# $\mathbb {Z}_{84}/(7) \cong \mathbb {Z}_{7}$

Prove $$\mathbb {Z}_{84}/(7) \cong \mathbb {Z}_{7}$$ using each of the three isomorphism theorems for rings.

For the first isomorphism theorem I defined a homomorphism $$\phi: \mathbb {Z}_{84} \to \mathbb {Z}_{7}$$ defined by $$\phi(x+84\mathbb {Z}):=x+7\mathbb {Z}$$ and found that $$ker(\phi) = (7)$$ and said that $$\phi$$ is clearly surjective. Is this thinking correct?

Moving onto the other two isomorphism theorems I'm having trouble with these.

For the second theorem, I think that using the homomorphism $$\phi(x+7\mathbb {Z}):=x+7\mathbb {Z}$$ and $$S = \mathbb{Z}/84\mathbb{Z}$$ and $$I = 7\mathbb{Z}$$ might work, but I'm having trouble showing $$S+I=\mathbb{Z}$$ and $$S ∩ I = 7\mathbb{Z}$$

For the third theorem, I think that using the homomorphism $$\phi(x+7\mathbb {Z}):=x+84\mathbb {Z} + 7\mathbb {Z}$$ and $$R = \mathbb{Z}$$ and $$I=84\mathbb{Z}$$ and $$J = 7\mathbb{Z}$$ might work, but I'm having trouble showing $$J/I=7\mathbb{Z}$$

Also would a general form of this $$\mathbb {Z}_{m}/(n) \cong \mathbb {Z}_{n}$$ if $$n|m$$?

Your answer for the first isomorphism theorem is correct, but you would have to check that $$\phi$$ is well-defined upto choice of representative.
For the second isomorphism theorem, $$S$$ must be a subring of $$\mathbb{Z}$$; $$\mathbb{Z}/84\mathbb{Z}$$ is not a subring of $$\mathbb{Z}$$.
For the third theorem, $$J/I$$ is an ideal of $$\mathbb{Z}/84\mathbb{Z}$$ not $$\mathbb{Z}$$, and in fact $$J/I = 7\mathbb{Z}/84\mathbb{Z} \simeq 7(\mathbb{Z}/84\mathbb{Z})$$. The result is then an immediate application of the third isomorphism theorem.