Finding the order of an element in a group $mathbb{Z}_{12}$
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I'm trying to calculate the order of an element in a group. I saw a simple example: find the order of an element $3$ in $mathbb{Z}_{12}$. First they wrote that $langle 3 rangle = {0,3,6,9}$ but I could not understand why. As I understand, we are looking for all the number $ain mathbb{Z}_{12}$ so there exists an $nin mathbb{N}$ so $3^n=a$. So it should be $langle 3 rangle = {3,9}$ because there is no $nin mathbb{N}$ so $3^n=0$ or $3^n=6$. Am I missing something?
Also, is there a faster way to determine the order? What if I would like to calculate the order of $7$ when $mathbb{Z}_{66}$?
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I'm trying to calculate the order of an element in a group. I saw a simple example: find the order of an element $3$ in $mathbb{Z}_{12}$. First they wrote that $langle 3 rangle = {0,3,6,9}$ but I could not understand why. As I understand, we are looking for all the number $ain mathbb{Z}_{12}$ so there exists an $nin mathbb{N}$ so $3^n=a$. So it should be $langle 3 rangle = {3,9}$ because there is no $nin mathbb{N}$ so $3^n=0$ or $3^n=6$. Am I missing something?
Also, is there a faster way to determine the order? What if I would like to calculate the order of $7$ when $mathbb{Z}_{66}$?
group-theory
In group theory notation. There is one binary operation which may be written as $*$ or $cdot$ or $times$ or $+$ or $circ$ or not written at all. Then the notation $3^n$ means to apply that operation $n$ times. In $mathbb Z_{12}$ the operation is $*=+$. So $3^0 =$ the identity equals $0$. $3^1 = 3$. $3^2 = 3*3=3+3 = 6$ and $3^3 = 3*3*3 = 3+3+3=9$ and $3^4 = 3*3*3*3= 3+3+3+3 = 0$. We can go one step weirder and not $3^k = ktimes 3$. Because the group is about addition multiplication ... doesn't exist... not in the current universe.
– fleablood
Nov 18 at 0:25
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up vote
2
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up vote
2
down vote
favorite
I'm trying to calculate the order of an element in a group. I saw a simple example: find the order of an element $3$ in $mathbb{Z}_{12}$. First they wrote that $langle 3 rangle = {0,3,6,9}$ but I could not understand why. As I understand, we are looking for all the number $ain mathbb{Z}_{12}$ so there exists an $nin mathbb{N}$ so $3^n=a$. So it should be $langle 3 rangle = {3,9}$ because there is no $nin mathbb{N}$ so $3^n=0$ or $3^n=6$. Am I missing something?
Also, is there a faster way to determine the order? What if I would like to calculate the order of $7$ when $mathbb{Z}_{66}$?
group-theory
I'm trying to calculate the order of an element in a group. I saw a simple example: find the order of an element $3$ in $mathbb{Z}_{12}$. First they wrote that $langle 3 rangle = {0,3,6,9}$ but I could not understand why. As I understand, we are looking for all the number $ain mathbb{Z}_{12}$ so there exists an $nin mathbb{N}$ so $3^n=a$. So it should be $langle 3 rangle = {3,9}$ because there is no $nin mathbb{N}$ so $3^n=0$ or $3^n=6$. Am I missing something?
Also, is there a faster way to determine the order? What if I would like to calculate the order of $7$ when $mathbb{Z}_{66}$?
group-theory
group-theory
asked Nov 17 at 23:59
vesii
635
635
In group theory notation. There is one binary operation which may be written as $*$ or $cdot$ or $times$ or $+$ or $circ$ or not written at all. Then the notation $3^n$ means to apply that operation $n$ times. In $mathbb Z_{12}$ the operation is $*=+$. So $3^0 =$ the identity equals $0$. $3^1 = 3$. $3^2 = 3*3=3+3 = 6$ and $3^3 = 3*3*3 = 3+3+3=9$ and $3^4 = 3*3*3*3= 3+3+3+3 = 0$. We can go one step weirder and not $3^k = ktimes 3$. Because the group is about addition multiplication ... doesn't exist... not in the current universe.
– fleablood
Nov 18 at 0:25
add a comment |
In group theory notation. There is one binary operation which may be written as $*$ or $cdot$ or $times$ or $+$ or $circ$ or not written at all. Then the notation $3^n$ means to apply that operation $n$ times. In $mathbb Z_{12}$ the operation is $*=+$. So $3^0 =$ the identity equals $0$. $3^1 = 3$. $3^2 = 3*3=3+3 = 6$ and $3^3 = 3*3*3 = 3+3+3=9$ and $3^4 = 3*3*3*3= 3+3+3+3 = 0$. We can go one step weirder and not $3^k = ktimes 3$. Because the group is about addition multiplication ... doesn't exist... not in the current universe.
– fleablood
Nov 18 at 0:25
In group theory notation. There is one binary operation which may be written as $*$ or $cdot$ or $times$ or $+$ or $circ$ or not written at all. Then the notation $3^n$ means to apply that operation $n$ times. In $mathbb Z_{12}$ the operation is $*=+$. So $3^0 =$ the identity equals $0$. $3^1 = 3$. $3^2 = 3*3=3+3 = 6$ and $3^3 = 3*3*3 = 3+3+3=9$ and $3^4 = 3*3*3*3= 3+3+3+3 = 0$. We can go one step weirder and not $3^k = ktimes 3$. Because the group is about addition multiplication ... doesn't exist... not in the current universe.
