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

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