Is the multiplicative group $Bbb Z_{36}^times$ cyclic?
$begingroup$
I'm trying to answer this question but also understanding a smart method to find if a group like the one mentioned has a cyclic generator or not. I know that similar questions have already been asked but honestly, I did not understand the explanations given. So I decided to ask hoping to better understand with this exercise that I'm trying to solve. Thank you for your understanding.
According to Wikipedia, $Bbb Z_{36}^times$ is not cyclic. Now, the multiplciatove group has 12 elements which are 1, 5, 7, 11, 13, 17, 19, 23, 25, 29, 31, 35. Of course I need to prove that no one of them could be a generator of $Bbb Z_{36}^times$. How can I proceed?
group-theory discrete-mathematics ring-theory totient-function
edited Jan 22 at 19:19
Ethan Bolker
44k552117
asked Jan 22 at 19:17
PCNFPCNF
1338
$endgroup$
add a comment |
$begingroup$
I'm trying to answer this question but also understanding a smart method to find if a group like the one mentioned has a cyclic generator or not. I know that similar questions have already been asked but honestly, I did not understand the explanations given. So I decided to ask hoping to better understand with this exercise that I'm trying to solve. Thank you for your understanding.
According to Wikipedia, $Bbb Z_{36}^times$ is not cyclic. Now, the multiplciatove group has 12 elements which are 1, 5, 7, 11, 13, 17, 19, 23, 25, 29, 31, 35. Of course I need to prove that no one of them could be a generator of $Bbb Z_{36}^times$. How can I proceed?
group-theory discrete-mathematics ring-theory totient-function
edited Jan 22 at 19:19
Ethan Bolker
44k552117
asked Jan 22 at 19:17
PCNFPCNF
1338
$endgroup$
2
$begingroup$
Note that the order of $x$ and $x^{-1}$ will agree, so you need only check roughly half the elements.
$endgroup$
– Randall
Jan 22 at 19:18
2
$begingroup$
There are so few you can try them one at a time. You will find shortcuts along the way that speed things up.
$endgroup$
– Ethan Bolker
Jan 22 at 19:19
2
$begingroup$
I'd prove that none of them have order of $12$. For this, I'd consider them mod $3$ and mod $4$
$endgroup$
– Jakobian
Jan 22 at 19:22
$begingroup$
I would also consider showing that there are four elements with squares equal to $1$. At both ends and near the midpoint. In a cyclic group there can be at most two. Warning: this assumes that you have covered relevant pieces abouot the structure of cyclic groups (not necessarily done in a first course on algberaic structures).
$endgroup$
– Jyrki Lahtonen
Jan 22 at 19:24
1
$begingroup$
CRT : as a ring $mathbb{Z}_{36} = 9mathbb{Z}_{4} + (-8)mathbb{Z}_{9} cong mathbb{Z}_{4} times mathbb{Z}_{9}$. The 3 subgroups of order $2$ are then clear from there.
$endgroup$
– reuns
Jan 22 at 19:31
add a comment |
$begingroup$
I'm trying to answer this question but also understanding a smart method to find if a group like the one mentioned has a cyclic generator or not. I know that similar questions have already been asked but honestly, I did not understand the explanations given. So I decided to ask hoping to better understand with this exercise that I'm trying to solve. Thank you for your understanding.
According to Wikipedia, $Bbb Z_{36}^times$ is not cyclic. Now, the multiplciatove group has 12 elements which are 1, 5, 7, 11, 13, 17, 19, 23, 25, 29, 31, 35. Of course I need to prove that no one of them could be a generator of $Bbb Z_{36}^times$. How can I proceed?
group-theory discrete-mathematics ring-theory totient-function
edited Jan 22 at 19:19
Ethan Bolker
44k552117
asked Jan 22 at 19:17
PCNFPCNF
1338
$endgroup$
I'm trying to answer this question but also understanding a smart method to find if a group like the one mentioned has a cyclic generator or not. I know that similar questions have already been asked but honestly, I did not understand the explanations given. So I decided to ask hoping to better understand with this exercise that I'm trying to solve. Thank you for your understanding.
