There is a subgroup H of a group G, and pairs of elements $(x,y)$ of $G$. Determine if they belong to the same left $H$-cosets. I'm simply to provide a yes or no answer.
So I know what cosets are and I understand how to compute but I'm not sure how with these groups specifically.
For the first one, $\mathbb{Z}, H = 10\mathbb{Z}$, I'm not sure what H is exactly. Is it just every member of the integers left multiplied by 10 i.e $\{10, 20, 30, 40, ...\}$. If that's the case how can any of the pairs belong to the same left coset as they never contain the same integers?
And for the second, I'm not sure how the definition of $a^6 = b^2 = e$ and $ba = a^{6-1}b$ comes into play. $H = <a^1b^0>$, but how do we determine the left coset with these elements? Better yet, how would I determine what cyclic group that even defines...?
Would appreciate some help with these computational questions.
$10\mathbb{Z} = \{\ldots,-20,-10,0,10,20\ldots\}$ and the coset associated with $x$ would look like $\{\ldots,x-20, x-10, x, x+10, x+20,\ldots\}$. So for example, $4$ and $14$ are in the same coset $\{\ldots,-6, 4, 14, 24, \ldots\}$, while $3$ and $5$ are not. (Can you give a plain English description for what it means to be in the same coset in this example?)
For the dihedral group, it seems like you are struggling with the definition of the dihedral group. You should probably review your notes about it. For this particular dihedral group, you can think of it as isometries of a regular hexagon. Then $a$ can represent the counterclockwise rotation by $60^\circ$, while $b$ can be the reflection across one of its diagonals. Then $H$ is the cyclic group generated by the rotations.