"A Book of Abstract Algebra" presents the following introduction to Coset exercises:
In each of the following, $H$ is a subgroup of $G$.
$G = \mathbb{Z}_4$, $H = \lbrace 0, 2 \rbrace$
The cosets of $H$ in this example are:
$$H = H + 0 = H + 2 = \lbrace 0,2\rbrace$$
and
$$H + 1 = H + 3 = \lbrace 1,3\rbrace$$
Given the book's definition of coset:
Let $G$ be a group, and $H$ a subgroup of $G$. For any element $a$ in $G$, the symbol: $$aH$$
denotes the set of all products $ah$, as $a$ remains fixed and $h$ ranges over $H$. $aH$ is called aleft cosetof $H$ in $G$.
What is $a$ in both the left and right cosets? I don't understand, given that definition, how $H+0=H+2=\lbrace0,2\rbrace$.
Best Answer
Remember that the notation $aH$ assumes that the operation on the group is multiplication, in which case we denote the operation by juxtaposition.
In the case that the operation is addition (which is what we call it usually when the operation is abelian) then the cosets are of the form $a+H$.
As for why $H+0=H+2$, just think that $H+0=\{0,2\}$ and $H+2=\{2,4\}=\{2,0\}=H+0$ (remember the order of a set doesn't matter).
as an aside, remember that we are not really talking about integers here, but equivalence classes of integers. we should technically write $H=\{[0],[2]\}$ where $[0]=\{\cdots -8, -4, 0, 4, 8, \cdots\}$ and $[2]=\{\cdots -10, -6, -2, 2, 6, 10, \cdots\}$