## Zolotarev’s lemma

Posted: December 17, 2009 in Elementary Algebra; Problems & Solutions, Groups and Fields
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Problem. Let $p$ be a prime number and let $G$ be the multiplicative group of a field of order $p.$ Let $a \in G$ and define the bijection $f: G \longrightarrow G$ by $f(g)=ag, \ \forall g \in G.$ Prove that $\displaystyle \text{sgn}(f)=\left ( \frac{a}{p} \right).$

Solution. Let $A=\{f_a: \ \ a \in G \}.$ Then $A$ is a group under composition and $|A|=p-1.$ Let

$B=\{f_a \in A: \ \ \text{sgn}(f_a)=1 \}.$

Clearly $B$ is a subgroup of $A.$ Since $G$ is cyclic (see the theorem in this post), we can choose a generator $c$ for $G.$ Then $f_c$ is a cycle of length $p-1$ and therefore $f_c \notin B.$ Thus $B \neq A.$ Now if $a=b^2,$ then $f_a=f_{b^2}=f_b \circ f_b$ and hence $\text{sgn}(f_a)=(\text{sgn}(f_b))^2=1,$ that is $f_a \in B.$ Therefore $\displaystyle |B| \geq \frac{p-1}{2}$ and since $|B| \mid |A|=p-1$ and $|B| < p-1,$ we have $\displaystyle |B|=\frac{p-1}{2}. \ \Box$