– fleablood
Nov 18 at 0:25
In group theory notation. There is one binary operation which may be written as $*$ or $cdot$ or $times$ or $+$ or $circ$ or not written at all. Then the notation $3^n$ means to apply that operation $n$ times. In $mathbb Z_{12}$ the operation is $*=+$. So $3^0 =$ the identity equals $0$. $3^1 = 3$. $3^2 = 3*3=3+3 = 6$ and $3^3 = 3*3*3 = 3+3+3=9$ and $3^4 = 3*3*3*3= 3+3+3+3 = 0$. We can go one step weirder and not $3^k = ktimes 3$. Because the group is about addition multiplication ... doesn't exist... not in the current universe.
– fleablood
Nov 18 at 0:25
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3 Answers
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up vote
4
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I think your mistake with the theory is notational. In some group theory settings, "$3^n$" would represent the group operation combining the element with itself n times. This is just for convenience, as it looks cleaner to talk about elements in an abstract group like "$a^2b^3$", sort of how you would imagine integers being factored like primes.
Therefore, when the operation over the set $mathbb{Z}_k$ is + , "$3^n$" really means "$3n$," where you use the standard quotient to take remainders (over k).
You might try and use the greatest common divisor to figure out how large the set is. You want to see how large the set is in question, but the set should be linear. It only "wraps around" after a certain point, which I believe occurs at $frac{k}{text{gcd}(a, k)}$ for an element "a" and a modulus "k."
add a comment |
up vote
2
down vote
In a group we have an operation. We don't know what the operation actually IS but it is an operation.
Now maybe we shouldn't but we tend to use the notation of multiplication. So we'll often write $ain G; b in G$ and $a*b = c$ as $ab = c$. That doesn't actually mean the operation is arithmetic multiplication. It could be any group operation.
We use the notation $a^k; a in G; k in mathbb N$ to mean $underbrace{a*a*a*a....*a}_{ktext{ times}}$. This is similar to the concept of multiplication but the operation isn't necessarily arthemetical multiplication. It is the group operation.
Perhaps we shouldn't. If $*$ is not multiplication the $a^k$ is not multiplicative exponents. But we are doing ABSTRACT algebra. $*$ is not multiplication; it is any operation we want. Aid $a^k$ is not mmultiplicative exponents. It is exponents on our operation whatever it is.
And if our operation is ADDITION, nothing changes. $3^k = underbrace{3 + 3+ .... + 3}_{k text{ times}}$.
So
$3^2 = 3*3 = 3+3pmod {12} = 6$.
And $3^3 = 3*3*3 = 3+3+3pmod {12} = 9$. And $3^4 = 3*3*3*3 = 3+3+3+3pmod {12} = 0$.
In fact. If $a*b = a + bpmod {12}$ and $a^k = a+a+a..... + a pmod {12} = ktimes a pmod {12}$.
The thing is... As for as the additive group $mathbb Z_{12}$ goes the only operation we have is addition and multiplication does not exist. So thinking of "powers" as exponents over multiplication just... doesn't make sense. If anything they are "exponents of addition".
......
Also, is there a faster way to determine the order? What if I would like to calculate the order of 7 when Z66?
Okay $7^2 = 7+7 = 2times 7 = 14$ and ....
we want to find the smallest $k$ where $7^k = ktimes 7 equiv 0 pmod {66}$.
Well, trial and error shows us $7^9 = 9times 7 =63$ and $7^{10} = 10times 7 = 70 equiv 4 pmod {66}$ and.... and $7^{19} =19times 7 19 = 133equiv 1 pmod{66}$ and $7^{20} = 20times 7 = 140 equiv 8pmod{66}$ and... sheesh, how long do we have to go?!
Oh, heck... here's a secret.
We want $7^k = ktimes 7 equiv 0 pmod {66}$.
That means there is a $N$ so that $7k = 66N$. So $k = frac {66N}{7}$. Now $7$ is relatively prime to $66$ so $frac m7$ must be an integer. so let $frac N7 = v$.
Then $k = 66*v$. So $frac {k}{66} = v$. The smallest possible $k$ is if $j = 66$ and indeed $7^{66} = 66times 7 equiv 0 pmod{66}$ and we know that no smaller such number exists because.... well, $7$ and $66$ are relatively prime....
In general the formula if if $|a| = k$ in $mathbb Z_m$ so that $a^k = 0 implies ktimes a = mtimes N$ for some $N$ the smallest such $k$ that can do this is if $ktimes a = mtimes N$ is the least common multiple of $a$ and $m$. ANd that would be when $k = frac {m}{gcd(a,m)}$.
Hence the formula you saw. $|a| = frac {|mathbb Z_m|}{gcd(a, |mathbb Z_m|)}$. (because $|mathbb Z_m| = m$).
====
Actually John Nash's answer has a good insight for this.
$mathbb Z_n$ is a cyclic group generated by $1$. $a= 1+1+1....= 1^a$ and $a^k = 1^{ak}$ and $0 = 1^{nm}$ fr some $m$.
