Definition 7.2.1.
Let \(G\) be a finite abelian group. A character of \(G\) is a homomorphism \(\chi : G \to \bbC^*\text{,}\) the multiplicative group of nonzero complex numbers.
Section 7.2 Characters of finite abelian groups
Subsection 7.2.1 Definition
Remark 7.2.2.
The definition makes sense for infinite abelian groups, but we will not get into that in this course.
Remark 7.2.3.
The idea also makes sense for nonabelian groups, but the definition is slightly different, or, more precisely, the general definition of character happens to simplify in the abelian case. In representation theory, a representation of a group \(G\) is a homomorphism from \(G\) to a (multiplicative) group of matrices. The character of the representation is the map \(G \to \bbC\) taking each group element to the trace of the corresponding matrix in the representation. This is generally not a homomorphism; for matrices \(A,B\text{,}\) \(\operatorname{tr}(A)\operatorname{tr}(B)\) is generally not equal to \(\operatorname{tr}(AB)\text{.}\) Also, an invertible matrix can have trace zero, so characters can take the value zero.
However, if it happens that the matrices are \(1 \times 1\text{,}\) then the trace of the matrix is simply its entry. In this case the character is essentially the same thing as the representation, and it is a homomorphism.
It turns out that for abelian groups we can essentially reduce to the \(1 \times 1\) case. So we can more or less conflate characters with representations, and define characters as homomorphisms, with nonzero values.
Proposition 7.2.4.
Let \(G\) be a finite abelian group of order \(n\) and let \(\chi\) be a character on \(G\text{.}\) For every \(g \in G\text{,}\) \(\chi(g)\) is an \(n\)th root of unity.Proof.
By Lagrange’s theorem, \(g^n = 1\text{,}\) the identity in \(G\text{.}\) Then \(\chi(g)^n = \chi(g^n) = \chi(1) = 1\text{.}\)
Definition 7.2.5.
The trivial character which takes every element of \(G\) to \(1\) is denoted \(\chi_0\) and called the principal character of \(G\text{.}\)
The set of all characters of \(G\) is denoted \(\hat{G}\) (or \(\widehat{G}\)) and called the dual group of \(G\text{.}\) We define a multiplication on \(\hat{G}\text{,}\) pointwise: for \(\chi_1, \chi_2 \in \hat{G}\text{,}\) the product \(\chi_1 \chi_2\) is defined by
\begin{equation}
(\chi_1 \chi_2)(g) = (\chi_1(g))(\chi_2(g))\text{.}\tag{7.2.1}
\end{equation}
Proposition 7.2.6.
For every finite abelian group \(G\text{,}\) the set \(\hat{G}\) is a group with the multiplication defined above. The identity element is the principal character. For \(\chi \in \hat{G}\text{,}\) the inverse is the character denoted \(\chi^{-1}\) or \(\overline{\chi}\text{,}\) defined by \(\chi^{-1}(g) = (\chi(g))^{-1}\) or \(\overline{\chi}(g) = \overline{\chi(g)}\) (these are equal).
Proof.
It is left as an exercise to check that the pointwise multiplication is associative and that the principal character is an identity element. The fact that \((\chi(g))^{-1} = \overline{\chi(g)}\) follows from observing that since \(\chi(g)\) is a root of unity, then it has norm \(1\text{;}\) \(z^{-1} = \overline{z}\) holds for any complex number of norm \(1\) (exercise).
Our next goal is to prove that in fact \(\hat{G} \cong G\text{.}\) We start with cyclic groups.
Definition 7.2.7.
For a real number \(t\) we define \(e(t) = \exp(2 \pi i t) = \cos(2 \pi t) + i \sin(2 \pi t)\text{.}\)
Proposition 7.2.8.