According to Wikipedia, $Bbb Z_{36}^times$ is not cyclic. Now, the multiplciatove group has 12 elements which are 1, 5, 7, 11, 13, 17, 19, 23, 25, 29, 31, 35. Of course I need to prove that no one of them could be a generator of $Bbb Z_{36}^times$. How can I proceed?
group-theory discrete-mathematics ring-theory totient-function
group-theory discrete-mathematics ring-theory totient-function
edited Jan 22 at 19:19
Ethan Bolker
44k552117
asked Jan 22 at 19:17
PCNFPCNF
1338
edited Jan 22 at 19:19
Ethan Bolker
44k552117
asked Jan 22 at 19:17
PCNFPCNF
1338
edited Jan 22 at 19:19
Ethan Bolker
44k552117
edited Jan 22 at 19:19
Ethan Bolker
44k552117
edited Jan 22 at 19:19
Ethan Bolker
44k552117
44k552117
asked Jan 22 at 19:17
PCNFPCNF
1338
asked Jan 22 at 19:17
PCNFPCNF
1338
asked Jan 22 at 19:17
PCNFPCNF
1338
1338
2
$begingroup$
Note that the order of $x$ and $x^{-1}$ will agree, so you need only check roughly half the elements.
$endgroup$
– Randall
Jan 22 at 19:18
2
$begingroup$
There are so few you can try them one at a time. You will find shortcuts along the way that speed things up.
$endgroup$
– Ethan Bolker
Jan 22 at 19:19
2
$begingroup$
I'd prove that none of them have order of $12$. For this, I'd consider them mod $3$ and mod $4$
$endgroup$
– Jakobian
Jan 22 at 19:22
$begingroup$
I would also consider showing that there are four elements with squares equal to $1$. At both ends and near the midpoint. In a cyclic group there can be at most two. Warning: this assumes that you have covered relevant pieces abouot the structure of cyclic groups (not necessarily done in a first course on algberaic structures).
$endgroup$
– Jyrki Lahtonen
Jan 22 at 19:24
1
$begingroup$
CRT : as a ring $mathbb{Z}_{36} = 9mathbb{Z}_{4} + (-8)mathbb{Z}_{9} cong mathbb{Z}_{4} times mathbb{Z}_{9}$. The 3 subgroups of order $2$ are then clear from there.
$endgroup$
– reuns
Jan 22 at 19:31
add a comment |
2
$begingroup$
Note that the order of $x$ and $x^{-1}$ will agree, so you need only check roughly half the elements.
$endgroup$
– Randall
Jan 22 at 19:18
2
$begingroup$
There are so few you can try them one at a time. You will find shortcuts along the way that speed things up.
$endgroup$
– Ethan Bolker
Jan 22 at 19:19
2
$begingroup$
I'd prove that none of them have order of $12$. For this, I'd consider them mod $3$ and mod $4$
$endgroup$
– Jakobian
Jan 22 at 19:22
$begingroup$
I would also consider showing that there are four elements with squares equal to $1$. At both ends and near the midpoint. In a cyclic group there can be at most two. Warning: this assumes that you have covered relevant pieces abouot the structure of cyclic groups (not necessarily done in a first course on algberaic structures).
$endgroup$
– Jyrki Lahtonen
Jan 22 at 19:24
1
$begingroup$
CRT : as a ring $mathbb{Z}_{36} = 9mathbb{Z}_{4} + (-8)mathbb{Z}_{9} cong mathbb{Z}_{4} times mathbb{Z}_{9}$. The 3 subgroups of order $2$ are then clear from there.
$endgroup$
– reuns
Jan 22 at 19:31
2
2
$begingroup$
Note that the order of $x$ and $x^{-1}$ will agree, so you need only check roughly half the elements.
$endgroup$
– Randall
Jan 22 at 19:18
$begingroup$
Note that the order of $x$ and $x^{-1}$ will agree, so you need only check roughly half the elements.
$endgroup$
– Randall
Jan 22 at 19:18
2
2
$begingroup$
There are so few you can try them one at a time. You will find shortcuts along the way that speed things up.
$endgroup$
– Ethan Bolker
Jan 22 at 19:19
$begingroup$
There are so few you can try them one at a time. You will find shortcuts along the way that speed things up.
$endgroup$
– Ethan Bolker
Jan 22 at 19:19
2
2
$begingroup$
I'd prove that none of them have order of $12$. For this, I'd consider them mod $3$ and mod $4$
$endgroup$
– Jakobian
Jan 22 at 19:22
$begingroup$
I'd prove that none of them have order of $12$. For this, I'd consider them mod $3$ and mod $4$
$endgroup$
– Jakobian
Jan 22 at 19:22
$begingroup$
I would also consider showing that there are four elements with squares equal to $1$. At both ends and near the midpoint. In a cyclic group there can be at most two. Warning: this assumes that you have covered relevant pieces abouot the structure of cyclic groups (not necessarily done in a first course on algberaic structures).
$endgroup$
– Jyrki Lahtonen
Jan 22 at 19:24
$begingroup$
I would also consider showing that there are four elements with squares equal to $1$. At both ends and near the midpoint. In a cyclic group there can be at most two. Warning: this assumes that you have covered relevant pieces abouot the structure of cyclic groups (not necessarily done in a first course on algberaic structures).
