What is the set $\mathbb{Z}\backslash 2 \mathbb{Z}$? I have not taken group theory and was confused by a notation
$$\mathbb{Z}\backslash 2 \mathbb{Z}$$
Can someone please help me to understand this and its possible generalization?
 A: Fix $n\in\Bbb N$.

The set
$$\Bbb Z\setminus n\Bbb Z=\{x\in\Bbb Z\mid x\notin n\Bbb Z\},$$
where
$$n\Bbb Z=\{ny\in\Bbb Z\mid y\in\Bbb Z\}.$$

The group
$$\Bbb Z/ n\Bbb Z=\{x+n\Bbb Z\mid x\in \Bbb Z\},$$
where $n\Bbb Z$ is as above, is understood as a group under the operation
$$(a+n\Bbb Z)+(b+n\Bbb Z):=(a+b)+n\Bbb Z.$$
It is a quotient group. Each of its elements is what is known as a coset of $n\Bbb Z$.

Your case, of course, is when $n=2$.
A: $\mathbb{Z}$ be the set of all integers.
We want to put all integers into $ n$  distinct buckets and two integers belong to same bucket if they are similar.
Now, it is meaningless until we define what does it mean by the word "similar " and "bucket".
Two integers $a$ and $b$ are similar $(a\thicksim b)$ if $a-b$ is divisible by $n$.
It is easy to check that the relation is an equivalence relation.
Then the equivalence class(bucket) of $a$,
$[a]=\{b\in \mathbb{Z} : b\thicksim a \}$
$a\in \mathbb{Z} $ and $n>1$ by division algorithm,
$\begin{align}
a&=nq+r
&(q: quotient, r:reminder,0\le r\le ( n-1) ) \end{align}$
$a-r=nq$ and hence, $a\thicksim r$
$\implies [a]= [r] $
Hence, there is $n$ distinct equivalence classes $[0], [1],..., [n-1]$.
Then the set of all equivalence classes $\{ [0], [1],..., [n-1]
\} =\mathbb{Z}/ \mathbb{nZ}$
In your question, $n=2$
$\mathbb{Z}/ \mathbb{2Z} = \{ [0], [1] \}$
Two integers belongs to same equivalence class iff they leave the same reminder upon division by $2$.
Like, $2\in [0] $ because both are divisible by 2.
