The Set of Square Elements in the Multiplicative Group $(\Zmod{p})^*$

by Yu · 11/27/2017

Problem 616

Suppose that $p$ is a prime number greater than $3$.
Consider the multiplicative group $G=(\Zmod{p})^*$ of order $p-1$.

(a) Prove that the set of squares $S=\{x^2\mid x\in G\}$ is a subgroup of the multiplicative group $G$.

(b) Determine the index $[G : S]$.

(c) Assume that $-1\notin S$. Then prove that for each $a\in G$ we have either $a\in S$ or $-a\in S$.

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Contents

Problem 616
Proof.

Proof.

(a) Prove that $S=\{x^2\mid x\in G\}$ is a subgroup of $G$.

Consider the map $\phi:G \to G$ defined by $\phi(x)=x^2$ for $x\in G$.
Then $\phi$ is a group homomorphism. In fact, for any $x, y \in G$, we have
\begin{align*}
\phi(xy)=(xy)^2=x^2y^2=\phi(x)\phi(y)
\end{align*}
as $G$ is an abelian group.

By definition of $\phi$, the image is $\im(\phi)=S$.
Since the image of a group homomorphism is a group, we conclude that $S$ is a subgroup of $G$.

(b) Determine the index $[G : S]$.

By the first isomorphism theorem, we have
\[G/\ker(\phi)\cong S.\]

If $x\in \ker(\phi)$, then $x^2=1$.
It follows that $(x-1)(x+1)=0$ in $\Zmod{p}$.
Since $\Zmod{p}$ is an integral domain, it follows that $x=\pm 1$ and $\ker(\phi)=\{\pm 1\}$.

Thus, $|S|=|G/\ker(\phi)|=(p-1)/2$ and hence the index is
\[[G:S]=|G|/|S|=2.\]

(c) Assume that $-1\notin S$. Then prove that for each $a\in G$ we have either $a\in S$ or $-a\in S$.

Since $-1\notin S$ and $[G:S]=2$, we have the decomposition
\[G=S\sqcup (-S).\] Suppose that an element $a$ in $G$ is not in $S$.

Then, we have $a\in -S$.
Thus, there exists $b\in S$ such that $a=-b$.
It follows that $-a=b\in S$. Therefore, we have either $a\in S$ or $-a\in S$.

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