$endgroup$
– Jyrki Lahtonen
Jan 22 at 19:24
1
1
$begingroup$
CRT : as a ring $mathbb{Z}_{36} = 9mathbb{Z}_{4} + (-8)mathbb{Z}_{9} cong mathbb{Z}_{4} times mathbb{Z}_{9}$. The 3 subgroups of order $2$ are then clear from there.
$endgroup$
– reuns
Jan 22 at 19:31
$begingroup$
CRT : as a ring $mathbb{Z}_{36} = 9mathbb{Z}_{4} + (-8)mathbb{Z}_{9} cong mathbb{Z}_{4} times mathbb{Z}_{9}$. The 3 subgroups of order $2$ are then clear from there.
$endgroup$
– reuns
Jan 22 at 19:31
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Clearly $1$ is not a generator. To check whether $5$ is a generator, compute a few powers;
$$5^1equiv5,qquad 5^2equiv25,qquad 5^3equiv17,qquad5^4equiv13,qquad5^5equiv29,qquad5^6equiv1.$$
This already tells you that $5$, $25$, $17$, $13$ and $29$ are not generators. Next do the same for $7$...
answered Jan 22 at 19:46
ServaesServaes
26k33997
$endgroup$
add a comment |
$begingroup$
It is a classical fact that $mathbb Z_m^times$ is cyclic if and only if $min{1,2,4,p^alpha,2p^alpha}$, where $p$ is an odd prime and $alpha$ is a positive integer.
A simple way to see that $mathbb Z_{36}^times$ is not cyclic (which also gives a hint about the proof of the general result) is to observe that for any integer $g$ coprime with $36$, one has $g^6equiv 1pmod 9$ and also $g^6equiv 1pmod 4$ (in fact, already $g^2equiv 1pmod 4$), as it follows by a direct computation or by using Euler's theorem. Therefore $g^6equiv 1pmod {36}$, showing that $mathbb Z_{36}^times$ does not contain elements of order $|mathbb Z_{36}^times|=12$.
answered Jan 22 at 20:36
W-t-PW-t-P
1,287611
$endgroup$
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%2f3083582%2fis-the-multiplicative-group-bbb-z-36-times-cyclic%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
$begingroup$
Clearly $1$ is not a generator. To check whether $5$ is a generator, compute a few powers;
$$5^1equiv5,qquad 5^2equiv25,qquad 5^3equiv17,qquad5^4equiv13,qquad5^5equiv29,qquad5^6equiv1.$$
This already tells you that $5$, $25$, $17$, $13$ and $29$ are not generators. Next do the same for $7$...
answered Jan 22 at 19:46
ServaesServaes
26k33997
$endgroup$
add a comment |
$begingroup$
Clearly $1$ is not a generator. To check whether $5$ is a generator, compute a few powers;
$$5^1equiv5,qquad 5^2equiv25,qquad 5^3equiv17,qquad5^4equiv13,qquad5^5equiv29,qquad5^6equiv1.$$
This already tells you that $5$, $25$, $17$, $13$ and $29$ are not generators. Next do the same for $7$...
answered Jan 22 at 19:46
ServaesServaes
26k33997
$endgroup$
add a comment |
$begingroup$
Clearly $1$ is not a generator. To check whether $5$ is a generator, compute a few powers;
$$5^1equiv5,qquad 5^2equiv25,qquad 5^3equiv17,qquad5^4equiv13,qquad5^5equiv29,qquad5^6equiv1.$$
This already tells you that $5$, $25$, $17$, $13$ and $29$ are not generators. Next do the same for $7$...
answered Jan 22 at 19:46
ServaesServaes
26k33997
$endgroup$
Clearly $1$ is not a generator. To check whether $5$ is a generator, compute a few powers;
$$5^1equiv5,qquad 5^2equiv25,qquad 5^3equiv17,qquad5^4equiv13,qquad5^5equiv29,qquad5^6equiv1.$$
This already tells you that $5$, $25$, $17$, $13$ and $29$ are not generators. Next do the same for $7$...
answered Jan 22 at 19:46
ServaesServaes
26k33997
answered Jan 22 at 19:46
ServaesServaes
26k33997
answered Jan 22 at 19:46
ServaesServaes
26k33997
answered Jan 22 at 19:46
ServaesServaes
26k33997
26k33997
add a comment |
add a comment |
$begingroup$
It is a classical fact that $mathbb Z_m^times$ is cyclic if and only if $min{1,2,4,p^alpha,2p^alpha}$, where $p$ is an odd prime and $alpha$ is a positive integer.
