Here is an unanswered exercise from class notes.
For the group <{$i,-i,1,-1$},*> and a subgroup H where $|H|$ = 2. Find the left cosets of G induced by H. G={$i,-i,1,-1$}
Here is an unanswered exercise from class notes.
For the group <{$i,-i,1,-1$},*> and a subgroup H where $|H|$ = 2. Find the left cosets of G induced by H. G={$i,-i,1,-1$}
On
Besides to another post, you may think of $G$ as one of the subgroups of the quaternions of order $8$ ($Q_8$). Here is the lattice of $Q_8$:

We see that the only possible subset of $G$ of order $2$ is indeed the subgroup $H$. Since $|G|=4$ so $[G:H]=2$. There is nothing to do when we are finding $gH,~g\in G\cap H$. Indeed $$-1H=H, +1H=H$$ So try focusing on $gH,~ g\in (G-H)$ that are: $$+iH, ~-iH$$
It can be found that H={$1,-1$} is a subset of G and it's found that <{$1,-1$},*> is a subgroup of G. An example of such that is not a subgroup is when $\underset{2}{H}$ = {$1,i$}. Although $\underset{2}{H}$ is a subset of G, it is not closed under the group operation.
Def: Coset - "set composed of all the products obtained by multiplying each element of a subgroup by one element of the group containing the subgroup"
A Subgroup of <{$i,-i,1,-1$},*> is <{$1,-1,*$}>
Left cosets are:
1*{1,-1} = {1,-1}
-1*{1,-1} = {-1,1}
i*{1,-1} = {i,-i}
-i*{1,-1} = {-i,i}
I'm sorry if it's hard to read. I'm not familiar with the formatting.