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How shall I show that all non-zero element of $\mathbb Z_3[i]$ form an abelian group of group of order $8$ under multiplication...

Please any hint how shall I show this result?

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2 Answers 2

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$\mathbb (Z_3[i])^\times$ = $(a+bi)$ s. t. $(a,b)\in\mathbb Z_3\times \mathbb Z_3-(0,0)$ and $(a+bi)^{-1}=(a-bi)\cdot(a^2+b^2)^{-1}$ this it's true because $a^2+b^2\neq 0$ and this element have an inverse in $\mathbb Z_3 ^\times$ .Finally this group is obviusly abelian whit 8 elements.

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Hint :

$\mathbb{Z}_{3}[i]$ = $\{a+ib\ |\ a,b\in \mathbb{Z_3}\}$ and $|\mathbb{Z_3}|$ = $3$, so $|\mathbb{Z_{3}}[i]|$ = $9$. Can you argue from this point ?

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