So the order $k$ of $a$ will be occur with $ak=nm$ is the least common multiple of $n$ and $a$. I.E. if $k = frac{n}{gcd(n,a)}$.
(If you are curious what rule $m = frac {a}{gcd(n,a)}$ plays... well it doesn't really play any role. It's how many times $a$ has to "double back" because its remainder doesn't divide "nicely" into $n$. It's ... only as significant as we want it to be.)
I saw the use of the following equation: $|x|=frac{|mathbb{Z_{m}}|}{gcd(|mathbb{Z_{m}}|,x)}$. But for this question it will give us: $|7|=frac{66}{gcd(66,7)}=66$ and not $19$. Am I missing something again?
– vesii
Nov 18 at 20:08
Argh... No! You are right and I am wrong! I made a very foolish mistake! $1$ is not the identity for addition. $0$ is the identity. (I was even going to point that out but I thought that would confuse things). SO we want $7^k = e$ or $ktimes 7 equiv 0 mod 66$ not $ktimes 7 equiv 1 mod 66$. !!!! BIG difference! ... I'm sorry and I'll fix it.
– fleablood
Nov 18 at 21:56
It's funny beecause in making the mistake $7^{19} =1$ I made the exact same mistake you did. I forgot for a moment we were dealing with ADDITION and not multiplication. Because this is addition we need $7^k = ktimes 0 equiv 0 pmod{66}$.
– fleablood
Nov 18 at 22:11
I'm surprised no-one else caught me on this!
– fleablood
Nov 18 at 22:15
that's ok! thanks for the answer. Is there a similar formula for $mathbb{Z}_{n}^{times}$?
– vesii
Nov 19 at 16:29
|
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up vote
1
down vote
$Z$₁₂ is a cyclic group under addition. So the generator 'a' of the form ={na | for some n€ $Z$ }.
In order to find order of $3$ we need to do as follows:
<3>={3,6,9,0}={0,3,6,9}
here n=4 for which <3> is zero. Hence order of $3$ is $4$.
This answer has a very good insight. I think the OP would do well to consider it.
– fleablood
Nov 19 at 16:58
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
I think your mistake with the theory is notational. In some group theory settings, "$3^n$" would represent the group operation combining the element with itself n times. This is just for convenience, as it looks cleaner to talk about elements in an abstract group like "$a^2b^3$", sort of how you would imagine integers being factored like primes.
Therefore, when the operation over the set $mathbb{Z}_k$ is + , "$3^n$" really means "$3n$," where you use the standard quotient to take remainders (over k).
You might try and use the greatest common divisor to figure out how large the set is. You want to see how large the set is in question, but the set should be linear. It only "wraps around" after a certain point, which I believe occurs at $frac{k}{text{gcd}(a, k)}$ for an element "a" and a modulus "k."
add a comment |
up vote
4
down vote
I think your mistake with the theory is notational. In some group theory settings, "$3^n$" would represent the group operation combining the element with itself n times. This is just for convenience, as it looks cleaner to talk about elements in an abstract group like "$a^2b^3$", sort of how you would imagine integers being factored like primes.
Therefore, when the operation over the set $mathbb{Z}_k$ is + , "$3^n$" really means "$3n$," where you use the standard quotient to take remainders (over k).
You might try and use the greatest common divisor to figure out how large the set is. You want to see how large the set is in question, but the set should be linear. It only "wraps around" after a certain point, which I believe occurs at $frac{k}{text{gcd}(a, k)}$ for an element "a" and a modulus "k."
add a comment |
up vote
4
down vote
up vote
4
down vote
I think your mistake with the theory is notational. In some group theory settings, "$3^n$" would represent the group operation combining the element with itself n times. This is just for convenience, as it looks cleaner to talk about elements in an abstract group like "$a^2b^3$", sort of how you would imagine integers being factored like primes.
Therefore, when the operation over the set $mathbb{Z}_k$ is + , "$3^n$" really means "$3n$," where you use the standard quotient to take remainders (over k).
You might try and use the greatest common divisor to figure out how large the set is. You want to see how large the set is in question, but the set should be linear. It only "wraps around" after a certain point, which I believe occurs at $frac{k}{text{gcd}(a, k)}$ for an element "a" and a modulus "k."
I think your mistake with the theory is notational. In some group theory settings, "$3^n$" would represent the group operation combining the element with itself n times. This is just for convenience, as it looks cleaner to talk about elements in an abstract group like "$a^2b^3$", sort of how you would imagine integers being factored like primes.
Therefore, when the operation over the set $mathbb{Z}_k$ is + , "$3^n$" really means "$3n$," where you use the standard quotient to take remainders (over k).
You might try and use the greatest common divisor to figure out how large the set is. You want to see how large the set is in question, but the set should be linear. It only "wraps around" after a certain point, which I believe occurs at $frac{k}{text{gcd}(a, k)}$ for an element "a" and a modulus "k."
edited Nov 18 at 0:17
answered Nov 18 at 0:11
theREALyumdub
900617
900617
add a comment |
add a comment |
up vote
2
down vote
In a group we have an operation. We don't know what the operation actually IS but it is an operation.
