# Factor group of quaternion group

I have $Q_8=\{I,A,A^2,A^3,B,AB,A^2B,A^3B\}$ where $A=\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$ and $B=\begin{pmatrix} 0 & i \\ i & 0 \end{pmatrix}$

Now I take a look at $Q_8/N$ for all $N\unlhd Q_8$ (i.e for which G is there an isomorphism $Q_8/N\cong G$ ??)

I know that $|N|\in\{2,4\}$. If $|N|=2$ then $N=\{I,-I\}$ and clearly $Q_8/N\cong V$ where V is Klein-Four group.

What if $|N|=4$ ?

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The quotient group in that case is of order $2$. – Lubin Dec 10 '12 at 20:40
So they probably will be isomorphic to $\mathbb Z_2$ and $\{I\}$ but how do the $Q_8/N$ look explicitly ? – Voyage Dec 10 '12 at 20:45

Let's work out explicitly what the quotient looks like by a subgroup of order 4. Specifically, let's quotient by $\left<A\right> = \{1,A,A^2,A^3\}$.
$Q/\left<A\right>$ is, by definition, the set of cosets of $\left<A\right>$ in $Q$. As pointed out, there must be two cosets because 8/4=2. One of these cosets, $S_1$, contains the elements of $\left<A\right>$, and the other, $S_2$, contains the leftovers: $\{B,AB,A^2B,A^3B\}$.
Strictly speaking, these two cosets are the elements of $Q/\left<a\right>$. The group operation is defined as the usual product for two subsets of a group. If $S$ and $T$ are subsets of group $G$, their product $ST$ is said to be the set $\{st|s\in S, t\in T\}$.
Under this product, the two cosets of $\left<A\right>$ in $Q$ form a group. You can verify directly that $S_1S_1 = S_2S_2 = S_1$ and $S_2S_1=S_1S_2 = S_2$. These relations demonstrate that $Q/\left<A\right>$ is cyclic of order 2.
This is a relatively unexciting example. If there were more than two cosets, you might have to compute several pairwise products before discovering the structure of the quotient group. Because $|Q/\left<A\right>|=2$, there was only one possibility for a group structure. Also, if you were to take $G/H$ with $H$ not normal in $G$, you would find that the cosets would not form a group under the product we defined.