Existence of a Subgroup of Order 4 in $(\mathbb{Z}/p\mathbb{Z})^{\times}$

Question

I'm stuck on a detail in the proof of Lemma 17 in Chapter 8 of Dummit and Foote's Abstract Algebra. It's claimed that the quotient group of $(\mathbb{Z}/p\mathbb{Z})^{\times}$, where $p$ is prime and congruent to 1 mod 4, by $\{\pm 1\}$ contains an order 2 subgroup, whose preimage under the quotient map is then an order 4 subgroup of $(\mathbb{Z}/p\mathbb{Z})^{\times}$. I don't see why the existence of this order 2 subgroup in the quotient is guaranteed.

I would like an explanation that avoids relying on the fact that $(\mathbb{Z}/p\mathbb{Z})^{\times}$ is cyclic, which hasn't yet been proven at this point in the book.

Answer 1

Let $G$ be the quotient of $\Bbb{Z}/p\Bbb{Z}^{\times}$ by $\{\pm1\}$.

$\left\vert \Bbb{Z}/p\Bbb{Z}^{\times}\right\vert=p-1$, so $\vert G\vert=\dfrac{p-1}{2}$.

Since $p\equiv1$ modulo $4$, it follows that $\dfrac{p-1}{2}$ is even. Hence the order of $G$ is even.

At this point, Dummit and Foote are probably using Cauchy's theorem, which says that if a prime $q$ divides the order of a finite group, then the group has an element of order $q$. Since the order of $G$ is even, $G$ has an element of order $2$. Hence $G$ has a subgroup of order $2$.

Answer 2

This also follows from quadratic reciprocity law, which says that, among other things, $$\left( \frac{-1}{p} \right)=(-1)^{(p-1)/2}. $$ In particular, if $p \equiv 1\mod 4$, then $(p-1)/2$ is even, and thus $\left( \frac{-1}{p} \right)=1 $. This implies that there exists an element $x\in (\mathbb Z/p\mathbb Z)^\times$ such that $$x^2\equiv -1.$$ Thus the subgroup of $(\mathbb Z/p\mathbb Z)^\times$ generated by $x$ has order 4.