Now maybe we shouldn't but we tend to use the notation of multiplication. So we'll often write $ain G; b in G$ and $a*b = c$ as $ab = c$. That doesn't actually mean the operation is arithmetic multiplication. It could be any group operation.
We use the notation $a^k; a in G; k in mathbb N$ to mean $underbrace{a*a*a*a....*a}_{ktext{ times}}$. This is similar to the concept of multiplication but the operation isn't necessarily arthemetical multiplication. It is the group operation.
Perhaps we shouldn't. If $*$ is not multiplication the $a^k$ is not multiplicative exponents. But we are doing ABSTRACT algebra. $*$ is not multiplication; it is any operation we want. Aid $a^k$ is not mmultiplicative exponents. It is exponents on our operation whatever it is.
And if our operation is ADDITION, nothing changes. $3^k = underbrace{3 + 3+ .... + 3}_{k text{ times}}$.
So
$3^2 = 3*3 = 3+3pmod {12} = 6$.
And $3^3 = 3*3*3 = 3+3+3pmod {12} = 9$. And $3^4 = 3*3*3*3 = 3+3+3+3pmod {12} = 0$.
In fact. If $a*b = a + bpmod {12}$ and $a^k = a+a+a..... + a pmod {12} = ktimes a pmod {12}$.
The thing is... As for as the additive group $mathbb Z_{12}$ goes the only operation we have is addition and multiplication does not exist. So thinking of "powers" as exponents over multiplication just... doesn't make sense. If anything they are "exponents of addition".
......
Also, is there a faster way to determine the order? What if I would like to calculate the order of 7 when Z66?
Okay $7^2 = 7+7 = 2times 7 = 14$ and ....
we want to find the smallest $k$ where $7^k = ktimes 7 equiv 0 pmod {66}$.
Well, trial and error shows us $7^9 = 9times 7 =63$ and $7^{10} = 10times 7 = 70 equiv 4 pmod {66}$ and.... and $7^{19} =19times 7 19 = 133equiv 1 pmod{66}$ and $7^{20} = 20times 7 = 140 equiv 8pmod{66}$ and... sheesh, how long do we have to go?!
Oh, heck... here's a secret.
We want $7^k = ktimes 7 equiv 0 pmod {66}$.
That means there is a $N$ so that $7k = 66N$. So $k = frac {66N}{7}$. Now $7$ is relatively prime to $66$ so $frac m7$ must be an integer. so let $frac N7 = v$.
Then $k = 66*v$. So $frac {k}{66} = v$. The smallest possible $k$ is if $j = 66$ and indeed $7^{66} = 66times 7 equiv 0 pmod{66}$ and we know that no smaller such number exists because.... well, $7$ and $66$ are relatively prime....
In general the formula if if $|a| = k$ in $mathbb Z_m$ so that $a^k = 0 implies ktimes a = mtimes N$ for some $N$ the smallest such $k$ that can do this is if $ktimes a = mtimes N$ is the least common multiple of $a$ and $m$. ANd that would be when $k = frac {m}{gcd(a,m)}$.
Hence the formula you saw. $|a| = frac {|mathbb Z_m|}{gcd(a, |mathbb Z_m|)}$. (because $|mathbb Z_m| = m$).
====
Actually John Nash's answer has a good insight for this.
$mathbb Z_n$ is a cyclic group generated by $1$. $a= 1+1+1....= 1^a$ and $a^k = 1^{ak}$ and $0 = 1^{nm}$ fr some $m$.
So the order $k$ of $a$ will be occur with $ak=nm$ is the least common multiple of $n$ and $a$. I.E. if $k = frac{n}{gcd(n,a)}$.
(If you are curious what rule $m = frac {a}{gcd(n,a)}$ plays... well it doesn't really play any role. It's how many times $a$ has to "double back" because its remainder doesn't divide "nicely" into $n$. It's ... only as significant as we want it to be.)
I saw the use of the following equation: $|x|=frac{|mathbb{Z_{m}}|}{gcd(|mathbb{Z_{m}}|,x)}$. But for this question it will give us: $|7|=frac{66}{gcd(66,7)}=66$ and not $19$. Am I missing something again?
– vesii
Nov 18 at 20:08
Argh... No! You are right and I am wrong! I made a very foolish mistake! $1$ is not the identity for addition. $0$ is the identity. (I was even going to point that out but I thought that would confuse things). SO we want $7^k = e$ or $ktimes 7 equiv 0 mod 66$ not $ktimes 7 equiv 1 mod 66$. !!!! BIG difference! ... I'm sorry and I'll fix it.
– fleablood
Nov 18 at 21:56
It's funny beecause in making the mistake $7^{19} =1$ I made the exact same mistake you did. I forgot for a moment we were dealing with ADDITION and not multiplication. Because this is addition we need $7^k = ktimes 0 equiv 0 pmod{66}$.
– fleablood
Nov 18 at 22:11
I'm surprised no-one else caught me on this!
– fleablood
Nov 18 at 22:15
that's ok! thanks for the answer. Is there a similar formula for $mathbb{Z}_{n}^{times}$?
– vesii
Nov 19 at 16:29
|
show 4 more comments
up vote
2
down vote
In a group we have an operation. We don't know what the operation actually IS but it is an operation.
