# Doubt on showing that when $p$ is prime, the only unipotent class is $p-1$?

Mathematics Asked by Billy Rubina on November 25, 2020

I am trying to understand the solution of the following exercise:

A number $$a$$ is unipotent if $$aneq1$$ and $$a^2equiv 1 pmod{p}$$.

• Show that when $$p$$ is prime, the only unipotent class is $$p-1$$.

• $$(p-1)xequiv1 pmod{p}$$ has a unique solution in $$Bbb{Z}/pBbb{Z}$$. This solution is $$xequiv -1 pmod{p}$$. It is the same as $$x=p-1$$.

I may be doing some very silly mistake: We want to prove that $$a=p-1$$ is the only number such that $$a^2equiv 1 pmod{p}$$ but why does the above proof proves it? Aren’t we proving only that $$a$$ is unipotent without – somehow – checking it for $$(p-1),(p-2),(p-3),dots$$?

Frankly I'm not quite sure what the answer's proof is trying to say, but the easiest way to see this result is to note that $$Bbb Z/pBbb Z$$ is a field, and thus the polynomial $$x^2-1$$ can only have at most two roots (i.e., as many roots as its degree). As $$x=pm1$$ both solve this polynomial, these are the only roots. As we're excluding $$x=1$$, this shows that there is a unique unipotent element modulo $$p$$.

That being said, in the case $$p=2$$ there are actually no unipotent elements since $$1=-1mod 2$$ is the only element that squares to $$1$$ in $$Bbb Z/2Bbb Z$$.

Correct answer by shibai on November 25, 2020

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