Let $G$ be finite abelian group and $\hat G$ be its character group.

I need hint proving that if $a\in G$ and $\chi(a)=1$ for all $\chi\in\hat G$ then $a=0$ (the identity element).

I can prove it for cyclic groups but not in general.

  • $\begingroup$ Do you mean $a=e$ the identity? $\endgroup$ – Ethan Jan 10 '14 at 10:41
  • $\begingroup$ @Ethan Yes, since it is abelian group I write $0$ $\endgroup$ – Ashot Jan 10 '14 at 10:41
  • $\begingroup$ Why? $\mathbb{R}^{\times}$ is an abelian group under multiplication and it has identity $1$. Surely only for additive groups do we generally write $0$? $\endgroup$ – fretty Jan 10 '14 at 10:44
  • $\begingroup$ For your question note that all finite abelian groups have their cyclic decomposition and the character group decomposes too. This should answer your question. $\endgroup$ – fretty Jan 10 '14 at 10:45
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    $\begingroup$ Hint: Find a proper non-trivial subgroup $H$ not containing $a$ (show such exists when the group is not cyclic), look at $G/H$ and do induction on the order of $G$. $\endgroup$ – Tobias Kildetoft Jan 10 '14 at 10:45

If you can prove it for the special case of cyclic groups, then you can reduce the general case to the special one.

Let us say you want to prove that if $a \ne 0$, then there is a character $\chi$ of $G$ such that $\chi(a) \ne 1$.

Show that there is a subgroup $N$ of $G$ such that $a \notin N$ (that is, $a N \ne 0$ in $G/N$) and $G/N$ is cyclic.

Then take a character $\psi$ of $G/N$ such that $\psi(aN) \ne 1$, and lift $\psi$ to the character $\chi$ of $G$ defined by $$ \chi(x) = \psi (x N). $$

  • $\begingroup$ What is an easy argument for why such a subgroup exists (when $G$ is not cyclic of course)? $\endgroup$ – Tobias Kildetoft Jan 10 '14 at 10:56
  • $\begingroup$ @TobiasKildetoft I think it can be done by interpreting elements of $G$ as vectors and consider a subgroup of $G$ where a fixed coordinate is $0$(which is not $0$ in $a$) and other coordinates are arbitrary. $\endgroup$ – Ashot Jan 10 '14 at 11:04
  • $\begingroup$ @Ashot as vectors over what? Why should it be possible to include $a$ in a "basis"? $\endgroup$ – Tobias Kildetoft Jan 10 '14 at 11:48
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    $\begingroup$ Actually, I just realized that it follows by induction once any non-trivial subgroup not containing $a$ is found (and this is indeed easy to show). $\endgroup$ – Tobias Kildetoft Jan 10 '14 at 11:50
  • $\begingroup$ @TobiasKildetoft, I had only thought of using the structure theorem for finite abelian groups (which is, I believe, the argument Ashot is also suggesting), but I was sure that there is an elementary argument, and yours (+1) is straight from The Book. $\endgroup$ – Andreas Caranti Jan 10 '14 at 11:53

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