Let \(G = \langle g_0\) be a cyclic group of order \(n\text{.}\) There are exactly \(n\) characters of \(G\text{,}\) given as follows. For each \(m\) let \(\chi_m\) be defined by \(\chi_m(g_0^k) = e(mk/n)\text{.}\) Then \(\hat{G} = \{ \chi_0 = 1, \chi_1, \dotsc, \chi_{n-1} \}\text{.}\) In particular, \(\hat{G} = \langle \chi_1 \rangle\) is cyclic of order \(n\text{.}\)
Proof.
First we verify that each \(\chi_m\) is a character. We have to check that \(\chi_m(ab) = \chi_m(a)\chi_m(b)\text{.}\) This holds because
\begin{equation}
\chi_m(g_0^k g_0^\ell) = \chi_m(g_0^{k+\ell}) = e(m(k+\ell)/n)
= e(mk/n) e(m\ell/n) = \chi_m(g_0^k) \chi_m(g_0^\ell)\text{.}\tag{7.2.2}
\end{equation}
Next we check that they are pairwise distinct. Observe that \(\chi_m(g_0^1) = e(m/n)\text{.}\) These are different for different values of \(m\) in the range \(0 \leq m \lt n\text{.}\) (It is true that \(e(m/n) = e((m+n)/n)\text{,}\) and indeed, \(\chi_m = \chi_{m+n}\text{.}\) This is why we stop our list at \(\chi_{n-1}\text{;}\) continuing would simply repeat the characters already on our list.) Since the functions have different values at the point \(g_0\text{,}\) they are different functions.
Now we check that every character is one of the \(\chi_m\text{.}\) Let \(\chi\) be any character of \(G\text{.}\) The value \(\chi(g_0)\) must be an \(n\)th root of unity. So, it is equal to \(e(m/n)\) for some \(m\text{,}\) with \(0 \leq m \lt n\text{.}\) Then
\begin{equation}
\chi(g_0^k) = \chi(g_0)^k = e(m/n)^k = e(mk/n) = \chi_m(g_0^k)\tag{7.2.3}
\end{equation}
for every \(k\text{.}\) Therefore \(\chi = \chi_m\text{.}\)
Finally, we check that \(\hat{G} = \langle \chi_1 \rangle\) is cyclic. Indeed, for every \(m\) and \(k\text{,}\)
\begin{equation}
(\chi_1^m)(g_0^k) = (\chi_1(g_0^k))^m = (e(k/n))^m = e(mk/n) = \chi_m(g_0^k)\text{.}\tag{7.2.4}
\end{equation}
The fact that this holds for all \(k\) means \(\chi_1^m = \chi_m\text{.}\) So, the powers of \(\chi_1\) give all the elements of \(\hat{G}\text{.}\)
Proposition 7.2.9.
For finite abelian groups \(G\) and \(H\text{,}\) \(\widehat{G \times H} \cong \hat{G} \times \hat{H}\text{.}\)
Proof.
We define a mapping \(\hat{G} \times \hat{H} \to \widehat{G \times H}\) as follows. Let \(\chi_1 \in \hat{G}\) and \(\chi_2 \in \hat{H}\text{.}\) Define a character \((\chi_1,\chi_2)\) on \(G \times H\) by
\begin{equation}
(\chi_1,\chi_2)(g,h) = \chi_1(g) \chi_2(h)\text{.}\tag{7.2.5}
\end{equation}
This is a homomorphism from \(\hat{G} \times \hat{H}\) to \(\widehat{G \times H}\) (exercise). It is injective. (If \(\chi_1\) is nonprincipal on \(G\) or \(\chi_2\) is nonprincipal on \(H\text{,}\) then \(\chi_1(g_1) \neq 1\) for some \(g_1 \in G\) or likewise \(\chi_2(h_1) \neq 1\) for some \(h_1 \in H\text{;}\) hence \((\chi_1,\chi_2)(g_1,1_H) = \chi_1(g_1) \neq 1\) or similarly \((\chi_1,\chi_2)(1_G,h_1) = \chi_2(h_1) \neq 1\text{,}\) and either way, \((\chi_1,\chi_2)\) is nonprincipal on \(G \times H\text{.}\))
The map is surjective. For any character \(\chi\) on \(G \times H\text{,}\) define \(\chi_1 : G \to \bbC^*\) by \(\chi_1(g) = \chi(g,1_H)\) and define \(\chi_2 : H \to \bbC^*\) by \(\chi_2(h) = \chi(1_G,h)\text{.}\) Then \(\chi_1\) is a character on \(G\text{,}\) \(\chi_2\) is a character on \(H\text{,}\) and \(\chi = (\chi_1,\chi_2)\) (exercise).
