Why is $\left(\mathbb F_l^{\times}\right)^2<\mathbb F_l^{\times}$

Question

Why is $\left(\mathbb F_l^{\times}\right)^2<\mathbb F_l^{\times}$ when $l$ is odd prime

I.e. Left side is a strict subgroup of the right side ?

What does the squaring mean, is every element squared ?

And then we must have an self-inverse element to fulfill the strict inequality ?

Answer 1

It means that $\{x^2, x\in \mathbb{F}_l^\times\}$ is a strict subgroup of $\mathbb{F}_l^\times$.

This is the case if and only if $l$ is odd. Indeed the kernel of $x\mapsto x^2$ (which is a morphism) is $\{1,-1\}$, and this is a nontrivial subgroup if and only if $l$ is odd (if and only if $1\neq -1$).