"A Book of Abstract Algebra" presents this exercise:
In each of the following, $H$ is a subgroup of $G$. List the cosets of $H$. For each coset, list the elements of the coset.
$G=\mathbb{Z}_{15}, H=\langle 5 \rangle $
My attempt follows to calculate the Right and Left Cosets:
$$H + 5 = \langle 10 \rangle $$
Is this correct? If not, please let me know how to figure out the cosets of $H$ in this problem.
The cosets are going to be $1+H,2+H,3+H,4+H,H$.
Notice that $H=\left<5\right>={0,5,10}$, so the cosets are: $$\{0,5,10\}\\\{1,6,11\}\\\{2,7,12\}\\\{3,8,13\}\\\{4,9,14\}$$ Because $x+H=\{x+h:h\in H\}$
EDIT: Also, you need not worry about $6+H$ et al because the higher elements cycle back down nicely.
The left cosets of $H$ are $1+H$, $2+H$, $3+H$, $4+H$ and $H$ itself. Because $\mathbb{Z}_{15}$ is Abelian, these are the same as the right cosets.
As for what's in the cosets... $$x+H = \{x+h:h \in H\}$$