A simple way to see that $mathbb Z_{36}^times$ is not cyclic (which also gives a hint about the proof of the general result) is to observe that for any integer $g$ coprime with $36$, one has $g^6equiv 1pmod 9$ and also $g^6equiv 1pmod 4$ (in fact, already $g^2equiv 1pmod 4$), as it follows by a direct computation or by using Euler's theorem. Therefore $g^6equiv 1pmod {36}$, showing that $mathbb Z_{36}^times$ does not contain elements of order $|mathbb Z_{36}^times|=12$.
answered Jan 22 at 20:36
W-t-PW-t-P
1,287611
$endgroup$
add a comment |
$begingroup$
It is a classical fact that $mathbb Z_m^times$ is cyclic if and only if $min{1,2,4,p^alpha,2p^alpha}$, where $p$ is an odd prime and $alpha$ is a positive integer.
A simple way to see that $mathbb Z_{36}^times$ is not cyclic (which also gives a hint about the proof of the general result) is to observe that for any integer $g$ coprime with $36$, one has $g^6equiv 1pmod 9$ and also $g^6equiv 1pmod 4$ (in fact, already $g^2equiv 1pmod 4$), as it follows by a direct computation or by using Euler's theorem. Therefore $g^6equiv 1pmod {36}$, showing that $mathbb Z_{36}^times$ does not contain elements of order $|mathbb Z_{36}^times|=12$.
answered Jan 22 at 20:36
W-t-PW-t-P
1,287611
$endgroup$
add a comment |
$begingroup$
It is a classical fact that $mathbb Z_m^times$ is cyclic if and only if $min{1,2,4,p^alpha,2p^alpha}$, where $p$ is an odd prime and $alpha$ is a positive integer.
A simple way to see that $mathbb Z_{36}^times$ is not cyclic (which also gives a hint about the proof of the general result) is to observe that for any integer $g$ coprime with $36$, one has $g^6equiv 1pmod 9$ and also $g^6equiv 1pmod 4$ (in fact, already $g^2equiv 1pmod 4$), as it follows by a direct computation or by using Euler's theorem. Therefore $g^6equiv 1pmod {36}$, showing that $mathbb Z_{36}^times$ does not contain elements of order $|mathbb Z_{36}^times|=12$.
answered Jan 22 at 20:36
W-t-PW-t-P
1,287611
$endgroup$
It is a classical fact that $mathbb Z_m^times$ is cyclic if and only if $min{1,2,4,p^alpha,2p^alpha}$, where $p$ is an odd prime and $alpha$ is a positive integer.
A simple way to see that $mathbb Z_{36}^times$ is not cyclic (which also gives a hint about the proof of the general result) is to observe that for any integer $g$ coprime with $36$, one has $g^6equiv 1pmod 9$ and also $g^6equiv 1pmod 4$ (in fact, already $g^2equiv 1pmod 4$), as it follows by a direct computation or by using Euler's theorem. Therefore $g^6equiv 1pmod {36}$, showing that $mathbb Z_{36}^times$ does not contain elements of order $|mathbb Z_{36}^times|=12$.
answered Jan 22 at 20:36
W-t-PW-t-P
1,287611
answered Jan 22 at 20:36
W-t-PW-t-P
1,287611
answered Jan 22 at 20:36
W-t-PW-t-P
1,287611
answered Jan 22 at 20:36
W-t-PW-t-P
1,287611
1,287611
add a comment |
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.
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%2f3083582%2fis-the-multiplicative-group-bbb-z-36-times-cyclic%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
2
$begingroup$
Note that the order of $x$ and $x^{-1}$ will agree, so you need only check roughly half the elements.
$endgroup$
– Randall
Jan 22 at 19:18
2
$begingroup$
There are so few you can try them one at a time. You will find shortcuts along the way that speed things up.
$endgroup$
– Ethan Bolker
Jan 22 at 19:19
2
$begingroup$
I'd prove that none of them have order of $12$. For this, I'd consider them mod $3$ and mod $4$
$endgroup$
– Jakobian
Jan 22 at 19:22
$begingroup$
I would also consider showing that there are four elements with squares equal to $1$. At both ends and near the midpoint. In a cyclic group there can be at most two. Warning: this assumes that you have covered relevant pieces abouot the structure of cyclic groups (not necessarily done in a first course on algberaic structures).
$endgroup$
– Jyrki Lahtonen
Jan 22 at 19:24
1
$begingroup$
CRT : as a ring $mathbb{Z}_{36} = 9mathbb{Z}_{4} + (-8)mathbb{Z}_{9} cong mathbb{Z}_{4} times mathbb{Z}_{9}$. The 3 subgroups of order $2$ are then clear from there.
$endgroup$
– reuns
Jan 22 at 19:31