Finding $|K^\times/\ker(s)|$ and isomorphism $K^\times/S\cong\mathbb{Z}/2\mathbb{Z}$ for finite field K Let $K$ be a finite field with $q$ elements and $K^\times := K\setminus\{0\}$ be the multiplicative group. Assume that the characteristic of $K$ is not $2$, and let $s:K^\times\to K^\times$ given by $x\mapsto x^2$ be a group homomorphism.
First of all, I want to find the number of elements of $S := \operatorname{im}(s)$.
The kernel of $s$ must contain only two elements, namely $1$ and $n-1$, thus $S \cong K^\times/\ker(s)$ and so $$|S|=|K^\times/\ker(s)|=\frac{q-1}{2}.$$ My only problem here is that I found the fact that $|\ker(s)|=2$ by trial and error $$(s(x)=1\iff x=n-1 \lor x=1)$$ however I am lacking formal proof for this. I would appreciate any help in this regard.
My second problem lies in proving that there is an isomorphism $K^\times/S\cong\mathbb{Z}/2\mathbb{Z}$. Because $$|S|=\frac{q-1}{2} \quad \textrm{and} \quad |K^\times|=|K|-1=q-1$$ we have that $\;|K^\times/S|=2=|\mathbb{Z}/2\mathbb{Z}|$. Therefore they have the same dimension.
To finish this, I believe I would need to show that $K^\times/S$ is a cyclic group just like $\mathbb{Z}/2\mathbb{Z}$ but here as well I am stuck and require help. Thank you in advance.

The elements of the kernel are exactly the roots of the polynomial $x^2-1=(x-1)(x+1)\in K[x]$. Clearly its roots are $1$ and $-1$, and since the characteristic of $K$ is not $2$ they are two different elements. Thus $|Ker(s)|=2$.
As for the second question, all groups of order $2$ are isomorphic to each other. Each group of order $2$ must have the form $G=\{e,g\}$ with the multiplication $e^2=e, eg=ge=g$ and $g^2=e$. So it is always the same group up to how you call the elements and the operation.
