Calculating the order of an element in the group $U_{27}$
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Recently I asked how to calculate the order of an element in $U_{27}=mathbb{Z}_{27}^*$ (Multiplicative group of integers modulo $27$) (link). Problem is I still don't understand the material but I would like to explain what I know and what I don't. Tried to find some similar thread on the same topic and I found the following tread (link). Although it does not answer my question directly, it points into that direction. I know the hard way to find the order of an element in $U_{27}$. For example in order to find the order of $5$ in $U_{27}$ I would do:
$$begin{align*}
5&=5bmod 27=5\
5^2&=25bmod 27=25\
5^3&=125bmod 27=17
end{align*}$$
And so on, until I find $ninmathbb{N}$ so $5^n=1$. It could take awhile, in fact I know that the order of $5$ in $U_{27}$ is $18$ ($5^{18}=3814697265625(mod27)=1$), so I will have to calculate $18$ times and facing some big numbers. I think that there is a fast way using the euler function. How can I use it in my advantage? is there a formula?
abstract-algebra group-theory
add a comment |
up vote
2
down vote
favorite
Recently I asked how to calculate the order of an element in $U_{27}=mathbb{Z}_{27}^*$ (Multiplicative group of integers modulo $27$) (link). Problem is I still don't understand the material but I would like to explain what I know and what I don't. Tried to find some similar thread on the same topic and I found the following tread (link). Although it does not answer my question directly, it points into that direction. I know the hard way to find the order of an element in $U_{27}$. For example in order to find the order of $5$ in $U_{27}$ I would do:
$$begin{align*}
5&=5bmod 27=5\
5^2&=25bmod 27=25\
5^3&=125bmod 27=17
end{align*}$$
And so on, until I find $ninmathbb{N}$ so $5^n=1$. It could take awhile, in fact I know that the order of $5$ in $U_{27}$ is $18$ ($5^{18}=3814697265625(mod27)=1$), so I will have to calculate $18$ times and facing some big numbers. I think that there is a fast way using the euler function. How can I use it in my advantage? is there a formula?
abstract-algebra group-theory
1
Hint: If you know $5^3 equiv 125 mod 27 = 17$, does that help you compute $5^6 = (5^3)^2 mod 27$?
– David G. Stork
Nov 20 at 20:43
$|mathbb{U}_27|=18=2*3^2$ So the order of element is eigher $2,3,6,9,18$ by Lagrange theorem. So check $a^2(mod 27)$, if it is not $1$, then check $a^3$... until you get 1.
– mathnoob
Nov 20 at 20:46
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Recently I asked how to calculate the order of an element in $U_{27}=mathbb{Z}_{27}^*$ (Multiplicative group of integers modulo $27$) (link). Problem is I still don't understand the material but I would like to explain what I know and what I don't. Tried to find some similar thread on the same topic and I found the following tread (link). Although it does not answer my question directly, it points into that direction. I know the hard way to find the order of an element in $U_{27}$. For example in order to find the order of $5$ in $U_{27}$ I would do:
$$begin{align*}
5&=5bmod 27=5\
5^2&=25bmod 27=25\
5^3&=125bmod 27=17
end{align*}$$
And so on, until I find $ninmathbb{N}$ so $5^n=1$. It could take awhile, in fact I know that the order of $5$ in $U_{27}$ is $18$ ($5^{18}=3814697265625(mod27)=1$), so I will have to calculate $18$ times and facing some big numbers. I think that there is a fast way using the euler function. How can I use it in my advantage? is there a formula?
