Compute the left and right cosets of $H$ and $N$ in $D_8$ For $D_8$, the dihedral group of symmetries of the square with generators $r,s,$. Consider the subgroups $H = \langle s \rangle = \{e,s\}$ and $N = \{e,r^2,s,sr^2\}$ (you may assume without proof that $H$ and $N$ are indeed groups)
I already have the solution for this problem, but I just want an explanation for why they're the solution. 
This is the solution provided:
Left cosets of $H$ in $D_8$:
$eH = sH = H$
$rH = sr^3H = \{r,sr^3\}$
$r^2H = sr^2H = \{r^2,sr^2\}$
$r^3H = srH = \{r^3,sr\}$
Right cosets of $H$ in $D_8$:
$He = Hs = H$
$Hr = Hsr = \{r,sr\}$
$Hr^2 = Hsr^2 = \{r^2,sr^2\}$
$Hr^3 = Hsr^3 = \{r^3,sr^3\}$
Left cosets of $N$ in $D_8$:
$eN = r^2N = sN = sr^2N = N$
$rN = r^3N = srN = sr^3N = \{r,r^3,sr^3,sr\}$
Right cosets of $N$ in $D_8$:
$Ne = Nr^2 = Ns = Nsr^2 = N$
$Nr = Nr^3 = Ndr = Nsr^3 = \{r,r^3,sr,sr^3\}$
An explanation of this would be amazing, thanks.
 A: If $K\subseteq G$ is a subgroup, then we define $gK=\{gk\mid k\in K\}$ to be a left coset of $K.$
Since $K\subseteq G$ is a subgroup, then the order of $K$ divides the order of $G$ (by Lagrange's theorem). That is to say, the number of distinct cosets determined by $K\subseteq G$ is precisely $\frac{|G|}{|K|}.$ The rest, as you have in your problem statement, is a matter of sitting down and writing out some direct computation. What I have stated about the number of distinct cosets is what tells you've written down all of the cosets you can.
A: Using the relations between $r$ and $s$ ($r^4=e, s^2=e, rs=sr^3$), you first derive: 
$$r^2s=rrs=rsr^3=sr^3r^3=sr^2$$
$$r^3s=rr^2s=rsr^2=sr^3r^2=sr$$
Now it all follows directly. For example, picked randomly from the middle: $rN=\{re, rr^2, rs, rsr^2\}=\{r, r^3, sr^3, sr^3r^2\}=\{r, r^3, sr^3, sr\}$. The rest is equally tedious but straightforward. 
A: In case you have not seen Lagrange's theorem yet: as Chickenmancer says, a left coset of $H$ in $G$ is just the set consisting of all the elements of $G$ of the form $\tilde{g}h$ for some fixed $\tilde{g} \in G$ and for every $h \in H$ (this, for instance, would just be the left coset denoted $\tilde{g}H$).
So you just write down all the possibilities and see which ones are equal. For instance, $eH=sH$ because $eH = \{ee,es\}=\{e,s\}$ and $sH=\{se,ss\}=\{s,e\}$, so as sets they are equal (they contain exactly the same elements). Since $G$ has only 8 elements, you have at most 8 left cosets of $H$ (but it turns out that there are really just 4 possibilities, which you each get twice).
