$)

*$\pm i \mapsto N$ 

*$\pm j \mapsto (\pm j, \pm k)$ 

*$\pm k \mapsto (\pm j, \pm k)$


So, $H/N$ is $(N \cup (\pm j, \pm k))$ i.e. $H$, isn't it?
So I'm confused, kernel is $N$ very well but how is  $\lvert H/N\rvert = 2$ ? What am I doing wrong?
 A: $H/N$ is not $H$. The reason you are confused is probably strongly related to how you write things. Especially when you are at the elementary stage, you should really take care to write things properly.
It does not make sense (formally) to write $N=(\pm 1, \pm i)$. First, you are using an ad hoc shorthand to avoid writing $1$ and $-1$ and $i$ and $-i$ separately, but more importantly, you are using parentheses, which in mathematics denote ordered tuples, while $N$ is a set (or a group, depending on which aspect you are focusing on -- but certainly not an ordered one), so you should write it as $\{1,-1,i,-i\}$ or, if you really want to be concise, $\langle i \rangle$ (i.e. the subgroup generated by $i$) -- but again, you should refrain from even that kind of shorthand at this stage (though it is entirely “kosher”, formally speaking, unlike the ones you have been using).
Likewise, $j$ and $k$ both map to $\{j,-j,k,-k\}$, and thus $H/N=\{\{1,-1,i,-i\},\{j,-j,k,-k\}\}\neq  H$. Clearly, $H/N$ has two elements: $N=-1N=iN=-iN$ and $kN=-kN=jN=-jN$.
Sure, $\bigcup H/N=H$, but that should come as no surprise. Indeed, given any group $G$ and any subgroup $H\leq G$ we have by definition $\bigcup G/H=G$, but we never have $G/H=G$, even if $H$ is trivial -- though with enough (mathematical) maturity to avoid the various related pitfalls, it is eventually acceptable to identify the two in the trivial case.
