Cosets of D4, Cycle Notation What are the left and right cosets of D₄?
I know that there are 24 elements in S₄ and 8 in D₄, which gives 24÷8=3 distinct left and right cosets.
I'm having trouble multiplying permutations and have no idea if I'm doing them correctly.
For example, (1234)(1234)=13, (1234)(13)(24)=(1432), (1234)(1432)=e
Are any of these even right?
 A: "What are the left and right cosets of $D_4$?" does not make sense. You have to ask "what are the left and right cosets of H in G?", where H is some subgroup of G. NasuSama's answer has given you a good way of calculating the cosets of $\{e, y^2\}$ in $D_4$, but I suspect you wanted to know the cosets of $D_4$ in $S_4$.
First of all, cycle notation. You should think about $(1234)$ as a function, let's call it $f$, such that


*

*$f(1) = 2$

*$f(2) = 3$

*$f(3) = 4$

*$f(4) = 1.$


Similarly, $(13)(24)$ is a function, let's call it $g$, such that


*

*$g(1) = 3$

*$g(2) = 4$

*$g(3) = 1$

*$g(4) = 2.$


Do you see why? Now, when you "multiply" f by g, you simply find the function $fg$ (or $f\circ g$ if you prefer). So $fg(1) = f(g(1)) = f(3) = 4$, and so on. (Warning: some people write $fg$ to mean $gf$. Both notations are in use - check with your lecturer or textbook.)
Do you now understand how to form cosets? To find the left coset of $D_4$ in $S_4$ corresponding to the element $(123)$, just left-multiply everything in $D_4$ by $(123)$.
Here are a few helpful facts about cosets of $H$ in $G$:


*

*Any two left cosets are either exactly the same, or completely disjoint.

*If $h\in H$, then $hH = H$.

*If $g\in G$ but $g\not\in H$, then $gH \neq H$.

*If $g_2 \in g_1H$, then $g_1H = g_2H$.


Everything I said above works with left cosets replaced by right cosets. Be careful when you mix left with right, though:


*

*It is not generally true that $g_1H$ and $Hg_2$ must be disjoint or equal.

*It is not generally true that $g_1H = Hg_1$.

