Cardinality and Elements of a Quotient Module I'm stuck in the computation of the isomorphism that gives:
 $$\mathbb{Z}/(7)\cong \frac{\mathbb{Z}/(21)}{(7)/(21)}$$ where $(a)$ denotes the ideal (aka $\mathbb{Z}$-module) generated by the integer $a\in\mathbb{Z}$ and the fraction of the right side is a quotient $\mathbb{Z}$-module.   

What is the cardinality of $(7)/(21)$, how is it computed, which are its elements?

Thank you in advanced :)
 A: Writing out the elements explicitly, $$ (7) = \{ -14, -7, 0 , 7 , 14, 21, \cdots \}$$ and $$ (21) = \{ -21, 0 , 21, 42, \cdots \}.$$ Consider what the elements of $(7)$ are modulo $(21).$ Some examples: $$-14 + (21) = 7 + (21), \ \ \ -7+(21) = 14+(21).$$
In general, all the residues of $(7)$ modulo $(21)$ are $$ \frac{(7)}{(21)} = \{ 0+(21), 7+(21), 14+(21)\}.$$ 
So $\dfrac{(7)}{(21)} \cong \mathbb{Z}/(3)$ by the isomorphism $\phi(7n+(21)) = n\mod 3.$ Check this is infact an isomorphism. 
Now consider $$\mathbb{Z}/(21)=\{ 0+(21), 1+(21), \cdots, \ 20+(21)\}.$$
So computing its residues modulo $(7)/(21)$ (which we computed above) we see $$ \frac{\mathbb{Z}/(21)}{(7)/(21)} = \left\{ 0+(21) + \frac{(7)}{(21)}, 1+ (21)+\frac{(7)}{(21)}, \cdots, \  6+(21)+\frac{(7)}{(21)} \right\}.$$
Just take a look at the elements on the right hand side, I'm sure you can suspect how that group behaves - just like $\mathbb{Z}/(7),$ since the extra $(21)+\dfrac{(7)}{(21)}$ terms are just kind of "tagged along" in the arithmetic. So one strongly suspects that $$ \frac{\mathbb{Z}/(21)}{(7)/(21)} \cong \mathbb{Z}/(7)$$ by the isomorphism $$\varphi\left(n+ (21)+\dfrac{(7)}{(21)}\right) = n\mod 7.$$ You should check for yourself that this is indeed an isomorphism.
