Group theory
Date:
Every element of a finite field is the sum of two squares
Def:
Let $ G $ be a group and $S$ and $T$ subsets
(not necessarily subgroups) of $G$. We define
$$ST := \{st \mid s \in S, t \in T\}$$
Note that even if $S$ and $T$ are subgroups, $ST$ doesn't have to be one.
This is because the composition of elements of the from $st$ do not have to
be of that form.
Proposition[1]:
Let $G$ be a finite group and $S$ and $T$ subsets of $G$.
If $\#(S) + \#(T) > \#(G)$ then $ST=G$.
Proof:
Fix $g \in G$. Consider the map $\rho_g: T \to G$ with $t \mapsto gt^{-1}$.
Then $\rho_g$ is clearly injective. Thus $\#(\rho_g(T)) = \#(T)$.
Hence $\#(\rho_g(T)) + \#(S) > \#(G)$ meaning that $\rho_g(T) \cap S \neq
\varnothing$. Hence there exists and $s \in S$ with $s \in \rho_g(T)$.
We have then that $s = gt^{-1} \Leftrightarrow st = g$. Thus $g \in ST$.
Since $g$ was arbitrary, we have $G = ST$ as desired. $\blacksquare$
Corollary:
Every element in a finite field is the sum of two squares.
Proof:
Let $p$ be a prime number. Let $U := \mathbb{F}_{p^n}^*$. Consider the group
homomorphism $\varphi: U \to U$ given by $u \mapsto u^2$ ($\varphi$ is a
homomorphism only because $U$ is commutative!).
Now if $p = 2$, we get that
$\varphi$ is an automorphism (the Frobenius field automorphism restricted on the
unit group), making every non zero element in $\mathbb{F}_{p^n}$
a square and hence trivially a sum of squares. Since zero is also trivially a
sum of squares, we are done.
If $p \neq 2$, then $\ker(\varphi) = \{\pm 1\}$,
since $u^2 = 1 \Leftrightarrow (u-1)(u+1) = 0$ in $\mathbb{F}_{p^n}$.
Hence the subgroup of squares of $U$ has index $2$ in $U$. Since $\#(U) = p^n-1$,
we get that for the set $S$ of perfect squares in $\mathbb{F}_{p^n}$,
$\#(S) = \frac{\#(U)}{2} + 1 = \frac{p^n+1}{2}$. Now let $G$ be the
additive group of $\mathbb{F}_{p^n}$. Since $2\#(S) = p^n+1 > p^n = \#(G)$
we can apply the above proposition to the group $G$ with $S = S$ and $T = S$.
This delivers $G = S+S$, meaning that every element of $\mathbb{F}_{p^n}$ is a
sum of squares. $\blacksquare$
- The proposition and corollary are Exercise 2.29 in Rotman's
Introduction to the Theory of Groups
Date:
Every element of a finite field is the sum of two squares
Def: Let $ G $ be a group and $S$ and $T$ subsets (not necessarily subgroups) of $G$. We define $$ST := \{st \mid s \in S, t \in T\}$$
Note that even if $S$ and $T$ are subgroups, $ST$ doesn't have to be one. This is because the composition of elements of the from $st$ do not have to be of that form.
Proposition[1]:
Let $G$ be a finite group and $S$ and $T$ subsets of $G$.
If $\#(S) + \#(T) > \#(G)$ then $ST=G$.
Proof:
Fix $g \in G$. Consider the map $\rho_g: T \to G$ with $t \mapsto gt^{-1}$.
Then $\rho_g$ is clearly injective. Thus $\#(\rho_g(T)) = \#(T)$.
Hence $\#(\rho_g(T)) + \#(S) > \#(G)$ meaning that $\rho_g(T) \cap S \neq
\varnothing$. Hence there exists and $s \in S$ with $s \in \rho_g(T)$.
We have then that $s = gt^{-1} \Leftrightarrow st = g$. Thus $g \in ST$.
Since $g$ was arbitrary, we have $G = ST$ as desired. $\blacksquare$
Corollary:
Every element in a finite field is the sum of two squares.
Proof:
Let $p$ be a prime number. Let $U := \mathbb{F}_{p^n}^*$. Consider the group
homomorphism $\varphi: U \to U$ given by $u \mapsto u^2$ ($\varphi$ is a
homomorphism only because $U$ is commutative!).
Now if $p = 2$, we get that
$\varphi$ is an automorphism (the Frobenius field automorphism restricted on the
unit group), making every non zero element in $\mathbb{F}_{p^n}$
a square and hence trivially a sum of squares. Since zero is also trivially a
sum of squares, we are done.
If $p \neq 2$, then $\ker(\varphi) = \{\pm 1\}$,
since $u^2 = 1 \Leftrightarrow (u-1)(u+1) = 0$ in $\mathbb{F}_{p^n}$.
Hence the subgroup of squares of $U$ has index $2$ in $U$. Since $\#(U) = p^n-1$,
we get that for the set $S$ of perfect squares in $\mathbb{F}_{p^n}$,
$\#(S) = \frac{\#(U)}{2} + 1 = \frac{p^n+1}{2}$. Now let $G$ be the
additive group of $\mathbb{F}_{p^n}$. Since $2\#(S) = p^n+1 > p^n = \#(G)$
we can apply the above proposition to the group $G$ with $S = S$ and $T = S$.
This delivers $G = S+S$, meaning that every element of $\mathbb{F}_{p^n}$ is a
sum of squares. $\blacksquare$
- The proposition and corollary are Exercise 2.29 in Rotman's Introduction to the Theory of Groups