abstract-algebra group-theory
Recently I asked how to calculate the order of an element in $U_{27}=mathbb{Z}_{27}^*$ (Multiplicative group of integers modulo $27$) (link). Problem is I still don't understand the material but I would like to explain what I know and what I don't. Tried to find some similar thread on the same topic and I found the following tread (link). Although it does not answer my question directly, it points into that direction. I know the hard way to find the order of an element in $U_{27}$. For example in order to find the order of $5$ in $U_{27}$ I would do:
$$begin{align*}
5&=5bmod 27=5\
5^2&=25bmod 27=25\
5^3&=125bmod 27=17
end{align*}$$
And so on, until I find $ninmathbb{N}$ so $5^n=1$. It could take awhile, in fact I know that the order of $5$ in $U_{27}$ is $18$ ($5^{18}=3814697265625(mod27)=1$), so I will have to calculate $18$ times and facing some big numbers. I think that there is a fast way using the euler function. How can I use it in my advantage? is there a formula?
abstract-algebra group-theory
abstract-algebra group-theory
edited Nov 21 at 3:21
Chinnapparaj R
4,6591825
4,6591825
asked Nov 20 at 20:17
vesii
635
635
1
Hint: If you know $5^3 equiv 125 mod 27 = 17$, does that help you compute $5^6 = (5^3)^2 mod 27$?
– David G. Stork
Nov 20 at 20:43
$|mathbb{U}_27|=18=2*3^2$ So the order of element is eigher $2,3,6,9,18$ by Lagrange theorem. So check $a^2(mod 27)$, if it is not $1$, then check $a^3$... until you get 1.
– mathnoob
Nov 20 at 20:46
add a comment |
1
Hint: If you know $5^3 equiv 125 mod 27 = 17$, does that help you compute $5^6 = (5^3)^2 mod 27$?
– David G. Stork
Nov 20 at 20:43
$|mathbb{U}_27|=18=2*3^2$ So the order of element is eigher $2,3,6,9,18$ by Lagrange theorem. So check $a^2(mod 27)$, if it is not $1$, then check $a^3$... until you get 1.
– mathnoob
Nov 20 at 20:46
1
1
Hint: If you know $5^3 equiv 125 mod 27 = 17$, does that help you compute $5^6 = (5^3)^2 mod 27$?
– David G. Stork
Nov 20 at 20:43
Hint: If you know $5^3 equiv 125 mod 27 = 17$, does that help you compute $5^6 = (5^3)^2 mod 27$?
– David G. Stork
Nov 20 at 20:43
$|mathbb{U}_27|=18=2*3^2$ So the order of element is eigher $2,3,6,9,18$ by Lagrange theorem. So check $a^2(mod 27)$, if it is not $1$, then check $a^3$... until you get 1.
– mathnoob
Nov 20 at 20:46
$|mathbb{U}_27|=18=2*3^2$ So the order of element is eigher $2,3,6,9,18$ by Lagrange theorem. So check $a^2(mod 27)$, if it is not $1$, then check $a^3$... until you get 1.
– mathnoob
Nov 20 at 20:46
add a comment |
2 Answers
2
active
oldest
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up vote
0
down vote
First of all note that,
if $G$ is group and $ain G$ then $o(a)=o(a^{-1})$.
if $G$ is group and $ain G$ be an element of order $m$ then $o(a^k)=frac{m}{gcd(m,k)}$ where $kin mathbb{N}$
By using these two results you can avoid half of calculation needed to compute order of elements.
Now let's start,
$o(1)= 1$ since $1$ is identity in $U_{27}$.
