How to derive the relation between $k$ and $l$ given $langle g^k rangle = langle g^l rangle$ in a cyclic...

Where to find order of arguments for default functions

How does practicing restraint and performing actions of merit purify the mind?

Term for the "extreme-extension" version of a straw man fallacy?

Whats the best way to handle refactoring a big file?

How to get regions to plot as graphics

How easy is it to start Magic from scratch?

Why didn't Theresa May consult with Parliament before negotiating a deal with the EU?

Inappropriate reference requests from Journal reviewers

How do I construct this japanese bowl?

Why didn't Khan get resurrected in the Genesis Explosion?

Why is there a PLL in CPU?

Anatomically Correct Strange Women In Ponds Distributing Swords

How to be diplomatic in refusing to write code that breaches the privacy of our users

Rotate a column

Why do professional authors make "consistency" mistakes? And how to avoid them?

Opposite of a diet

How to write papers efficiently when English isn't my first language?

Implement the Thanos sorting algorithm

How to write the block matrix in LaTex?

Increase performance creating Mandelbrot set in python

How to use tikz in fbox?

How to Reset Passwords on Multiple Websites Easily?

How can I open an app using Terminal?

What's the point of interval inversion?



How to derive the relation between $k$ and $l$ given $langle g^k rangle = langle g^l rangle$ in a cyclic group $C_n = langle g rangle$?



The Next CEO of Stack OverflowProve that the order of the cyclic subgroup $langle g^krangle $ is $n/{operatorname{gcd}(n,k)}$Is $mathbb{Z}/langle nrangle$ cyclic and of order $n$?If an element $c$ of order $d$ belongs to both $langle arangle$ and $langle brangle$, then $langle arangle=langle brangle=langle crangle$?Cyclic group, automorphism, and isomorphism$G$ cyclic group of order $24$, and $H=langle x^6rangle$, find the order of each element of $G/H$Prove that the presentation of a cyclic group of infinite order is $langle amid rangle$Show that ${C_infty }/leftlangle {{c^n}} rightrangle simeq {C_n}$Efficient solution finding quotient group and is there generator in quotient group $mathbb{Z}_{15}^*/langle 11 rangle$??Prove that the order of the cyclic subgroup $langle g^krangle $ is $n/{operatorname{gcd}(n,k)}$What is the difference between cyclic subgroup $left langle aright rangle$ and $left langle a^{k} right rangle$?Prove $langle k rangle / langle km rangle simeq mathbb {Z}_m,$ where $langle k rangle$ is the subgroup generated by $k$ and $kin mathbb{Z}$












2












$begingroup$


It is known that




For a cyclic group $C_n = langle g rangle$ of order $n$, we have $langle g^k rangle = langle g^{(k, n)} rangle$, where $k in mathbb{Z}$.





I am able to verify this result.
Now, I try to derive it by showing something like




If $langle g^k rangle = langle g^l rangle$, then we have $l = (k, n)$ (in terms of "mod n").




$langle g^k rangle = langle g^l rangle$ means that $g^k in langle g^l rangle$ and $g^l in langle g^k rangle$.



That is, $exists t in mathbb{Z}: g^k = g^{lt}$ and $exists s in mathbb{Z}: g^l = g^{ks}$.



Therefore, $k equiv lt ; (text{mod}; n)$ and $l equiv ks ; (text{mod}; n)$.




How to proceed with this argument?






abstract-algebra group-theory modular-arithmetic greatest-common-divisor cyclic-groups







share|cite|improve this question











$endgroup$












  • $begingroup$
    You can proceed like here.
    $endgroup$
    – Dietrich Burde
    Mar 16 at 9:09










  • $begingroup$
    @DietrichBurde Thanks. Because $frac{n}{(k,n)} = frac{n}{((k, n), n)}$, we can guess that $l = (k, n)$. However, how to derive that from $langle g^k rangle = langle g^l rangle$? Could you please show me more hints?
    $endgroup$
    – hengxin
    Mar 16 at 9:41


















2












$begingroup$


It is known that




For a cyclic group $C_n = langle g rangle$ of order $n$, we have $langle g^k rangle = langle g^{(k, n)} rangle$, where $k in mathbb{Z}$.





I am able to verify this result.
Now, I try to derive it by showing something like




If $langle g^k rangle = langle g^l rangle$, then we have $l = (k, n)$ (in terms of "mod n").




$langle g^k rangle = langle g^l rangle$ means that $g^k in langle g^l rangle$ and $g^l in langle g^k rangle$.



That is, $exists t in mathbb{Z}: g^k = g^{lt}$ and $exists s in mathbb{Z}: g^l = g^{ks}$.



Therefore, $k equiv lt ; (text{mod}; n)$ and $l equiv ks ; (text{mod}; n)$.




