Let $f$ be a map from quotient ring from $\mathbb{Z}/8\mathbb{Z}$ to $\mathbb{Z}/4\mathbb{Z}$ defined by $f(z+8\mathbb{Z}) = z+4\mathbb{Z}$. Check that whether or not that $f$ is isomorphism.
My attempt: (Edited)
For homomorphism, let $x+8\mathbb{Z},y+8\mathbb{Z} \in \mathbb{Z}/8\mathbb{Z}$. Then, \begin{align*} f((x+8\mathbb{Z})+(y+8\mathbb{Z})) &= f((x+y)+8\mathbb{Z}) \\ &= (x+y)+4\mathbb{Z} \\ &= (x+4\mathbb{Z})+(y+8\mathbb{Z}) \\ &= f(x+8\mathbb{Z}) + f(y+8\mathbb{Z}) \end{align*}
and \begin{align*} f((x+8\mathbb{Z})(y+8\mathbb{Z})) &= f(xy + 8\mathbb{Z}) \\ &= xy+4\mathbb{Z} \\ &= (x+4\mathbb{Z})(y+4\mathbb{Z}) \\ &= f(x+8\mathbb{Z})f(y+8\mathbb{Z}) \end{align*}
Thus, $f$ is a homomorphism.
Next, $f$ is injective iff $Ker(f) = \{\bar{0}\}$.
If $x+8\mathbb{Z} \in Ker(f)$, then $0+4\mathbb{Z} = f(x+8\mathbb{Z}) = x+4\mathbb{Z}$. This implies that $4|a$. If $4|a$, then $f(x+8\mathbb{Z})=x+4\mathbb{Z} = 0+4\mathbb{Z}$ and $x+8\mathbb{Z} \in Ker(f)$.
It follows that $Ker(f) = \{8\mathbb{Z},4+8\mathbb{Z}\} \neq \{\bar{0}\}$.
And then, $f$ is not isomorphism.
Am I true? Where is the mistakes of mine above? Any idea? Thanks in advanced.