$o(2)= 18$ since $2^{18}≡1mod 27$
and $2^m ≢ 1mod 27$ for $m<18 in mathbb{N}$. Now using our above results:
$o(4)=o(2^2)=frac{o(2)}{gcd(o(2),2)}=frac{18}{gcd(18,2)}=frac{18}{2}=9$
$o(8)=o(2^3)= frac{o(2)}{gcd(o(2),3)}=frac{18}{gcd(18,3)}=frac{18}{3}=6$
Similarly you can compute easily, $o(16)=o(2^4)$, $o(5)=o(2^5)$, $o(10)=o(2^6)$, $o(20)=o(2^7)$, $o(13)=o(2^8)$ etc. in fact order of all elements in $U_{27}$ can be computed just by using above second result.(because as $2$ is generator in $U_{27}$ and hence it will generate all elements)
To see how first result work: as we know $[2•14]_{27}=[14•2]_{27}=[1]_{27}$ so $2$ and $14$ are inverses in $U_{27}$ and hence $o(14)=o(2)=18$
add a comment |
up vote
0
down vote
Here $U_{27}$ is a cyclic group of order $18=2 times 3^2$ . Note that if $g$ generates $U(p^2)$ then $g$ generates $U(p^k),k geq 2$ as well! where $p$ is an odd prime. [For a proof of this, refer this paper, Lemma $3(2)$]
Here $langle5 rangle= U(9) $ and so $langle 5 rangle= U(3^k) ,k geq 2$ and in particular $langle 5 rangle= U(3^3)=U(27) $, so $vert 5 vert =18$
Are you writing $x = langle Grangle$ to mean that $x$ generates $G$? I have never seen it done that way around before.
– Tobias Kildetoft
Nov 21 at 9:37
@TobiasKildetoft: oh! sorry! I change my notation!
– Chinnapparaj R
Nov 21 at 9:40
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
First of all note that,
if $G$ is group and $ain G$ then $o(a)=o(a^{-1})$.
if $G$ is group and $ain G$ be an element of order $m$ then $o(a^k)=frac{m}{gcd(m,k)}$ where $kin mathbb{N}$
By using these two results you can avoid half of calculation needed to compute order of elements.
Now let's start,
$o(1)= 1$ since $1$ is identity in $U_{27}$.
$o(2)= 18$ since $2^{18}≡1mod 27$
and $2^m ≢ 1mod 27$ for $m<18 in mathbb{N}$. Now using our above results:
$o(4)=o(2^2)=frac{o(2)}{gcd(o(2),2)}=frac{18}{gcd(18,2)}=frac{18}{2}=9$
$o(8)=o(2^3)= frac{o(2)}{gcd(o(2),3)}=frac{18}{gcd(18,3)}=frac{18}{3}=6$
Similarly you can compute easily, $o(16)=o(2^4)$, $o(5)=o(2^5)$, $o(10)=o(2^6)$, $o(20)=o(2^7)$, $o(13)=o(2^8)$ etc. in fact order of all elements in $U_{27}$ can be computed just by using above second result.(because as $2$ is generator in $U_{27}$ and hence it will generate all elements)
To see how first result work: as we know $[2•14]_{27}=[14•2]_{27}=[1]_{27}$ so $2$ and $14$ are inverses in $U_{27}$ and hence $o(14)=o(2)=18$
add a comment |
up vote
0
down vote
First of all note that,
if $G$ is group and $ain G$ then $o(a)=o(a^{-1})$.
if $G$ is group and $ain G$ be an element of order $m$ then $o(a^k)=frac{m}{gcd(m,k)}$ where $kin mathbb{N}$
By using these two results you can avoid half of calculation needed to compute order of elements.
Now let's start,
$o(1)= 1$ since $1$ is identity in $U_{27}$.
$o(2)= 18$ since $2^{18}≡1mod 27$
and $2^m ≢ 1mod 27$ for $m<18 in mathbb{N}$. Now using our above results:
$o(4)=o(2^2)=frac{o(2)}{gcd(o(2),2)}=frac{18}{gcd(18,2)}=frac{18}{2}=9$
$o(8)=o(2^3)= frac{o(2)}{gcd(o(2),3)}=frac{18}{gcd(18,3)}=frac{18}{3}=6$
Similarly you can compute easily, $o(16)=o(2^4)$, $o(5)=o(2^5)$, $o(10)=o(2^6)$, $o(20)=o(2^7)$, $o(13)=o(2^8)$ etc. in fact order of all elements in $U_{27}$ can be computed just by using above second result.(because as $2$ is generator in $U_{27}$ and hence it will generate all elements)
To see how first result work: as we know $[2•14]_{27}=[14•2]_{27}=[1]_{27}$ so $2$ and $14$ are inverses in $U_{27}$ and hence $o(14)=o(2)=18$
add a comment |
up vote
0
down vote
up vote
0
down vote
First of all note that,
if $G$ is group and $ain G$ then $o(a)=o(a^{-1})$.
