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Let $p$ be a prime. Let $A = \{a,2a,3a,\ldots,(p-1)a \}$, where $\gcd(a,p)=1$. Show that there is exactly one element in $A$ congruent to $1$ modulo $p$.

My attempt: It is suffices to show that for each $a \in \{1,2,3,\ldots,p-1\}$, there is exist exactly one $b \in \{1,2,3,\ldots,p-1\}$ such that $ab \equiv 1 \pmod{p}$. Suppose there is exist an element in $A$, say $c$, such that $ab \equiv ac \equiv 1 \pmod{p}$. Then, \begin{equation*} b \equiv 1\cdot b \equiv (ac)b \equiv (ab)c \equiv 1\cdot c \equiv c \pmod{p} \end{equation*} Hence, $b$ is unique and we're done.

Am i true? If not, how to show it?

Bill Dubuque
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lap lapan
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2 Answers2

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Yes, you’re essentially correct. On the one hand, if there are two such solutions, say $m,n$ and $$ma\equiv na\equiv 1\pmod p\,$$ then you’d have that that $m\equiv n\pmod p$, which forces $m=n$. On the other hand, because $\gcd(a,p)=1$, then $a$ has an inverse modulo $p$, say $b$, hence $ab\equiv 1\pmod p$.

Jack LeGrüß
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By Fermat's little theorem $a^{(p-1)} \equiv 1 (mod \ p)$ if $\gcd(a,p)=1$ therefore $a a^{(p-2)} \equiv 1 (mod \ p)$ thus by selecting the smallest positive integer congruent to $a^{(p-2)} (mod \ p)$ you have the solution.

yugikaiba
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