I have a quick question that is bugging me.
Let $D_{2n}$ be the dihedral group of the $n-$gon (write $\rho$ the "elementary" rotation, and $\sigma$ a reflection), and let $\mathbb{Z}_2$ be the multiplicative group of order $2$.
I am trying to prove that the function $\phi : D_2n \to \mathbb{Z}_2 : \rho^i \sigma^j \mapsto (-1)^i$ is a morphism if and only if $n$ is even.
Here's where it bugs me though:
In my head, to prove that a function was a group morphism, the only thing needed was to prove it preserved structure (ie $\forall x,y \in D_{2n} : \phi(xy) = \phi(x)\phi(y)$).
Now, I believe that this can be proven in both cases. Indeed:
Pick $x = \rho^i \sigma^j, y = \rho^a\sigma^b \in D_{2n}$. $xy = (\rho^i \sigma^j)(\rho^a\sigma^b) = \rho^{i +(-1)^ja}\sigma^{b+j} \implies \phi(xy) = (-1)^{i+a}$.
That sounds right to me (maybe it isn't though, hence why I am asking).
However, in the case where $n$ is odd, the problem arises in the fact that $\rho^n$ = $1_{D_{2n}}$, whereas $\phi(\rho)\phi(\rho^{n-1}) = (-1)^n = -1$. Which means the identity doesn't get sent to $1$, and hence $\phi$ can't be a morphism.
I understand all the steps. The only thing bugging me is that: before coming up with that contradiction in the case where $n$ is odd, $\phi$ seemed to me like it satisfied the "structure-preserving" property all fine.
So, unless we happen to think of that specific example, the function did seem to check all the cases to be a morphism.
So what went wrong? Did I miss something that was meant to tell me "beware of the odd case"? Because had the question not been oriented to tell me that something was going to go wrong with $n$ odd, I would have just checked the structure property seemed to be alright for both cases, and wouldn't have given it much thought.
You can simplify this problem by removing $\sigma$ completely, and just look at the group $C_n$ of rotations of the regular $n$-gon. The same issue appears. So that's what I will do.
The short answer is that your proof that the morphism preserves the multiplicative structure fails if $|i + a|\geq n$. In fact, if we're not careful in how we define $\phi$, it ends up not being a function at all.
If we just say $\phi(\rho^i) = (-1)^i$, then this is not well-defined, because $\rho = \rho^{n+1}$, but $(-1)$ and $(-1)^{n+1}$ might not be the same. We have to specify something along the lines of "$\phi(\rho^i) = (-1)^i$ for $0\leq i<n$". Once we do that, we also see that we have to take care in showing $$\phi(\rho^{i+a}) = \phi(\rho^i\rho^a) = \phi(\rho^i)\phi(\rho^a)$$ if $i+a\geq n$, because the very definition of $\phi$ requires that extra care. And in taking this extra care, we get railroaded into looking at whether $n$ is even or odd, because the above equality boils down to $(-1)^{i+a-n} = (-1)^i(-1)^a$.
If your exercise set told you that $\phi$ was a function, but it was defined exactly as you mention here, then they have made a mistake, as that's not always a function. It's nice if you catch that, but I wouldn't say it's your fault if that's what tripped you up, especially as a beginner student in the subject.
And if they had included the specification I mentioned above, that alone would've reminded you to check what happens if $i$ and $a$ get too large, which immediately would have led to the even-versus-odd issue. So "beware of the odd case" is not really something you have to remember to check explicitly every time, but it appears naturally most of the times where it is relevant.