Why $f(x)=x$ for any $x\in f(X)$?

Question

1. The set $S(X,X)$ of all mappings of a set $X$ to itself with the composition of mappings in the role of multiplication, where $|X|>1$. Why is not it a group?
2. Let $X$ be a nonempty set. Then the idempotents of the semigroup $S(X,X)$ of all mappings of $X$ to itself are precisely the mappings $f: X \to X$ satisfying the condition $f(x)=x$ for every $x \in f(X)$. An element $x$ of a semigoup is callled an idemotent if $xx=x$. My question is this: Here why $f(x)=x$ for any $x\in f(X)$?

Thanks a Lot!

Answer 1

Hints:

1. If $|X|>1$, then there exist functions $f:X\to X$ without inverses.

2. Write out what it would mean to have $f\circ f=f$.