Finite Abelian grouos and their Charactera The Chapter starts with Postulates of a group.

• A group is a three tuple $\langle G,., e \rangle$, where $G$ is a non-empty set of elements and $.$ is the group operator, $e$ is the identity.

Postulates of a Group

• Closure
• Associativity
• Existence of Identity
• Inverse.

An Abelian Group is a commutative group.

$\forall a,b \in G : ab=ba$

Finite Groups : If G is a finite set. Subgroup : A non empty Subset of $G$, $G'$, which itself is a group with the same group operator.

§ Thm 6.1:

If Elements $a,b,c \in G$ satisfy:

$$ ac=bc \lor ca=cb $$

then $a=b$.

If $G'$ is a subgroup of $G$, then for any element $a \in G$ we call $n \in \mathbb N$ to be an Indicator of $a$ if $a^n \in G'$, and $n$ is the smallest positive integer $[\neq 0]$.

§ Thm 6.6

Let $G'$ be a subgroup of a finite abelian group $G$, where $G \neq G'$. Choose an element $a \in G / G'$ and let $h$ be it's indicator. Then the set of products:

is a subgroup of $G$ which contains $G'$. Order of $G''$ is $h$ times that of $G'$.

§ Charaters of Finite abelian groups

§ Definition of Character:

Let $G$ be an arbitrary finite group. A complex valued function $f: G \rightarrow \mathbb C^\times$ defined on $G$ is called a character if $f$ is a group homomorphism. That is, it has the multiplicative property:

§ Theorem 6.7

If f defined over FAG $G$ which has indentity $e$, then $f(e)=1$ and each function value f(a) is a root of unity.

Hence, to be more accurate, we can write $f: G \rightarrow U(1)$ where $U(1) \equiv \{ c \in C : |c| = 1 \}$

§ Theorem 6.8

A FAG of order $n$ has exactly $n$ distinct characters.

$G'$ is a proper subgroup of $G$.. $G''$ is constructed from $G'$ using

• $G_1=\{ e \}$
• $G_2 = \langle G_1,a_1 \rangle$
• $G_3 = \langle G_2,a_2 \rangle$
• $G_{r+1} = \langle G_r,a_r \rangle$

• $G_1$ cleary has 1 distinct character: $f(e) = 1$.
• assume for induction that $G_r$ has $r$ distinct characters.
• Let $h$ be the indicator of $a_r$ in $G_{r+1} = \langle G_r, a_r \rangle$

• From $G_r$ to $G_r+1$ there are exactly $h$ different ways to extend each character of $G_r$

• We'll have $mh$ characters of $G_{r+1}$ which is the same as it's order.

§ How do we extend?

• $f$ is a character on $G_{r+1}$,
• any element in $G_{r+1}$ is of the form $x.a_r^k$
• Let's we extend some character $f$ to $f'$.
• $f'(x \cdot a_r^k)= f'(x)f'(a_r)^k$.
• since $x \in G_r, f'(x) = f(x)$.
• How many choices do we have for $f'(a_r)$?
• Let $h$ be the indicator of $a_r \in G_{r+1}$ relative to $G_r$. That is, $a_r \not \in G_r$, but $a_r^h \in G_r$.

If we have an element $c =a_r^h$ $c \in G_r$, $f'(c) = f(c)$

$f'(a_r)^h = f(c)$ $f'(a_r)^h = 1 \cdot f(c)$ $f'(a_r) = e^{2\pi k/h} \cdot f(c)^{1/h}$

So now for the value of $f'(a_r)$ we have h choices. So we can extend f into h new f'.

TODO: show that this is a legit group hom. [I believe this ] $f'(xa_r^k.ya_r^j) = f'(xa_r^k)f'(ya_r^j)$ [Can be proved ]

§ The Charater Group

This section $G$ is a FAG of order n. The principle/principal character is called $f_1$, is the function $f_1(\_) = 1$. The others, denoted by $f_2,f_3, \dots f_n$ ar called non-principal charactes. They have the property that $f(a)\neq 1$ for some $a \in G$.

