I need to show that $$\mathbb Z_6/([3]) \simeq \mathbb Z_3$$ How would I use the first isomorphism theorem to show this?
2026-03-27 15:18:02.1774624682
Show that $\mathbb Z_6/([3]) \simeq \mathbb Z_3$
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Hint:
Consider the homomorphism: \begin{align} \mathbf Z/6\mathbf Z &\longmapsto \mathbf Z/3\mathbf Z \\ n+6\mathbf Z&\longmapsto n+3\mathbf Z \end{align} Show it is well defined. What is its kernel?