# Question to the proof of: Let $A$ be a finite abelian group and let $g \in A$. Suppose that $\chi(g)=1$ for every $\chi \in \hat A$. Then $g=1$.

Good day,

Currently I am working with the book "A First Course in Harmonic Analysis" by A. Deitmar and I am stuck in the beginning of Chapter 5 on the proof to Lemma 5.1.5. I am repeating the few definitons/theorems that came before.

Theorem 5.1.1 Any finite abelian group $A$ is isomorphic to a product $A_1 \times A_2 \times ... \times~ A_k$ of cyclic groups.

Let $A$ be a finite group. A character $\chi$ of A is a group homomorphism $\chi: A \to T$ to the unit torus. Let $\hat A$ be the set of all characters of A.

Lemma 5.1.2 The pointwise product $(\chi, \eta) \mapsto \chi \eta$ with $\chi \eta (a) = \chi(a) \eta(a)$ makes $\hat{A}$ an abelian group. We call $\hat A$ the dual group, or Pontryagin dual, of $A$.

Lemma 5.1.3 Let A be cyclic of order N. Fix a generator $\tau$ of A, i.e. $A= \{ 1, \tau, \tau^2, ..., \tau^{N-1} \}$ and $\tau^N=1$. The characters of the group $A$ are given by $\eta_l(\tau^k)=e^{2 \pi i k l /N}, k \in \mathbb{Z}$, for $l=0,1,...,N-1$. The group $\hat A$ is again cyclic of order N.

Lemma 5.1.5 Let $A$ be a finite abelian group and let $g \in A$. Suppose that $\chi(g)=1$ for every $\chi \in \hat A$. Then $g=1$.

Proof:

1. Lemma 5.1.3 show that the claim holds for cyclic groups. In the light of Theorem 5.1.1 it remains to show that if the claim holds for the groups $A$ and $B$, then is also holds for $A \times B$.

2. For this let $(a_0,b_0) \in A \times B$ with $\eta(a_0,b_0)=1$ for all $\eta \in \widehat{A \times B}$.

3. For every $\chi \in \hat A$ the map $\chi(a,b)=\chi(a)$ is a character of $A \times B$, and therefore $\chi(a_0)=1$, which implies $a_0=1$ and similarly $b_0=1$.

I have to say I am confused by this proof. My thoughts:

1. Is $A$ in the proof another one as in the stated Lemma? I think so. I would have thought like this: Let $\mathbb{A}$ be a finite abelian group, then by Theorem 5.1.1 $\mathbb{A}$ is isomorphic to $A \times B$ where $A$ and $B$ are cyclic groups. (I will take this notation.)
2. Okay I get this, I have to prove for $g \in \mathbb{A}$: $$\chi(g)=1 ~\forall \chi \in \hat{\mathbb{A}} \Rightarrow g=1$$ Since $\mathbb{A} \cong A \times B$, I can prove equivalently for $(a,b) \in A \times B$: $$\eta(a,b)=1 ~\forall \eta \in \widehat{A \times B} \Rightarrow (a,b)=(1,1)$$
3. Since I know that Lemma 5.1.5 holds for cyclic groups by Lemma 5.1.3, that means for all $\chi \in \hat A$ and $\beta \in \hat B$, I have the results $$\big( \chi(a)=1 \Rightarrow a=1 \big) ~\text{and}~ \big( \beta(b)=1 \Rightarrow b=1 \big).$$ I don't get the statement $\chi(a,b)=\chi(a)$ from the proof which concludes the proof. First why should this be a character on $A \times B$? I have for $(a_1,b_1), (a_2,b_2) \in A \times B$: $$\chi\left( (a_1, b_1) \cdot (a_2, b_2) \right)= \chi(a_1a_2, b_1b_2)=\chi(a_1a_2) = \chi(a_1) \cdot \chi(a_2) = \chi(a_1,b) \cdot \chi(a_1,b) \neq \chi(a_1, b_1) \cdot \chi(a_2, b_2)$$ And even if it is a homomorphism, I have to prove the result for ALL characters of $A \times B$. Why should every character represented by such a structure? If all this would be fulfilled, okay, then I could conclude: $$1=\chi(a,b)=\chi(a) \Rightarrow a=1$$

I hope that my problems are clear. If not just ask again. I am thankful for every help.

Thanks a lot, Marvin

The author is assuming that we have an abelian group $G$ that decomposes as $$G= A \times B,$$ where $A$ and $B$ are cyclic groups. Furthermore, we get to assume that for all characters $\chi$ we have $\chi((a_0,b_0))=1$, for some $(a_0,b_0) \in A \times B$. But since the map $$\varphi_{\chi}: A \times B \to T$$ given by $\varphi_{\chi}((a,b))=\chi(a)$ for some character $\chi$ is a homomorphism (check this yourself), we have that the map $\varphi_{\chi}$ is a character of $A \times B$. But we know that the theorem holds for cyclic groups, and since $\varphi_{\chi}((a_0,b_0))=\chi(a_0)=1$ for all characters $\chi$, it must be that $a_0=1$. Similarly for $B$. Hence $(a_0,b_0)=1$, the multiplicative identity in the direct product.

To conclude the proof, one may use induction on the number of factors in the decomposition of an abelian group. But this is simple now that we have the two factor case (try it!).

• Thank you for your answer. For the homomorphism I have to check $$\varphi( (a_1,b_1) \cdot (a_2,b_2) ) = \varphi (a_1,b1) \varphi(a_2,b_2)$$ $$LHS=\varphi(a_1a_2,b_1b_2)=\chi(a_1a_2)=\chi(a_1)\chi(a_2)$$ $$RHS=\chi(a_1) \chi(a_2)$$ Is this okay? By induction you mean I have to show this result still holds for $G=A_1 \times A_2 \times ... \times A_n$? I will try it. – Fritz Apr 24 '16 at 17:32
• Okay, the result is valid for $$A=A_1 \times A_2 \times ... \times A_n$$ and now I have to show it for $n+1$, that means $$A=A_1 \times A_2 \times ... \times A_n \times A_{n+1}.$$ Let us define $B:=A_1 \times A_2 ... \times A_n$. Then $A=B \times A_{n+1}$. Ah damn, $B$ need not be cyclic. Okay, I have to overthink this. – Fritz Apr 24 '16 at 17:37
• Okay, but by induction hypothesis the result is valid for $B$. So we are again in the case of two factors. Thank you :) – Fritz Apr 24 '16 at 17:43
• @MarvinF. In this case $A$ should be $A_{n+1}$, but otherwise you got it! – leibnewtz Apr 24 '16 at 23:58

Yes, there are two different $A$'s and two different $\chi$'s, which is confusing. You at least resolved the problem of the $A$'s correctly.

$A,B$ in the direct product need not be cyclic (in fact, we may not be able to make both cyclic). What we need is that for every abelian non-cyclic $\Bbb A$, there is a product decomposition as $A\times B$ with smaller factors (so we can perform induction).

Given a character $\chi\in \hat A$, we find a character (rather given the different name) $\eta\in\widehat{A\times B}$ given by $\eta(a,b):=\chi(a)$. As $\eta(a_0,b_0)=1$, we conclude $\chi(a_0)=1$. This holds for all $\chi\in \hat A$. By induction hypothesis (because $|A|<|\Bbb A|$), we conclude that $a_0=1$. The symmetrical argument for the other factor shows that $b_0=1$.