$\begingroup$Q:" G group, H subgroup. G=D6, D diedras group, and H=. Find the cosets of H."
I don't understand the method to find the cosets, I've searched for answers but somehow it stays confusing... Being aH the left coset, do I give an arbitrary value from $D_6$ to $a$? And how does it multiply by $H=\{e, r, r^2, r^3, r^4, r^5\}$?
$\endgroup$2 Answers
$\begingroup$There's no "method" to compute cosets, just the definition.
Given $a\in G$, the coset $aH$ is the subset of $G$ consisting of the elements of the form $ah$ as $h$ varies through $H$.
In the case under question $G=D_6$ and $H$ is the subgroup of rotations, so
if $a\in H$ the coset $aH$ is just $H$;
if $a\notin H$ the coset $aH$ consists of the 6 elements in $G$ not in $H$.
(These two claims should be checked)
A general property of cosets is that they define a partition of $G$. This can be easily verified in the above example.
$\endgroup$$\begingroup$When we write $aH$, it means that we multiply each element of $H$ by $a$ on the left. That is: $$aH = \{ae, ar, ar^2, ar^3, ar^4, ar^5\}$$
To find all the cosets of $H$, you need to do the above computation for every possible value of $a\in G$. (Note that two different values of $a$ may give the same coset.)
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