Prove that there exists no primitive character mod $k$ if $k=2m$, where $m$ is odd. This is from Apostol Chapter 8 Exercise 5 extended for all characters, real or not.

(Corrected using solution of Bruno Joyal)

Using Apostol Thm 8.16: $m|k$ so take $(a,k)=(b,k)=1$ (so that $a,b$ are odd) and take $a\equiv b \pmod m$.

Then $a\equiv b \pmod k$ as either

$a=b+rm\equiv b \pmod k$ if $r$ even or

$a\equiv b+m \pmod k$ if $r$ odd. But $b,m$ odd so $(b+m,k)\not=(a,k)=1$. A contradiction.

Hence whenever $(a,k)=(b,k)=1$ with $a\equiv b \pmod m$ then $a\equiv b \pmod k$ so $\chi(a)=\chi(b)$ for every character mod $k$.

Thus $m$ is an induced modulus for $\chi$ so there can be no primitive characters mod $k$.


There is a problem with your solution. You are right that $\chi(a)=-1$ for at least one $a$, but you seem to conclude (falsely) that this holds for all $a$ with $(a, 2m)=1$.

The idea is that $(\mathbb{Z}/2m\mathbb{Z})^\times \cong (\mathbb{Z}/m\mathbb{Z})^\times$ whenever $m$ is odd. Hence any character mod $2m$ actually lifts to a character mod $m$ by composing with this isomorphism. In fact, there are no primitive characters mod $2m$, real or not - all of them come from characters mod $m$.

  • $\begingroup$ Thanks - you were too quick for me - I saw that error but could not correct it any further. $\endgroup$ – apatch Nov 9 '11 at 22:35
  • $\begingroup$ You're welcome! Does my answer help? $\endgroup$ – Bruno Joyal Nov 9 '11 at 22:46

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