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Prove that $\mathbb{Z}_p^{\times}/(\mathbb{Z}_p^{\times})^2$ is isomorphic to $\{\pm1\}$, where $p$ is a prime integer.

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I assume that $\mathbb{Z}_p$ here means $\mathbb{Z}/p\mathbb{Z}$, instead of the $p$-adic integers (which are usually denoted by $\mathbb{Z}_p$)? –  Álvaro Lozano-Robledo Dec 13 '11 at 23:30
If so, the claim is false for $p=2$. –  Chris Eagle Dec 13 '11 at 23:31
Jason's earlier question: math.stackexchange.com/questions/91247/… –  Zev Chonoles Dec 13 '11 at 23:34
It's likely he means $\mathbb{Z}/p\mathbb{Z}$, given his recent question math.stackexchange.com/questions/91247/…. Also seems that this is homework. –  Brandon Carter Dec 13 '11 at 23:35
I would also assume that he does not mean the $p$-adics, since $[\mathbb{Z}_p^\times : (\mathbb{Z}_p^\times)^2] = 4$ when $p$ is odd, and $8$ when $p=2$. –  Brandon Carter Dec 13 '11 at 23:37

3 Answers 3

I take it that you mean to prove that $\mathbb{F}_p^\times/(\mathbb{F}_p^\times)^2 \cong \{\pm 1\}$, where $\mathbb{F}_p = \mathbb{Z}/p\mathbb{Z}$.

If so, use the fact that the map $(\mathbb{Z}/p\mathbb{Z})^\times \to \{\pm 1\}$ given by $a\bmod p\mapsto (\frac{a}{p})$ is a homomorphism of groups, where $(\frac{a}{p})$ is the Legendre symbol of $a$ over $p$.

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Mod $p$, there is an equal number of quadratic residues and nonresidues (ignoring 0 of course). Therefore, the index of $(\mathbb{Z}_p^\times)^2$ in $\mathbb{Z}_p^\times$ is 2, and so the quotient is isomorphic to $\{\pm 1\}$, as this is the only group of order 2.

Edited to only use words from group theory: Note that $\mathbb{Z}_p^\times$ is cyclic of order $p-1$. We'll show that the subgroup $(\mathbb{Z}_p^\times)^2$, which is the set of squares in $\mathbb{Z}_p^\times$, has index 2. Then the quotient is $\{\pm 1\}$, since this is the only group of order 2.

Let $a$ be a generator for $\mathbb{Z}_p^\times$. Then the elements of $\mathbb{Z}_p^\times$ are $a,a^2,\ldots,a^{p-1}=1$. If $k$ is even, then $a^k = (a^{k/2})^2$ is a square. However if $k$ is odd, then $a^k$ is not a square, for if $a^k = b^2$ for some $b \in \mathbb{Z}_p^\times$ and $k$ odd, then $b= a^j$ for some $j$, and so $a^k = a^{2j}$. But then $k -2j\equiv 0$ mod $p-1$, which is impossible since $k-2j$ is odd. Since $p-1$ is even, there are the same number of elements of $\mathbb{Z}_p^\times$ which are even and odd powers of $a$, so $(\mathbb{Z}_p^\times)^2$ has index 2.

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Is this Dane from sacramento? –  Jason Smith Dec 13 '11 at 23:40
Pretty close to Sacramento. I'm from Santa Rosa. –  Dane Dec 13 '11 at 23:47

I'll write $\mathbf F_p$ for $\mathbf Z/p\mathbf Z$. We need to assume that $p$ is odd, since all elements of $\mathbf F_2$ are squares. Maybe the easiest way to see the result is by looking at the squaring map $\mathbf F_p^* \to \mathbf F_p^*$, $x \mapsto x^2$. Since multiplication is abelian this map is a group homomorphism, and its image is the set of squares. You'd like to show that the kernel has order $2$. If $x$ is one of the integers $\{1, 2, \ldots, p - 1\}$ and $x^2 - 1$ is divisible by the prime $p$, then what?

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