Now maybe we shouldn't but we tend to use the notation of multiplication. So we'll often write $ain G; b in G$ and $a*b = c$ as $ab = c$. That doesn't actually mean the operation is arithmetic multiplication. It could be any group operation.
We use the notation $a^k; a in G; k in mathbb N$ to mean $underbrace{a*a*a*a....*a}_{ktext{ times}}$. This is similar to the concept of multiplication but the operation isn't necessarily arthemetical multiplication. It is the group operation.
Perhaps we shouldn't. If $*$ is not multiplication the $a^k$ is not multiplicative exponents. But we are doing ABSTRACT algebra. $*$ is not multiplication; it is any operation we want. Aid $a^k$ is not mmultiplicative exponents. It is exponents on our operation whatever it is.
And if our operation is ADDITION, nothing changes. $3^k = underbrace{3 + 3+ .... + 3}_{k text{ times}}$.
So
$3^2 = 3*3 = 3+3pmod {12} = 6$.
And $3^3 = 3*3*3 = 3+3+3pmod {12} = 9$. And $3^4 = 3*3*3*3 = 3+3+3+3pmod {12} = 0$.
In fact. If $a*b = a + bpmod {12}$ and $a^k = a+a+a..... + a pmod {12} = ktimes a pmod {12}$.
The thing is... As for as the additive group $mathbb Z_{12}$ goes the only operation we have is addition and multiplication does not exist. So thinking of "powers" as exponents over multiplication just... doesn't make sense. If anything they are "exponents of addition".
......
Also, is there a faster way to determine the order? What if I would like to calculate the order of 7 when Z66?
Okay $7^2 = 7+7 = 2times 7 = 14$ and ....
we want to find the smallest $k$ where $7^k = ktimes 7 equiv 0 pmod {66}$.
Well, trial and error shows us $7^9 = 9times 7 =63$ and $7^{10} = 10times 7 = 70 equiv 4 pmod {66}$ and.... and $7^{19} =19times 7 19 = 133equiv 1 pmod{66}$ and $7^{20} = 20times 7 = 140 equiv 8pmod{66}$ and... sheesh, how long do we have to go?!
Oh, heck... here's a secret.
We want $7^k = ktimes 7 equiv 0 pmod {66}$.
That means there is a $N$ so that $7k = 66N$. So $k = frac {66N}{7}$. Now $7$ is relatively prime to $66$ so $frac m7$ must be an integer. so let $frac N7 = v$.
Then $k = 66*v$. So $frac {k}{66} = v$. The smallest possible $k$ is if $j = 66$ and indeed $7^{66} = 66times 7 equiv 0 pmod{66}$ and we know that no smaller such number exists because.... well, $7$ and $66$ are relatively prime....
In general the formula if if $|a| = k$ in $mathbb Z_m$ so that $a^k = 0 implies ktimes a = mtimes N$ for some $N$ the smallest such $k$ that can do this is if $ktimes a = mtimes N$ is the least common multiple of $a$ and $m$. ANd that would be when $k = frac {m}{gcd(a,m)}$.
Hence the formula you saw. $|a| = frac {|mathbb Z_m|}{gcd(a, |mathbb Z_m|)}$. (because $|mathbb Z_m| = m$).
====
Actually John Nash's answer has a good insight for this.
$mathbb Z_n$ is a cyclic group generated by $1$. $a= 1+1+1....= 1^a$ and $a^k = 1^{ak}$ and $0 = 1^{nm}$ fr some $m$.
So the order $k$ of $a$ will be occur with $ak=nm$ is the least common multiple of $n$ and $a$. I.E. if $k = frac{n}{gcd(n,a)}$.
(If you are curious what rule $m = frac {a}{gcd(n,a)}$ plays... well it doesn't really play any role. It's how many times $a$ has to "double back" because its remainder doesn't divide "nicely" into $n$. It's ... only as significant as we want it to be.)
I saw the use of the following equation: $|x|=frac{|mathbb{Z_{m}}|}{gcd(|mathbb{Z_{m}}|,x)}$. But for this question it will give us: $|7|=frac{66}{gcd(66,7)}=66$ and not $19$. Am I missing something again?
– vesii
Nov 18 at 20:08
Argh... No! You are right and I am wrong! I made a very foolish mistake! $1$ is not the identity for addition. $0$ is the identity. (I was even going to point that out but I thought that would confuse things). SO we want $7^k = e$ or $ktimes 7 equiv 0 mod 66$ not $ktimes 7 equiv 1 mod 66$. !!!! BIG difference! ... I'm sorry and I'll fix it.
– fleablood
Nov 18 at 21:56
It's funny beecause in making the mistake $7^{19} =1$ I made the exact same mistake you did. I forgot for a moment we were dealing with ADDITION and not multiplication. Because this is addition we need $7^k = ktimes 0 equiv 0 pmod{66}$.
– fleablood
Nov 18 at 22:11
I'm surprised no-one else caught me on this!
– fleablood
Nov 18 at 22:15
that's ok! thanks for the answer. Is there a similar formula for $mathbb{Z}_{n}^{times}$?
– vesii
Nov 19 at 16:29
|
show 4 more comments
up vote
2
down vote
up vote
2
down vote
In a group we have an operation. We don't know what the operation actually IS but it is an operation.
