B2 - theorems

theorem 1: criterion for

a^{i} = a^{j}

$a_{i} = a_{j} iff n | i - j$

for $a \in G$ , if the order of $a$ is infinite, all distinct powers of $a$ will be distinct elements, so $a_{i} \neq a_{j}$ for $i \neq j$
if it is finite, $| a | = n$ , $⟨ a ⟩ = {e, a, a^{2}, \dots, a^{n - 1}}$ and $a^{n} = e$
so for any $k \in Z > n$ , $a^{k} = a^{k \mod n}$

theorem 2

for $a \in G$ such that $| a | = n :$

⟨ a^{k} ⟩ = ⟨ a^{gcd (n, k)} ⟩, | a^{k} | = \frac{n}{gcd (n, k)}

eg: consider $Z_{12}$
taking $⟨ a^{4} ⟩ = {0, 4, 8}$ , here the gcd is $4$ and its order is $(12 / 4 =) 3$
taking $⟨ a^{8} ⟩ = {0, 8, 4}$ , here the gcd is $4$ and its order is $(12 / 4 =) 3$
taking $⟨ a^{5} ⟩ = {0, 5, 10, 3, 8, 1, 6, 11, 4, 9, 2, 7,}$ , here the gcd is $1$ and its order is $(12 / 1 =) 12$
note that for relatively prime integers (gcd $= 1$ ), the entire set is generated

corollary 1

$G = ⟨ a^{k} ⟩$ iff $gcd (n, k) = 1$ , ie. relatively prime to both

corollary 2

$Z_{n} = ⟨ k ⟩$ iff $gcd (n, k) = 1$

theorem 3: the fundamental theorem of cyclic groups

every subgroup of a cyclic group is cyclic
furthermore, if $| ⟨ a ⟩ = n$ , the order of any subgroup is a divisor of $n$ , and for each positive integer $k$ that divides $n$ , $a$ has exactly one subgroup of order $k : ⟨ a^{n} / k ⟩$

eg: for $Z_{30}$ , there are is one subgroup each of order: $1, 2, 3, 5, 6, 10, 15, 30$ , and no others

corollary 1

for each positive integer $k$ that divides $n$ , $⟨ n / k ⟩$ is the unique subgroup of $Z_{n}$ of order $k$

using the previous theorems, the number of elements of each order in a finite subgroup can be easily counted

euler totient

the euler-phi function for $n$ , $ϕ (n)$ , denotes the number of integers less than $n$ that are relatively prime to it
it is equal to the order of the special unitary group

| U (n) | = ϕ (n)

for any prime $p :$

ϕ (p^{n}) = p^{n} - p^{n - 1}

for distinct primes $p_{1}, p_{2}, \dots, p_{m} :$

ϕ (p_{1}^{k_{1}} p_{2}^{k_{2}} \dots p_{m}^{k_{m}}) = ϕ (p_{1}^{k_{1}}) ϕ (p_{2}^{k_{2}}) \dots ϕ (p_{m}^{k_{m}})

theorem 4

if a non-zero positive integer $d$ divides $n$ , the number of elements of order $d$ in a cyclic group of order $n$ is $ϕ (d)$

an element of order $d$ generates a cyclic subgroup of order $d$ , which will have $ϕ (d)$ generators
eg: let $G = ⟨ a ⟩$ with $| G | = 12$ , how many elements of order $6$ ?
from theorem 1:

| a^{k} | = \frac{12}{gcd (12, k)} = 6

this is satisfied by $k = 2, 10$ , so $a^{2}$ and $a^{10}$ are of order $6$
this is the same as predicted by theorem 3, $ϕ (12) = 2$
in a cycle of length $6$ , steps of $1$ and $5$ are the only ones that generate the entire subgroup

corollary

in a finite group, the number of elements of order $d$ is divisible by $ϕ (d)$

to find the order of an element $a \in G$ :
- find the divisors of $| G |$ , say $k$
- check if the divisors satisfy $a^{k} = e$
- for any order $n k$ , if $a^{n k} \neq e$ , $a^{k} \neq e$ either
eg: consider $3$ in $U (50)$
$| U (50) | = ϕ (50) = 20$ , so $| 3 |$ must be $2, 4, 5, 10$ or $20$
$3^{4} = 31 ⟹$ not $4$ or $2$
$3^{10} = 3^{5} 3^{5} = 243 \cdot 243 = (- 7) (- 7) = 49 ⟹$ not $10$ or %5
therefore, $| 3 | = 20$