Mixing
![Is power distributable in product of two matrices? [closed]](https://lh5.googleusercontent.com/-coSJnxXLVJ0/AAAAAAAAAAI/AAAAAAAAA2I/FgA0685afNY/s72-c/photo.jpg?sz=32)
Calculating the order of an element in the group $U_{27}$
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
edited Nov 21 '18 at 3:21
Chinnapparaj R
5,2601826
asked Nov 20 '18 at 20:17
vesii
906
add a comment |
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
edited Nov 21 '18 at 3:21
Chinnapparaj R
5,2601826
asked Nov 20 '18 at 20:17
vesii
906
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 '18 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 '18 at 20:46
add a comment |
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
edited Nov 21 '18 at 3:21
Chinnapparaj R
5,2601826
asked Nov 20 '18 at 20:17
vesii
906
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 '18 at 3:21
Chinnapparaj R
5,2601826
asked Nov 20 '18 at 20:17
vesii
906
edited Nov 21 '18 at 3:21
Chinnapparaj R
5,2601826
asked Nov 20 '18 at 20:17
vesii
906
edited Nov 21 '18 at 3:21
Chinnapparaj R
5,2601826
edited Nov 21 '18 at 3:21
Chinnapparaj R
5,2601826
edited Nov 21 '18 at 3:21
Chinnapparaj R
5,2601826
5,2601826
asked Nov 20 '18 at 20:17
vesii
906
asked Nov 20 '18 at 20:17
vesii
906
asked Nov 20 '18 at 20:17
vesii
906
906
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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 at 20:46
add a comment |
2 Answers
2
active
oldest
votes
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 '18 at 8:22
Akash Patalwanshi
9821816
add a comment |
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 $vert 5 vert =18$
answered Nov 21 '18 at 3:19
Chinnapparaj R
5,2601826
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 '18 at 9:37
@TobiasKildetoft: oh! sorry! I change my notation!
– Chinnapparaj R
Nov 21 '18 at 9:40
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3006836%2fcalculating-the-order-of-an-element-in-the-group-u-27%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
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 '18 at 8:22
Akash Patalwanshi
9821816
add a comment |
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 '18 at 8:22
Akash Patalwanshi
9821816
add a comment |
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 '18 at 8:22
Akash Patalwanshi
9821816
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 '18 at 8:22
Akash Patalwanshi
9821816
answered Nov 21 '18 at 8:22
Akash Patalwanshi
9821816
answered Nov 21 '18 at 8:22
Akash Patalwanshi
9821816
answered Nov 21 '18 at 8:22
Akash Patalwanshi
9821816
9821816
add a comment |
add a comment |
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 $vert 5 vert =18$
answered Nov 21 '18 at 3:19
Chinnapparaj R
5,2601826
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 '18 at 9:37
@TobiasKildetoft: oh! sorry! I change my notation!
– Chinnapparaj R
Nov 21 '18 at 9:40
add a comment |
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 $vert 5 vert =18$
answered Nov 21 '18 at 3:19
Chinnapparaj R
5,2601826
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 '18 at 9:37
@TobiasKildetoft: oh! sorry! I change my notation!
– Chinnapparaj R
Nov 21 '18 at 9:40
add a comment |
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 $vert 5 vert =18$
answered Nov 21 '18 at 3:19
Chinnapparaj R
5,2601826
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 $vert 5 vert =18$
answered Nov 21 '18 at 3:19
Chinnapparaj R
5,2601826
edited Nov 21 '18 at 9:39
answered Nov 21 '18 at 3:19
Chinnapparaj R
5,2601826
answered Nov 21 '18 at 3:19
Chinnapparaj R
5,2601826
answered Nov 21 '18 at 3:19
Chinnapparaj R
5,2601826
5,2601826
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 '18 at 9:37
@TobiasKildetoft: oh! sorry! I change my notation!
– Chinnapparaj R
Nov 21 '18 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 '18 at 9:37
@TobiasKildetoft: oh! sorry! I change my notation!
– Chinnapparaj R
Nov 21 '18 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 '18 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 '18 at 9:37
@TobiasKildetoft: oh! sorry! I change my notation!
– Chinnapparaj R
Nov 21 '18 at 9:40
@TobiasKildetoft: oh! sorry! I change my notation!
– Chinnapparaj R
Nov 21 '18 at 9:40
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3006836%2fcalculating-the-order-of-an-element-in-the-group-u-27%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
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 '18 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 '18 at 20:46