Now maybe we shouldn't but we tend to use the notation of multiplication. So we'll often write $ain G; b in G$ and $a*b = c$ as $ab = c$. That doesn't actually mean the operation is arithmetic multiplication. It could be any group operation.
We use the notation $a^k; a in G; k in mathbb N$ to mean $underbrace{a*a*a*a....*a}_{ktext{ times}}$. This is similar to the concept of multiplication but the operation isn't necessarily arthemetical multiplication. It is the group operation.
Perhaps we shouldn't. If $*$ is not multiplication the $a^k$ is not multiplicative exponents. But we are doing ABSTRACT algebra. $*$ is not multiplication; it is any operation we want. Aid $a^k$ is not mmultiplicative exponents. It is exponents on our operation whatever it is.
And if our operation is ADDITION, nothing changes. $3^k = underbrace{3 + 3+ .... + 3}_{k text{ times}}$.
So
$3^2 = 3*3 = 3+3pmod {12} = 6$.
And $3^3 = 3*3*3 = 3+3+3pmod {12} = 9$. And $3^4 = 3*3*3*3 = 3+3+3+3pmod {12} = 0$.
In fact. If $a*b = a + bpmod {12}$ and $a^k = a+a+a..... + a pmod {12} = ktimes a pmod {12}$.
The thing is... As for as the additive group $mathbb Z_{12}$ goes the only operation we have is addition and multiplication does not exist. So thinking of "powers" as exponents over multiplication just... doesn't make sense. If anything they are "exponents of addition".
......
Also, is there a faster way to determine the order? What if I would like to calculate the order of 7 when Z66?
Okay $7^2 = 7+7 = 2times 7 = 14$ and ....
we want to find the smallest $k$ where $7^k = ktimes 7 equiv 0 pmod {66}$.
Well, trial and error shows us $7^9 = 9times 7 =63$ and $7^{10} = 10times 7 = 70 equiv 4 pmod {66}$ and.... and $7^{19} =19times 7 19 = 133equiv 1 pmod{66}$ and $7^{20} = 20times 7 = 140 equiv 8pmod{66}$ and... sheesh, how long do we have to go?!
Oh, heck... here's a secret.
We want $7^k = ktimes 7 equiv 0 pmod {66}$.
That means there is a $N$ so that $7k = 66N$. So $k = frac {66N}{7}$. Now $7$ is relatively prime to $66$ so $frac m7$ must be an integer. so let $frac N7 = v$.
Then $k = 66*v$. So $frac {k}{66} = v$. The smallest possible $k$ is if $j = 66$ and indeed $7^{66} = 66times 7 equiv 0 pmod{66}$ and we know that no smaller such number exists because.... well, $7$ and $66$ are relatively prime....
In general the formula if if $|a| = k$ in $mathbb Z_m$ so that $a^k = 0 implies ktimes a = mtimes N$ for some $N$ the smallest such $k$ that can do this is if $ktimes a = mtimes N$ is the least common multiple of $a$ and $m$. ANd that would be when $k = frac {m}{gcd(a,m)}$.
Hence the formula you saw. $|a| = frac {|mathbb Z_m|}{gcd(a, |mathbb Z_m|)}$. (because $|mathbb Z_m| = m$).
====
Actually John Nash's answer has a good insight for this.
$mathbb Z_n$ is a cyclic group generated by $1$. $a= 1+1+1....= 1^a$ and $a^k = 1^{ak}$ and $0 = 1^{nm}$ fr some $m$.
So the order $k$ of $a$ will be occur with $ak=nm$ is the least common multiple of $n$ and $a$. I.E. if $k = frac{n}{gcd(n,a)}$.
(If you are curious what rule $m = frac {a}{gcd(n,a)}$ plays... well it doesn't really play any role. It's how many times $a$ has to "double back" because its remainder doesn't divide "nicely" into $n$. It's ... only as significant as we want it to be.)
In a group we have an operation. We don't know what the operation actually IS but it is an operation.
Now maybe we shouldn't but we tend to use the notation of multiplication. So we'll often write $ain G; b in G$ and $a*b = c$ as $ab = c$. That doesn't actually mean the operation is arithmetic multiplication. It could be any group operation.
We use the notation $a^k; a in G; k in mathbb N$ to mean $underbrace{a*a*a*a....*a}_{ktext{ times}}$. This is similar to the concept of multiplication but the operation isn't necessarily arthemetical multiplication. It is the group operation.
Perhaps we shouldn't. If $*$ is not multiplication the $a^k$ is not multiplicative exponents. But we are doing ABSTRACT algebra. $*$ is not multiplication; it is any operation we want. Aid $a^k$ is not mmultiplicative exponents. It is exponents on our operation whatever it is.
And if our operation is ADDITION, nothing changes. $3^k = underbrace{3 + 3+ .... + 3}_{k text{ times}}$.
So
$3^2 = 3*3 = 3+3pmod {12} = 6$.
And $3^3 = 3*3*3 = 3+3+3pmod {12} = 9$. And $3^4 = 3*3*3*3 = 3+3+3+3pmod {12} = 0$.
In fact. If $a*b = a + bpmod {12}$ and $a^k = a+a+a..... + a pmod {12} = ktimes a pmod {12}$.