§ Thm 6.9

If multiplication of characters is defined by the relation:

• $(f_i \cdot f_j)(a) \equiv f_i(a)f_j(a)$ for each $a \in G$

then set of characters forms an Abelian group of order $n$. The group is dentoed by $\hat G$. The identity is $f_1$. The inverse is of $f_i$ is the reciprocal $1/f_i$.

Note: For each $f$ we have $|f(a)| = 1$, since it is the character of a FAG.

• $1/f(a)$ becomes congugate of f(a).
• $\overline{f}(a) \equiv \overline{f(a)}$ is also a character of G.

Does there exist a morphism $\phi: G \rightarrow \hat G$?

§ Orthogonality relations for characters

$G$ = FAG of order n with elements $a_1, a_2, ... a_n$ and let $f_1, f_2,...,f_n$ be the characters of G, with $f_1$ being principal character.

• We denote $A = A(G)$ the $n\times n$ matrix $[a_{ij}]$ whose element $a_{ij}$ in the $i$th row and $j$th column is

§ Thm 6.10

The sum of entries in the $i$th row of $A$ is given by

• For f i = f 1 every value is 1 so sum is n.

• For $f_i \neq f_1$, there is an element $b \in G$ such that $f(b) \neq 1$ as $a_r$ runs through elements of G so does product $b \cdot a_r$, hence:

• $S(1-f_i(b)) =0$
• $f_i(b) \neq 1$ so $S=0$ [slick how $f_i(b) \neq 1$ shows up. ]

§ Thm 6.11

Let $A^\star$ be the transpose conjugate of $A$. Then we have

• $AA^\star = nI$,

where $I$ is $n\times n$ identity matrix. Hence $n^{-1} A^\star$ is the Inverse of $A$.

Let $B = AA^\star$ The entry $b_{ij}$ in the ith row and jth column of B is

Where $f_k = f_i\overline{(f_j)} = f_i/f_j$. Now $f_i/f_j = f_1$ if and only if $i=j$, Hence by thm 6.10 we have:

$B = nI$

• TODO: this reminds me of a fourier transform; what's the relation?

§ Thm 6.12 --- Orhtogonatilty Relations for Characters

We have:

§ Proof:

Then the left side of the sum is $c_{ij}$. Now since $\overline{f_r}(a_i) = f_r(a_i)^{-1} = f_r(a_i^{-1})$

If $a_i = e$ then we have

§ Thm 6.13:

The sum of entries of jth column of A is given by:

16/05/2020

§ Dirichlet Characters

From here $G$ is group of reduced residue classes modulo a fixed positive integer $k$.

First we prove that $G$ is a group if a multiplication is suitably defined.

§ Reduced Residue system modulo $k$

A set of $\varphi(k)$ integers $\{a_{1},a_{2},\dots,a_{\varphi(k)}\}$ incongruent modulo $k$, each of which is relatively prime to $k$.

• $a_i \not \equiv a_j \mod k \quad \forall i \neq j$
• $\gcd(a_i, k) = 1 \quad \forall i$

For each integer $a$ the corresponding residue class $\hat{a}$ is the set of all integers congruent to $a$ modulo $k$:

We define multiplication of residue classes by the relation:

§ Thm 6.14

With multiplication defined as above, the set of reduced residue classes modulo k is finite abelian group of order $\varphi(k)$. Isn't this group $\mathbb Z/k \mathbb Z^\times$?

• Identity : $\hat{1}$
• Inverse of $\hat{a}$ : $\hat{b}$ such that $ab \equiv1\ mod\ k$.

§ Definition of Dirichlet Characters:

Let $G$ be the group of reduced residue classes modulo k. ( $G = \mathbb Z/k\mathbb Z^\times$) Coressponding to each character f of G, we define an arithmetic function $\chi = \chi_f$ as follows:

§ Thm 6.15

There are $\varphi(k)$ characters modulo k, each of which is completely multiplicative and periodic with period k. i.e.

