Group theory

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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. $\blacksquare$


  1. The proposition and corollary are Exercise 2.29 in Rotman's Introduction to the Theory of Groups