The thing is... As for as the additive group $mathbb Z_{12}$ goes the only operation we have is addition and multiplication does not exist. So thinking of "powers" as exponents over multiplication just... doesn't make sense. If anything they are "exponents of addition".
......
Also, is there a faster way to determine the order? What if I would like to calculate the order of 7 when Z66?
Okay $7^2 = 7+7 = 2times 7 = 14$ and ....
we want to find the smallest $k$ where $7^k = ktimes 7 equiv 0 pmod {66}$.
Well, trial and error shows us $7^9 = 9times 7 =63$ and $7^{10} = 10times 7 = 70 equiv 4 pmod {66}$ and.... and $7^{19} =19times 7 19 = 133equiv 1 pmod{66}$ and $7^{20} = 20times 7 = 140 equiv 8pmod{66}$ and... sheesh, how long do we have to go?!
Oh, heck... here's a secret.
We want $7^k = ktimes 7 equiv 0 pmod {66}$.
That means there is a $N$ so that $7k = 66N$. So $k = frac {66N}{7}$. Now $7$ is relatively prime to $66$ so $frac m7$ must be an integer. so let $frac N7 = v$.
Then $k = 66*v$. So $frac {k}{66} = v$. The smallest possible $k$ is if $j = 66$ and indeed $7^{66} = 66times 7 equiv 0 pmod{66}$ and we know that no smaller such number exists because.... well, $7$ and $66$ are relatively prime....
In general the formula if if $|a| = k$ in $mathbb Z_m$ so that $a^k = 0 implies ktimes a = mtimes N$ for some $N$ the smallest such $k$ that can do this is if $ktimes a = mtimes N$ is the least common multiple of $a$ and $m$. ANd that would be when $k = frac {m}{gcd(a,m)}$.
Hence the formula you saw. $|a| = frac {|mathbb Z_m|}{gcd(a, |mathbb Z_m|)}$. (because $|mathbb Z_m| = m$).
====
Actually John Nash's answer has a good insight for this.
$mathbb Z_n$ is a cyclic group generated by $1$. $a= 1+1+1....= 1^a$ and $a^k = 1^{ak}$ and $0 = 1^{nm}$ fr some $m$.
So the order $k$ of $a$ will be occur with $ak=nm$ is the least common multiple of $n$ and $a$. I.E. if $k = frac{n}{gcd(n,a)}$.
(If you are curious what rule $m = frac {a}{gcd(n,a)}$ plays... well it doesn't really play any role. It's how many times $a$ has to "double back" because its remainder doesn't divide "nicely" into $n$. It's ... only as significant as we want it to be.)
edited Nov 19 at 17:08
answered Nov 18 at 0:50
fleablood
66.6k22684
66.6k22684
I saw the use of the following equation: $|x|=frac{|mathbb{Z_{m}}|}{gcd(|mathbb{Z_{m}}|,x)}$. But for this question it will give us: $|7|=frac{66}{gcd(66,7)}=66$ and not $19$. Am I missing something again?
– vesii
Nov 18 at 20:08
Argh... No! You are right and I am wrong! I made a very foolish mistake! $1$ is not the identity for addition. $0$ is the identity. (I was even going to point that out but I thought that would confuse things). SO we want $7^k = e$ or $ktimes 7 equiv 0 mod 66$ not $ktimes 7 equiv 1 mod 66$. !!!! BIG difference! ... I'm sorry and I'll fix it.
– fleablood
Nov 18 at 21:56
It's funny beecause in making the mistake $7^{19} =1$ I made the exact same mistake you did. I forgot for a moment we were dealing with ADDITION and not multiplication. Because this is addition we need $7^k = ktimes 0 equiv 0 pmod{66}$.
– fleablood
Nov 18 at 22:11
I'm surprised no-one else caught me on this!
– fleablood
Nov 18 at 22:15
that's ok! thanks for the answer. Is there a similar formula for $mathbb{Z}_{n}^{times}$?
– vesii
Nov 19 at 16:29
|
show 4 more comments
I saw the use of the following equation: $|x|=frac{|mathbb{Z_{m}}|}{gcd(|mathbb{Z_{m}}|,x)}$. But for this question it will give us: $|7|=frac{66}{gcd(66,7)}=66$ and not $19$. Am I missing something again?
– vesii
Nov 18 at 20:08
Argh... No! You are right and I am wrong! I made a very foolish mistake! $1$ is not the identity for addition. $0$ is the identity. (I was even going to point that out but I thought that would confuse things). SO we want $7^k = e$ or $ktimes 7 equiv 0 mod 66$ not $ktimes 7 equiv 1 mod 66$. !!!! BIG difference! ... I'm sorry and I'll fix it.
– fleablood
Nov 18 at 21:56
It's funny beecause in making the mistake $7^{19} =1$ I made the exact same mistake you did. I forgot for a moment we were dealing with ADDITION and not multiplication. Because this is addition we need $7^k = ktimes 0 equiv 0 pmod{66}$.
– fleablood
Nov 18 at 22:11
I'm surprised no-one else caught me on this!
– fleablood
Nov 18 at 22:15
that's ok! thanks for the answer. Is there a similar formula for $mathbb{Z}_{n}^{times}$?
