Determine whether $\phi$ is a homomorphism

Question

1. Let $\phi: \mathbb{Z}_6 \rightarrow \mathbb{Z}_2$ be given by $\phi(x)=$the remainder of $x$ when divided by $2$, as in the division algorithm.
2. Let $\phi: \mathbb{Z}_9 \rightarrow \mathbb{Z}_2$ be given by $\phi(x)=$the remainder of $x$ when divided by $2$, as in the division algorithm.

They seem trivial but I cannot prove them in detail. I claim that 1 is a homomorphism while 2 is not.

Question 2: We simply note that $$\phi(1+8)=0 \neq 1 = \phi(1) + \phi(8).$$

For question 1, I write a proof like this:

For any $a, b\in \mathbb{Z}_6$, write $a+b=c$. Note that

1. $a$ is even, $b$ is even $\Rightarrow$ $c$ is even
2. $a$ is odd, $b$ is odd $\Rightarrow$ $c$ is even
3. $a$ & $b$ are of different parity $\Rightarrow$ $c$ is odd

We check that in all of the three cases, $\phi(a+b) = \phi(a)+\phi(b)$.

But I didn't make use of the assumption that $\phi$ is from $\mathbb{Z}_6$. In fact if I didn't take a careful look at the elements of $\mathbb{Z}_9$, I would just copy this proof to conclude that question 2 is a homomorphism! So, what is wrong with my proof? How can I improve it? Thanks in advance.

Answer 1

Your arguments are correct, and your observation and question are good. To lay your worries to rest you just need to realize that parity for elements in $\mathbb Z_n$ only makes sense when $n$ is even. Indeed, if $n$ is odd then for any $k\in \mathbb Z_n$, since $k=n+k$, but as integers they have different parities, it follows that you can't define the parity of elements in $\mathbb Z_n$, for $n$ odd. For $n$ even there is no problem and indeed your argument for $\mathbb Z_6$ generalizes easily.

It is possible to object to what I just said by defining parity in $\mathbb Z_n$ by simply looking at $\mathbb Z_n=\{0,1,2,3,\ldots \}$ and declaring these elements to be even or odd alternatively, starting with $0$ being even. This avoids referring to parity of integers. This is fine, but then the result that "the sum of two even elements in $\mathbb Z_n$ is even iff both elements have the same parity" is true iff $n$ is even. This again explains the behaviour you see.