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In Herstein's Topic in Algebra he writes

Let $F$ be a finite field having $q$ elements (if you wish, think of $J_p$ with its $p$ elements). Viewing $F$ merely as a group under addition, since $F$ has $q$ elements, by Corollary 2 to Theorem 2.4.1, $$ \underbrace{a + a + \dots + a}_{q \text{-times}} = qa = 0.$$

But he also writes

For simplicity of notation we shall henceforth drop the dot in $a \cdot b$ and merely write this product as $ab$.

But $q$ is not necessarily element of field $F$, so how does $qa = 0$ come?

Bill Dubuque
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Lazy Guy
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    This is defined earlier. In an abelian group, with an element $a$, for positive $n$, $na$ means "adding $a$ to itself $n$ times". It is the additive equivalent of writing $a^n$ when your operation is multiplication. – Arturo Magidin Oct 30 '24 at 14:15
  • Sorry but could you point to the relevant page – Lazy Guy Oct 30 '24 at 14:23
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    My edition is in Spanish. The meaning is literally in the quote you give. $qa$ is shorthand for "add $a$ to itself $q$ times." – Arturo Magidin Oct 30 '24 at 14:49
  • See the dupe on "abusive" notation for $n$'th multiples and powers. $\ \ $ – Bill Dubuque Oct 30 '24 at 22:29

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