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I want to prove that $(\mathbb{Z}/3\mathbb{Z})^\times/((\mathbb{Z}/3\mathbb{Z})^\times)^2$ is isomorphic to $\{\pm1\}$.

What I'm having trouble seeing is what the elements of $G=(\mathbb{Z}/3\mathbb{Z})^\times/((\mathbb{Z}/3\mathbb{Z})^\times)^2$ look like. When one writes $a \cdot ((\mathbb{Z}/3\mathbb{Z})^\times)^2$, $a\in (\mathbb{Z}/3\mathbb{Z})^\times$, do they mean $(a,1) \cdot ((\mathbb{Z}/3\mathbb{Z})^\times)^2$?

Also, am I correct in thinking that $(\mathbb{Z}/3\mathbb{Z})^\times=\{1,2\}$ is exactly the group $\{1,-1\}$?

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What is $\mathbb Z^x_3$? –  Peteris Dec 13 '11 at 21:35
    
Deleted my comment since the latex'ed edits change the question. –  sxd Dec 13 '11 at 21:39
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2 Answers

up vote 2 down vote accepted

$(\mathbb{Z}/3\mathbb{Z})^{\times}$ has two elements, say $\{1,2\}$ and the subgroup of squares is just $\{1^2,2^2\}=\{1\}$. so the quotient is a two element group (all of which are isomorphic).

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Yes, $(\mathbb{Z}/3\mathbb{Z})^{\times}$ is isomorphic to the multiplicative group $\{1,-1\}$.

In any group $G$ written multiplicatively, if $N$ is a subgroup and $a\in G$, then the coset $aN$, as a set, is $$aN = \{an\mid n\in N\}.$$ If $N$ is normal, then $G/N$, as a set consists of all cosets of $N$, with operation $aN\cdot bN = abN$.

So, in your example, $N= ((\mathbb{Z}/3\mathbb{Z})^{\times})^2$, so $$a \cdot( (\mathbb{Z}/3\mathbb{Z})^{\times})^2 = \left\{an\mid n\in(\mathbb{Z}/3\mathbb{Z}^{\times})^2\right\}.$$

What you write, $$(a,1) \cdot ((\mathbb{Z}/3\mathbb{Z})^\times)^2$$ doesn't make sense because $(a,1)$ is not an element of the group.

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