# For $Q$ the quaternion group, is $Q/Z(Q)$ a group? For which operation?

I'm new to group theory and am having trouble grasping the concept of cosets. For $N < G$, what exactly is $G/N$?

As an example consider the quaternion group $Q$ and it's center $Z = \{1, -1\}$. Then the left and right cosets are the same: $\{1, -1\}, \{i, -i\}, \{j, -j\}, \{k, -k\}$.

So cosets aren't groups, but is the collection of these cosets a group, and for what operation? (I'm pretty sure that it is in this case, because I read it's isomorphic with $V_4$, but I don't understand if it is general and how it works).

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This is explained in every textbook which deals with the subject. Have you tried any? – Mariano Suárez-Alvarez Jan 21 '13 at 16:22
Sorry I'm trying to study with lecture notes, which treat it rather summarily... – Mark Jan 21 '13 at 16:26
the correct term is "isomorphic" not "isomorphous" – user58512 Jan 21 '13 at 16:30
For some basic information about writing math at this site see e.g. here, here, here and here. – Julian Kuelshammer Jan 21 '13 at 16:40
@Mark, then I cannot suggest more emphatically to pick a good textbook on the subject! – Mariano Suárez-Alvarez Jan 21 '13 at 17:02

We know $Q=\{\pm 1,\pm i,\pm j, \pm k\}$ as an non abelian finite group of order $8$. As you noted correctly its center is $H=Z(Q)=\{+1,-1\}$. This is a very simple routine problem that the center of a group is a normal subset of it, so $Q/H$ is known as a group. To see why is it a group, let's write it as: $$Q/H=\{qH\mid q\in Q\}$$ From @user58512's answer, we have a well-known easy operetion which rules the elements in $Q$ as $$i^2=j^2=k^2=-1,ij=k,jk=i,ki=j,ji=-k,kj=-i,ik=-j$$ Now, you can easily compute $Q/H$ as $$\{+1H,-1H,+iH,-iH,+jH,-jH,+kH,-kH\}$$ but $gH=\{gh\mid h\in H\}$ so you get what @user58512 noted first as a set. Check it! Isn't it an independent group?

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+1 well-deserved! – amWhy Feb 11 '13 at 0:03

$\{\{1, -1\}, \{i, -i\}, \{j, -j\}, \{k, -k\}\}$ is the group.

before we had $ij = k = -ji$

but now we have $\{i, -i\}\cdot\{j, -j\} = \{ij, -ij, -ij, --ij\} = \{k,-k\}$, so it's an abelian group.

to see this is equivalent to V define a group homomorphism from V to it and check it's a bijective.

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