– vesii
Nov 19 at 16:29
I saw the use of the following equation: $|x|=frac{|mathbb{Z_{m}}|}{gcd(|mathbb{Z_{m}}|,x)}$. But for this question it will give us: $|7|=frac{66}{gcd(66,7)}=66$ and not $19$. Am I missing something again?
– vesii
Nov 18 at 20:08
I saw the use of the following equation: $|x|=frac{|mathbb{Z_{m}}|}{gcd(|mathbb{Z_{m}}|,x)}$. But for this question it will give us: $|7|=frac{66}{gcd(66,7)}=66$ and not $19$. Am I missing something again?
– vesii
Nov 18 at 20:08
Argh... No! You are right and I am wrong! I made a very foolish mistake! $1$ is not the identity for addition. $0$ is the identity. (I was even going to point that out but I thought that would confuse things). SO we want $7^k = e$ or $ktimes 7 equiv 0 mod 66$ not $ktimes 7 equiv 1 mod 66$. !!!! BIG difference! ... I'm sorry and I'll fix it.
– fleablood
Nov 18 at 21:56
Argh... No! You are right and I am wrong! I made a very foolish mistake! $1$ is not the identity for addition. $0$ is the identity. (I was even going to point that out but I thought that would confuse things). SO we want $7^k = e$ or $ktimes 7 equiv 0 mod 66$ not $ktimes 7 equiv 1 mod 66$. !!!! BIG difference! ... I'm sorry and I'll fix it.
– fleablood
Nov 18 at 21:56
It's funny beecause in making the mistake $7^{19} =1$ I made the exact same mistake you did. I forgot for a moment we were dealing with ADDITION and not multiplication. Because this is addition we need $7^k = ktimes 0 equiv 0 pmod{66}$.
– fleablood
Nov 18 at 22:11
It's funny beecause in making the mistake $7^{19} =1$ I made the exact same mistake you did. I forgot for a moment we were dealing with ADDITION and not multiplication. Because this is addition we need $7^k = ktimes 0 equiv 0 pmod{66}$.
– fleablood
Nov 18 at 22:11
I'm surprised no-one else caught me on this!
– fleablood
Nov 18 at 22:15
I'm surprised no-one else caught me on this!
– fleablood
Nov 18 at 22:15
that's ok! thanks for the answer. Is there a similar formula for $mathbb{Z}_{n}^{times}$?
– vesii
Nov 19 at 16:29
that's ok! thanks for the answer. Is there a similar formula for $mathbb{Z}_{n}^{times}$?
– vesii
Nov 19 at 16:29
|
show 4 more comments
up vote
1
down vote
$Z$₁₂ is a cyclic group under addition. So the generator 'a' of the form ={na | for some n€ $Z$ }.
In order to find order of $3$ we need to do as follows:
<3>={3,6,9,0}={0,3,6,9}
here n=4 for which <3> is zero. Hence order of $3$ is $4$.
This answer has a very good insight. I think the OP would do well to consider it.
– fleablood
Nov 19 at 16:58
add a comment |
up vote
1
down vote
$Z$₁₂ is a cyclic group under addition. So the generator 'a' of the form ={na | for some n€ $Z$ }.
In order to find order of $3$ we need to do as follows:
<3>={3,6,9,0}={0,3,6,9}
here n=4 for which <3> is zero. Hence order of $3$ is $4$.
This answer has a very good insight. I think the OP would do well to consider it.
– fleablood
Nov 19 at 16:58
add a comment |
up vote
1
down vote
up vote
1
down vote
$Z$₁₂ is a cyclic group under addition. So the generator 'a' of the form ={na | for some n€ $Z$ }.
In order to find order of $3$ we need to do as follows:
<3>={3,6,9,0}={0,3,6,9}
here n=4 for which <3> is zero. Hence order of $3$ is $4$.
$Z$₁₂ is a cyclic group under addition. So the generator 'a' of the form ={na | for some n€ $Z$ }.
In order to find order of $3$ we need to do as follows:
<3>={3,6,9,0}={0,3,6,9}
here n=4 for which <3> is zero. Hence order of $3$ is $4$.
answered Nov 18 at 0:20
John Nash
6818
6818
This answer has a very good insight. I think the OP would do well to consider it.
– fleablood
Nov 19 at 16:58
add a comment |
This answer has a very good insight. I think the OP would do well to consider it.
– fleablood
Nov 19 at 16:58
This answer has a very good insight. I think the OP would do well to consider it.
– fleablood
Nov 19 at 16:58
This answer has a very good insight. I think the OP would do well to consider it.
– fleablood
Nov 19 at 16:58
add a comment |
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In group theory notation. There is one binary operation which may be written as $*$ or $cdot$ or $times$ or $+$ or $circ$ or not written at all. Then the notation $3^n$ means to apply that operation $n$ times. In $mathbb Z_{12}$ the operation is $*=+$. So $3^0 =$ the identity equals $0$. $3^1 = 3$. $3^2 = 3*3=3+3 = 6$ and $3^3 = 3*3*3 = 3+3+3=9$ and $3^4 = 3*3*3*3= 3+3+3+3 = 0$. We can go one step weirder and not $3^k = ktimes 3$. Because the group is about addition multiplication ... doesn't exist... not in the current universe.
– fleablood
Nov 18 at 0:25