The following is group of Dirichlet characters taken from Wikipedia and I have a question in that.
Question: why in this table there are no terms involving $i,-i$? I think there must be as there would be 4 characters all of which are solutions of $x^{4} -1=0$ and so there should be $i,-i$. In Apostol's book characters have been given till $7$ and I applied the same rule and got right answer but I am stumped when I tried to use it for $8$.
Kindly tell what concept I am missing.
Thanks!!
I can sort of see your confusion : if $\phi(8) = 4$, then , like Apostol says in the case of $k=5$ where $\phi(5) = 4$, the possible values of $\chi_8(n)$ should be among $\{\pm 1 , \pm i\}$. So, why aren't the values $\pm i$ taken?
The answer , which may be discussed later in Apostol but is certainly avoided for the part we are discussing, is the following : even though it is true that $\chi_k(n)^{\phi(k)} = 1$ for all $(n,k) = 1$, it is NOT true that $\phi(k)$ is the smallest number which has this property. In fact, there is a different quantity for this, called the Carmichael $\lambda$ function.
So what is true, is that $\chi_k(n)^{\lambda(k)} = 1$ for all $(n,k) = 1$. There is a recursive formula for $\lambda(k)$ in terms of the prime factorization for $k$, available here. Using this formula, you can in fact check that $\lambda(8) = 2$. So, the point is that in fact, $\chi_8(n)^2 = 1$ for all $n \in \mathbb N$. This means that $\chi_8(n) = \pm 1$ for all $n$.
So why is there no element of order $4$? The reason why Apostol manages to use all the possible $\phi(k)$th roots of unity for small values of $k$, is because until $8$, it is in fact true that $\lambda(k) = \phi(k)$ : $k=8$ is the first counterexample. Therefore, until that point, there is always going to be at least one image $\chi_k(n)$ that is going to exactly be a $\phi(k)$th root of unity.
It is difficult to explain why $8$ is a counterexample, but the truth is that in general, powers of $2$ are the only ones that behave somewhat awkwardly. For example, in the case of $8$, if you wanted to approach it the "Apostol" way, then $\phi(8)= 4$, so we know that $\chi_8(n)^4 = 1$ for all $n$.
BUT, we also know that : $$ \chi_8(n)(3)^2 = \chi_8(n)(9) = \chi_8(n)(1) = 1 \\ \chi_8(n)(5)^2 = \chi_8(n)(25) = \chi_8(n)(1) = 1\\ \chi_8(n)(7)^2 = \chi_8(n)(49) = \chi_8(n)(1) = 1 $$
So we know that each $\chi_8(n)$ is equal to $\pm 1$ at each odd $n$. The only question is to now evaluate each of the characters. We can use the Apostol approach with plenty of variables as follows :
We need to check how many equations we have : the first is that row-wise sums are zero, therefore $a+b+c = d+e+f = g+h+i=j+k+l = -1$.
Then column-wise sums are zero, so $d+g+j = e+h+k = f+i+l = -1$.
We have intra-character multiplicativity $\chi_8(n)(3) \chi_8(n)(5) = \chi_8(n)(7)$ for all $n$, so $de=f,gh=i,jk=l$.
Note that $\chi_8(4) = \chi_8(3)\chi_8(2)$ because $\chi_8(3)\chi_8(2)$ is a character that cannot be equal to any of $\chi_8(1)$ or $\chi_8(2)$ or $\chi_8(3)$. From this, we get that $\chi_8(2)(l)\chi_8(3)(l) = \chi_8(4)(l)$ for all $l=3,5,7$. This leads to $dg = j, eh = k,fi = l$.
Now, we note that the $\chi_8(i)$ are all different. Let us analyze the quantities $d,e,g,h,j,k$. We can see , for example that both $d=g$ and $e=h$ cannot be true, because then $f=i$ so $\chi_8(2) = \chi_8(3)$ which we know is not true. Therefore, either $d \neq g$ OR $e \neq h$. Similarly, either $g \neq j$ OR $h \neq k$.
In a similar way, both $d=e = 1$ is not possible because then $f=1$ but then $\chi_8(2) = \chi_8(1)$ which is not possible.
So, let's look at the three ordered tuples $(d,e)$, $(g,h)$, $(j,k)$. These are distinct tuples, none of which can be $(1,1)$, but must contain either $\pm 1$ as each entry. That leaves exactly three possibilities, though. We MUST have : $$ \{(d,e),(g,h),(j,k)\} = \{(1,-1),(-1,1),(-1,-1)\} $$ From here, the exercise is quite easy to finish because any one of the tuples can take any of the possibilities, leading to the four characters as follows :
So Apostol's approach works, if a little lengthy.
Note that Wikipedia's approach is significantly different : their observation is that we can actually associate to each of $1,3,5,7$, a canonical character using a formula that assigns to each of the integers a power of $-1$ based on the fact that $\{(-1)^i5^j\}$ after reduction, forms a reduced residue system modulo $2^k$ for all $k$. So this approach is different and cannot be clubbed with Apostol's more primitive approach.
The basic reason why the Carmichael function is different for powers of $2$ versus powers of odd primes, can be explained in two ways : one is to notice that the $\lambda$ function is determined everywhere if it's determined at prime powers , and then to observe that $\lambda(8)=2$ actually lifts (as in , this fact gets used) into $\lambda(2^k) = 2^{k-2}$ for all $k \geq 3$. Meanwhile, for odd primes $p$, it is actually true that $\lambda(p^2) = \phi(p^2)$ which gets lifted onto $\lambda(p^k) = \phi(p^k)$ for all $k$.
Another reason is that one can use "Lifting the exponent" to actually prove the properties of the Carmichael function for prime powers, and over there, a key step that works for odd primes actually fails for $p=2$ (because a certain divisilibity step doesn't work). See page $3$ and the case $n=2$ at page $5$ of this document to see why $p=2$ requires separate treatment and what step fails.