Preferably explained in novice terms! I can start it off by having the multiplicative group modulo 8 with elements $[1], [3], [5], [7]$ and not sure where to go now.

I see there is a similar question on here but its answer is not simple enough for me!

I can provide the definition of a (non) primitive character:

Let $q \in \mathbb{N}$ and $\chi_{0}$, the principal character modulo q. A Dirichlet character $\chi$ mod q is called non-primitive if $\exists$ $q_{1}$ $< q$ with $q_{1}|q$ and a Dirichlet character $\chi_{1}$ mod $q_{1}$ which satisfies:

$\chi(n) = \chi_{0}(n) \chi_{1}(n) \, \, \forall n \in \mathbb{Z}$

Thus it's primitive if it's not non-primitive.


The group $(\mathbb Z/8\mathbb Z)^*$ has order $4$ and it is isomorphic to $C_2\times C_2$, where $C_2$ is the cyclic group with 2 elements. Now $\hom(C_2\times C_2,\mathbb C^*)\simeq \hom(C_2,\mathbb C^*)\times \hom(C_2,\mathbb C^*)$, which means that there are 4 characters of $(\mathbb Z/8\mathbb Z)^*$ because $|\hom(C_2,\mathbb C^*)|=2$. Now $\hom(C_2,\mathbb C^*)$ is the following set: there is the trivial morphism sending everything to $1$ and there is the morphism sending $t\mapsto -1$, if $C_2=\{1,t\}$. So we just need to describe an isomorphism $(\mathbb Z/8\mathbb Z)^*\to C_2\times C_2$ and then compose everything. For example one isomorphism is the map sending $[1]\mapsto (1,1)$, $[3]\mapsto (t,1)$, $[5]\mapsto (1,t)$ and $[7]\mapsto (t,t)$. This means that the four characters are the following: the trival one, the map sending $[1],[3]\mapsto 1$ and $[5],[7]\mapsto -1$, the map sending $[1],[5]\mapsto 1$ and $[3],[7]\mapsto -1$ and finally the one sending $[1],[7]\mapsto 1$ and $[3],[5]\mapsto -1$.

Now you have to check which one of them are primitive and which others are not. This is easy because the only nontrivial divisors of $8$ are $2,4$. Now clearly none of these characters can be defined modulo $2$ (apart from the principal one of course), so you just have to check modulo $4$. For example, the first nontrivial character that I mentioned cannot be defined modulo $4$ because $3\equiv 7\bmod 4$ but $[3]$ and $[7]$ have different images.

  • $\begingroup$ Thanks, I am not very good at analytic number theory. Could you please explain why the other morphism is sending t to -1? $\endgroup$ – Becca May 20 '14 at 12:40
  • $\begingroup$ Because $t$ has order $2$, so it must be sent to another element of order $2$, and the only such element in $\mathbb C^*$ is $-1$ $\endgroup$ – Ferra May 20 '14 at 12:43

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