Prove that $\mathbb{Z}_p^{\times}/(\mathbb{Z}_p^{\times})^2$ is isomorphic to $\{\pm1\}$, where $p$ is a prime integer.
Prove that $\mathbb{Z}_p^{\times}/(\mathbb{Z}_p^{\times})^2$ is isomorphic to $\{\pm1\}$.
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abstract-algebra
group-theory
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2I assume that $\mathbb{Z}_p$ here means $\mathbb{Z}/p\mathbb{Z}$, instead of the $p$-adic integers (which are usually denoted by $\mathbb{Z}_p$)? – 2011-12-13
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4If so, the claim is false for $p=2$. – 2011-12-13
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1Jason's earlier question: http://math.stackexchange.com/questions/91247/prove-that-mathbbz-3-mathbbz-times-mathbbz-3-mathbbz-times2 – 2011-12-13
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0It's likely he means $\mathbb{Z}/p\mathbb{Z}$, given his recent question http://math.stackexchange.com/questions/91247/prove-that-mathbbz-3-mathbbz-times-mathbbz-3-mathbbz-times2. Also seems that this is homework. – 2011-12-13
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0Is $(\mathbb{Z}_p^{\times})^2$ the set of quadratic residues in $\mathbb{Z}_p^\times$? Since precisely half the elements of $\mathbb{Z}_p^\times$ are quadratic residues, $\mathbb{Z}_p^{\times}/(\mathbb{Z}_p^{\times})^2$ is then a group of order $2$. – 2011-12-13
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1I 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 $p=2$. – 2011-12-13
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0What does $(\mathbb{Z}_p^x)^2$ mean. I was taking it to mean the cross product of $(\mathbb{Z}_p^x)$. Perhaps this is the confusion. – 2011-12-13