What are the elements of $D_{4}/\{e,r_{180}\}$? Let $D_{4}$ be the dihedral group of order $8$ consisting of the identity element, $e$; $r_{90}, r_{180}, r_{270}$, rotations of $90$, $180$, $270$ degrees, respectively; $h$ and $v$ a horizontal and vertical flip, respectively; and, $d$, $d^{\prime}$ a flip about the main diagonal (say, the line $y = x$) and a flip about the other diagonal, respectively.
My eventual goal here is to describe all homomorphic images of $D_{4}$ up to isomorphism.
To that effect, I have identified all the normal subgroups of $D_{4}$ to be $\{e\}$, $\{e, r_{180}\}$, $\{e, r_{90}, r_{180}, r_{270}\}$, $\{e, r_{180}, h, v\}$, $\{e,r_{180}, d, d^{\prime} \}$, and $D_{4}$ itself. 
Now, of course $D_{4}/\{e\} \simeq D_{4}$ and $D_{4}/D_{4} \simeq \{e\}$.
By Lagrange's Theorem, $\displaystyle |D_{4}/\{e, r_{180}\}| = \frac{|D_{4}|}{|\{ e, r_{180}\}|} = \frac{8}{2} = 4$, and I have proved that every group of order $4$ is isomorphic to either $Z_{4}$ (if it is cyclic) or $K_{4} \simeq Z_{2} \times Z_{2}$, the Klein-4 Group (if it is not cyclic).
In order to determine which one of these groups the quotient group $D_{4}/\{e,r_{180}\}$ is isomorphic to, I need to know what its elements are, and what the group operation is.  
I know that $D_{4}/\{e, r_{180}\}$ consists of left cosets of the form $eD_{4}$ and $r_{180}D_{4}$, but I don't know any more about what the elements look like and what the group operation is. My professor never explained this to us; could somebody please tell me? I'd prefer a direct answer rather than being given hints or asked further questions.
Thank you.
 A: By the way, you also know $D_4$ mod its normal subgroups of index two (order four), since the quotients' have size two they must be cyclic of order two.
Anyway, there are four elements of $D_4/\{e,r_{180}\}$, and they are cosets of $\{e,r_{180}\}$. (Not what you've written: you wrote $eD_4$ and $r_{180}D_4$, but those are both equal to $D_4$ itself and are not cosets of the subgroup we're quotienting by, $\{e,r_{180}\}$.) Write $H=\{e,r_{180}\}$. The four elements are


*

*$eH$ which equals $r_{180}H$,

*$r_{90}H$ which equals $r_{270}H$,

*$hH$ which equals $vH$,

*$dH$ which equals $d'H$.


Now compute the multiplication table. (Use $eH,r_{90}H,hH,dH$.) For instance, computing the product $(hH)(dH)$, we get $(hd)H$ which equals $r_{90}H$.
Or, you can actually sidestep computing the whole multiplication table. The only groups of order four up to isomorphism are $\mathbb{Z}_4$ and $\mathbb{Z}_2\times\mathbb{Z}_2$. One way to distinguish them is as follows: every nontrivial element of $\mathbb{Z}_2\times\mathbb{Z}_2$ has order two, whereas two elements in $\mathbb{Z}_4$ have order four and one has order two. So then just compute the squares of two nontrivial elements of $D_4/H$. If you get the identity twice, then it's $\mathbb{Z}_2\times\mathbb{Z}_2$, otherwise it's $\mathbb{Z}_4$.