if $G$ is group and $ain G$ be an element of order $m$ then $o(a^k)=frac{m}{gcd(m,k)}$ where $kin mathbb{N}$
By using these two results you can avoid half of calculation needed to compute order of elements.
Now let's start,
$o(1)= 1$ since $1$ is identity in $U_{27}$.
$o(2)= 18$ since $2^{18}≡1mod 27$
and $2^m ≢ 1mod 27$ for $m<18 in mathbb{N}$. Now using our above results:
$o(4)=o(2^2)=frac{o(2)}{gcd(o(2),2)}=frac{18}{gcd(18,2)}=frac{18}{2}=9$
$o(8)=o(2^3)= frac{o(2)}{gcd(o(2),3)}=frac{18}{gcd(18,3)}=frac{18}{3}=6$
Similarly you can compute easily, $o(16)=o(2^4)$, $o(5)=o(2^5)$, $o(10)=o(2^6)$, $o(20)=o(2^7)$, $o(13)=o(2^8)$ etc. in fact order of all elements in $U_{27}$ can be computed just by using above second result.(because as $2$ is generator in $U_{27}$ and hence it will generate all elements)
To see how first result work: as we know $[2•14]_{27}=[14•2]_{27}=[1]_{27}$ so $2$ and $14$ are inverses in $U_{27}$ and hence $o(14)=o(2)=18$
First of all note that,
if $G$ is group and $ain G$ then $o(a)=o(a^{-1})$.
if $G$ is group and $ain G$ be an element of order $m$ then $o(a^k)=frac{m}{gcd(m,k)}$ where $kin mathbb{N}$
By using these two results you can avoid half of calculation needed to compute order of elements.
Now let's start,
$o(1)= 1$ since $1$ is identity in $U_{27}$.
$o(2)= 18$ since $2^{18}≡1mod 27$
and $2^m ≢ 1mod 27$ for $m<18 in mathbb{N}$. Now using our above results:
$o(4)=o(2^2)=frac{o(2)}{gcd(o(2),2)}=frac{18}{gcd(18,2)}=frac{18}{2}=9$
$o(8)=o(2^3)= frac{o(2)}{gcd(o(2),3)}=frac{18}{gcd(18,3)}=frac{18}{3}=6$
Similarly you can compute easily, $o(16)=o(2^4)$, $o(5)=o(2^5)$, $o(10)=o(2^6)$, $o(20)=o(2^7)$, $o(13)=o(2^8)$ etc. in fact order of all elements in $U_{27}$ can be computed just by using above second result.(because as $2$ is generator in $U_{27}$ and hence it will generate all elements)
To see how first result work: as we know $[2•14]_{27}=[14•2]_{27}=[1]_{27}$ so $2$ and $14$ are inverses in $U_{27}$ and hence $o(14)=o(2)=18$
answered Nov 21 at 8:22
Akash Patalwanshi
9371816
9371816
add a comment |
add a comment |
up vote
0
down vote
Here $U_{27}$ is a cyclic group of order $18=2 times 3^2$ . Note that if $g$ generates $U(p^2)$ then $g$ generates $U(p^k),k geq 2$ as well! where $p$ is an odd prime. [For a proof of this, refer this paper, Lemma $3(2)$]
Here $langle5 rangle= U(9) $ and so $langle 5 rangle= U(3^k) ,k geq 2$ and in particular $langle 5 rangle= U(3^3)=U(27) $, so $vert 5 vert =18$
Are you writing $x = langle Grangle$ to mean that $x$ generates $G$? I have never seen it done that way around before.
