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I've read through the elementary proof of why there exists a solution $x$ to $x^2\equiv -1\pmod p$ iff $p\equiv 1 \pmod 4$ for $p$ an odd prime. Is there a group theory generalization for this fact as well?

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3 Answers

up vote 9 down vote accepted

I would sketch the group theoretic fact as follows: Let $G$ be a finite abelian group. Then the following are equivalent:

(1) For every $d$, there are at most $d$ solutions to $g^d=1$.

(2) $G$ is cyclic.

(3) For $n$ dividing $|G|$, there are precisely $\phi(n)$ elements of order $n$.

The only place that I know of where hypothesis (1) shows up naturally is considering the multiplicative groups of finite fields. From this we get:

There is an element of order $d$ in $\mathbb{F}_{p^r}$ if and only if $d| p^r-1$.

I'm not sure there is any more to say.

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Presumably you refer to the following proof, which contradicts Lagrange's or Fermat's little Theorem

Suppose $\rm\; p = 4n+3\;$ is prime. Then $\rm\ x^2 = -1 \;\Rightarrow\; x^{p-1} = (x^2)^{2n+1} = -1 \pmod{p}\;\Rightarrow\Leftarrow$

A finite field of order $\rm\:4n+3\:$ has subgroup of squares of order $\rm\:2n+1\:$. But one easily shows that in any group of odd order $\rm\; 2n+1\:$ the equation $\rm\; x^2 = a\;$ has the solution $\rm\: a^{n+1}\;$ (due to Lagrange in 1769). Specializing $\rm\; a = -1\:$ we conclude that $\rm\; x^2 = -1\:$ has no roots. On the other hand, the equation always has a root in a finite field of order $\rm 4n+1$ by the case $\rm k=4$ of Frobenius's theorem (1907) - which says that if $\rm k$ divides the order of a finite group $\rm G$ then $\rm k$ also divides the number of $\rm k$'th roots of $1$ in $\rm G\:$.

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See also here. –  Bill Dubuque May 1 '12 at 15:36
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Probably the simplest way to state a corresponding group theory result is the generalization of Frobenius's theorem mentioned by Bill Dubuque. The generalization says that if $G$ is a finite group, then the number of solutions in $G$ to $x^n=1$ is a multiple of $\gcd(|G|,n)$.

In the case of the congruence $x^2\equiv -1\pmod{p}$, the solutions to this congruence are exactly the solutions to $x^4\equiv 1 \pmod{p}$ that are not solutions to $x^2\equiv 1\pmod{p}$. The order of the group here is $p-1$, so the number of solutions to the first is a multiple of $\gcd(4,p-1)$; the number of solutions of the second is a multiple of $2$, and we know the two: $1$ and $-1$. Since every solution to $x^4\equiv 1\pmod{p}$ is a solution to $x^2\equiv 1 \pmod{p}$, there are solutions to $x^2\equiv -1\pmod{p}$ if and only if $\gcd(4,p-1) = 4$.

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It should be added, perhaps, that if there are exactly $n$ solutions in $G$, then they form a normal subgroup. This is known as Frobenius's conjecture, whose only known proof requires CFSG. –  user641 Sep 18 '10 at 15:15
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