In particular, if \(\hat{G} \cong G \) and \(\hat{H} \cong H \text{,}\) then \(\widehat{G \times H} \cong G \times H \text{.}\)
Theorem 7.2.10. Fundamental Theorem of Finite Abelian Groups.
Every finite abelian group is a direct product of cyclic groups. That is, any finite abelian group \(G\) is
\begin{equation}
G \cong \bbZ/m_1 \bbZ \times \dotsb \times \bbZ/m_k \bbZ\text{.}\tag{7.2.6}
\end{equation}
By the Chinese Remainder Theorem, if \(m = p_1^{a_1} \dotsm p_k^{a_k}\text{,}\) then
\begin{equation}
\bbZ / m \bbZ \cong \bbZ / p_1^{a_1} \bbZ \times \dotsb \times \bbZ / p_k^{a_k} \bbZ\text{.}\tag{7.2.7}
\end{equation}
So if we choose to, we can assume that the decomposition of an arbitrary finite abelian group is into cyclic groups of prime power order.
Corollary 7.2.11.
Let \(G\) be any finite abelian group. Then \(\hat{G} \cong G \text{.}\) In particular, \(|\hat{G}| = |G|\text{.}\)
Proof.
By induction on the number of cyclic factors in a decomposition of \(G\) as a direct product of cyclic groups.
Subsection 7.2.2 Sums of character values
We are also interested in taking sums of character values (just as earlier we considered sums of roots of unity).
Theorem 7.2.12.
Let \(G = \langle g_0 \rangle\) be a cyclic group of order \(n\text{.}\) Then the following hold.
- If a character \(\chi\) of \(G\) is given and we sum over elements of \(G\text{,}\) then\begin{equation} \sum_{g \in G} \chi(g) = \begin{cases} n, \amp \text{if } \chi = \chi_0 \\ 0, \amp \text{otherwise} \end{cases}\text{.}\tag{7.2.8} \end{equation}
- If an element \(g \in G\) is given and we sum over characters, then\begin{equation} \sum_{\chi \in \hat{G}} \chi(g) = \begin{cases} n, \amp \text{if } g = 1 \\ 0, \amp \text{otherwise} \end{cases}\tag{7.2.9} \end{equation}
Proof.
First suppose \(\chi = \chi_m\) is given. Then the sum is
\begin{equation}
\sum_{g \in G} \chi(g) = \sum_{k=0}^{n-1} \chi_m(g_0^k)
= \sum_{k=0}^{n-1} e(mk/n)\text{.}\tag{7.2.10}
\end{equation}
Since \(e(m/n)\) is an \(n\)th root of unity, the first claim follows.
Next, suppose \(g = g_0^k\) is given. Then the sum is
\begin{equation}
\sum_{\chi \in \hat{G}} \chi(g) = \sum_{m=0}^{n-1} \chi_m(g)
= \sum_{m=0}^{n-1} e(mk/n)\text{.}\tag{7.2.11}
\end{equation}
Here, \(e(k/n)\) is an \(n\)th root of unity, so once again the result follows.
(The textbook includes here the result that \(\hat{G}\) is cyclic of order \(n\) when \(G\) is. I omit that here because we already showed it earlier.)
We will generalize this to non-cyclic groups in the next section.