– Tobias Kildetoft
Nov 21 at 9:37
@TobiasKildetoft: oh! sorry! I change my notation!
– Chinnapparaj R
Nov 21 at 9:40
add a comment |
up vote
0
down vote
Here $U_{27}$ is a cyclic group of order $18=2 times 3^2$ . Note that if $g$ generates $U(p^2)$ then $g$ generates $U(p^k),k geq 2$ as well! where $p$ is an odd prime. [For a proof of this, refer this paper, Lemma $3(2)$]
Here $langle5 rangle= U(9) $ and so $langle 5 rangle= U(3^k) ,k geq 2$ and in particular $langle 5 rangle= U(3^3)=U(27) $, so $vert 5 vert =18$
Are you writing $x = langle Grangle$ to mean that $x$ generates $G$? I have never seen it done that way around before.
– Tobias Kildetoft
Nov 21 at 9:37
@TobiasKildetoft: oh! sorry! I change my notation!
– Chinnapparaj R
Nov 21 at 9:40
add a comment |
up vote
0
down vote
up vote
0
down vote
Here $U_{27}$ is a cyclic group of order $18=2 times 3^2$ . Note that if $g$ generates $U(p^2)$ then $g$ generates $U(p^k),k geq 2$ as well! where $p$ is an odd prime. [For a proof of this, refer this paper, Lemma $3(2)$]
Here $langle5 rangle= U(9) $ and so $langle 5 rangle= U(3^k) ,k geq 2$ and in particular $langle 5 rangle= U(3^3)=U(27) $, so $vert 5 vert =18$
Here $U_{27}$ is a cyclic group of order $18=2 times 3^2$ . Note that if $g$ generates $U(p^2)$ then $g$ generates $U(p^k),k geq 2$ as well! where $p$ is an odd prime. [For a proof of this, refer this paper, Lemma $3(2)$]
Here $langle5 rangle= U(9) $ and so $langle 5 rangle= U(3^k) ,k geq 2$ and in particular $langle 5 rangle= U(3^3)=U(27) $, so $vert 5 vert =18$
edited Nov 21 at 9:39
answered Nov 21 at 3:19
Chinnapparaj R
4,6591825
4,6591825
Are you writing $x = langle Grangle$ to mean that $x$ generates $G$? I have never seen it done that way around before.
– Tobias Kildetoft
Nov 21 at 9:37
@TobiasKildetoft: oh! sorry! I change my notation!
– Chinnapparaj R
Nov 21 at 9:40
add a comment |
Are you writing $x = langle Grangle$ to mean that $x$ generates $G$? I have never seen it done that way around before.
– Tobias Kildetoft
Nov 21 at 9:37
@TobiasKildetoft: oh! sorry! I change my notation!
– Chinnapparaj R
Nov 21 at 9:40
Are you writing $x = langle Grangle$ to mean that $x$ generates $G$? I have never seen it done that way around before.
– Tobias Kildetoft
Nov 21 at 9:37
Are you writing $x = langle Grangle$ to mean that $x$ generates $G$? I have never seen it done that way around before.
– Tobias Kildetoft
Nov 21 at 9:37
@TobiasKildetoft: oh! sorry! I change my notation!
– Chinnapparaj R
Nov 21 at 9:40
@TobiasKildetoft: oh! sorry! I change my notation!
– Chinnapparaj R
Nov 21 at 9:40
add a comment |
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1
Hint: If you know $5^3 equiv 125 mod 27 = 17$, does that help you compute $5^6 = (5^3)^2 mod 27$?
– David G. Stork
Nov 20 at 20:43
$|mathbb{U}_27|=18=2*3^2$ So the order of element is eigher $2,3,6,9,18$ by Lagrange theorem. So check $a^2(mod 27)$, if it is not $1$, then check $a^3$... until you get 1.
– mathnoob
Nov 20 at 20:46