Commentary (bollu, crypt)k = 3n:     0, 1,        2, 3, 4, 5, 6, 7chi n: 0, f(1),f(2),   0, f(1), f(2), 0this cannot  have period less than 3because f(_) is in nth roots of unity,cannot become 0.

Conversly, if $\chi$ is completely multiplicative and preiodic with period $k$ and if $\chi(n) = 0~\text{if}~ (n,k) >1$ then $\chi$ is one of the dirichlet character modulo k.

§ Proof Thm 6.15 (forward):

There are $\varphi(k)$ characters f of G, hence there are $\varphi(k)$ characters $\chi_f$ modulo $k$. The multiplicative property follows from f when both m,n are relatively prime to k. If one of them is not relatively prime, then neither is $mn$ hence both values become 0. The periodicity property follows from the fact that $\chi_f(n) = f(\hat{n})$ and that $a\equiv b \mod k$ implies $\gcd(a,k) = \gcd(b,k)$.

§ Thm 6.15: Proof of converse

To prove the converse, we note that the function f defined on the group G by the equation:

is a character of G, so \chi is a dirichlet character mod k.

The image of $\chi$ must be a root of unity because

§ Thm 6.16

Let $\chi_1, \chi_2 \dots \chi_{\varphi(k)}$ denote the $\varphi(k)$ dirichlet characters modulo k. Let m and n be two integers with $\gcd(n,k)=1$ Then we have:

§ Proof:

• If $\gcd(m,k)=1$ take $a_i = \hat{n}$ and $a_j = \hat{m}$ in the orthogonailty relation of theorem 6.12 and note that $\hat{m} = \hat{n}$ if and only if $m\equiv n \mod k$.
• If $\gcd(m,k) > 1$ each term in the sum vanishes and

$m \not \equiv n \mod k$.

§ Sums Involving Dirichlet characters:

§ Thm 6.17

Let $\chi$ be any non-principal character modulo k, let $f$ be a non-negative function which has a continuous negative derivative $f'(x)$ for all $x \ge x_0$. Then if $y \ge x \ge x_0$ we have:

If in addition $f(x) \rightarrow 0$ as $x \rightarrow \infty$ then the infinte series

converges and we have for $x\ge x_0$,

Proof: Let $A(x) = \sum_{n\le x} \chi(n)$. Since $\chi$ is non principal we have

Because sum of nth roots of unity is equal to 0 for n strictly greater than 1. $\chi$ evaluates to nth roots of unity plus some extra zeroes over [1..k]

By periodicity it follows that A(nk) = 0 for n=2,3.... hence $|A(x)| < \varphi(k)$ for all x. i.e A(x) = O(1).

Notice that we only need to care about the last unevaluated period. This period has to be of length less that $\phi(k)$. If it were equal to $\phi(k)$, the sum over this would be $0$. Now in this last period, which we shift to $[0\dots \texttt{leftover}]$, we get:

Note that $|\chi(n)| \leq 1$ since $\chi(n)$ evaluates to either 0 or a $\varphi(k)$ th root of unity whose absolute value is 1.

§ Chapter 4 thm 4.2:

Abel's Identity : For any arithmetical function $a(n)$ let

where $A(x) = 0$ if $x<1$. Assume $f$ has a continuous derivative on the interval $[y,x]$ where $0<y<x$. Then we have:

§ Proof:

Let $k =[x]$ and $m=[y]$ so that $A(x)=A(k)$ and $A(y)=A(m)$. Then:

• Link to proof on wikibooks

§ Proof of (7)

We note that

by using the fact that (1) $f'(t)$ does not change sign, (2) $|A(x)| \leq 1$.

If $f(x) \rightarrow 0$ as $x \rightarrow \infty$ then eqn 7 shows that the series

converges because of cauchy convergence (TODO) criterion. Mayhaps the proof is:

To prove eqn 8 we simply note that

Because of eqn 7, the limit on the right if O(f(x)). This completes the proof.

Mow we apply thm 6.17 successively with f(x) = 1/x, f(x) = (log(x))/x and f(x) = 1/\sq THm 6.18