How to proceed with this argument?






abstract-algebra group-theory modular-arithmetic greatest-common-divisor cyclic-groups







share|cite|improve this question











$endgroup$












  • $begingroup$
    You can proceed like here.
    $endgroup$
    – Dietrich Burde
    Mar 16 at 9:09










  • $begingroup$
    @DietrichBurde Thanks. Because $frac{n}{(k,n)} = frac{n}{((k, n), n)}$, we can guess that $l = (k, n)$. However, how to derive that from $langle g^k rangle = langle g^l rangle$? Could you please show me more hints?
    $endgroup$
    – hengxin
    Mar 16 at 9:41
















2












2








2


2



$begingroup$


It is known that




For a cyclic group $C_n = langle g rangle$ of order $n$, we have $langle g^k rangle = langle g^{(k, n)} rangle$, where $k in mathbb{Z}$.





I am able to verify this result.
Now, I try to derive it by showing something like




If $langle g^k rangle = langle g^l rangle$, then we have $l = (k, n)$ (in terms of "mod n").




$langle g^k rangle = langle g^l rangle$ means that $g^k in langle g^l rangle$ and $g^l in langle g^k rangle$.



That is, $exists t in mathbb{Z}: g^k = g^{lt}$ and $exists s in mathbb{Z}: g^l = g^{ks}$.



Therefore, $k equiv lt ; (text{mod}; n)$ and $l equiv ks ; (text{mod}; n)$.




How to proceed with this argument?






abstract-algebra group-theory modular-arithmetic greatest-common-divisor cyclic-groups







share|cite|improve this question











$endgroup$




It is known that




For a cyclic group $C_n = langle g rangle$ of order $n$, we have $langle g^k rangle = langle g^{(k, n)} rangle$, where $k in mathbb{Z}$.





I am able to verify this result.
Now, I try to derive it by showing something like




If $langle g^k rangle = langle g^l rangle$, then we have $l = (k, n)$ (in terms of "mod n").




$langle g^k rangle = langle g^l rangle$ means that $g^k in langle g^l rangle$ and $g^l in langle g^k rangle$.



That is, $exists t in mathbb{Z}: g^k = g^{lt}$ and $exists s in mathbb{Z}: g^l = g^{ks}$.



Therefore, $k equiv lt ; (text{mod}; n)$ and $l equiv ks ; (text{mod}; n)$.




How to proceed with this argument?






abstract-algebra group-theory modular-arithmetic greatest-common-divisor cyclic-groups




abstract-algebra group-theory modular-arithmetic greatest-common-divisor cyclic-groups






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 16 at 11:53







hengxin

















asked Mar 16 at 8:05









hengxinhengxin

1,6221429




1,6221429












  • $begingroup$
    You can proceed like here.
    $endgroup$
    – Dietrich Burde
    Mar 16 at 9:09










  • $begingroup$
    @DietrichBurde Thanks. Because $frac{n}{(k,n)} = frac{n}{((k, n), n)}$, we can guess that $l = (k, n)$. However, how to derive that from $langle g^k rangle = langle g^l rangle$? Could you please show me more hints?
    $endgroup$
    – hengxin
    Mar 16 at 9:41




















  • $begingroup$
    You can proceed like here.
    $endgroup$
    – Dietrich Burde
    Mar 16 at 9:09










  • $begingroup$
    @DietrichBurde Thanks. Because $frac{n}{(k,n)} = frac{n}{((k, n), n)}$, we can guess that $l = (k, n)$. However, how to derive that from $langle g^k rangle = langle g^l rangle$? Could you please show me more hints?
    $endgroup$
    – hengxin
    Mar 16 at 9:41


















$begingroup$
You can proceed like here.
$endgroup$
– Dietrich Burde
Mar 16 at 9:09




$begingroup$
You can proceed like here.
$endgroup$
– Dietrich Burde
Mar 16 at 9:09












$begingroup$
@DietrichBurde Thanks. Because $frac{n}{(k,n)} = frac{n}{((k, n), n)}$, we can guess that $l = (k, n)$. However, how to derive that from $langle g^k rangle = langle g^l rangle$? Could you please show me more hints?
$endgroup$
– hengxin
Mar 16 at 9:41






$begingroup$
@DietrichBurde Thanks. Because $frac{n}{(k,n)} = frac{n}{((k, n), n)}$, we can guess that $l = (k, n)$. However, how to derive that from $langle g^k rangle = langle g^l rangle$? Could you please show me more hints?
$endgroup$
– hengxin
Mar 16 at 9:41












1 Answer
1






active

oldest

votes


















1












$begingroup$

If $dmid n$, then the order of $langle g^drangle$ is $n/d$. Thus, the order of $langle g^krangle=langle g^{(n,k)}rangle$ is $n/(n,k)$, and similarly, the order of $langle g^lrangle=langle g^{(n,l)}rangle$ is $n/(n,l)$. However, a cyclic group has just one subgroup of every possible order. Therefore, $langle g^krangle=langle g^lrangle$ if and only if $n/(k,n)=n/(l,n)$; that is, $(k,n)=(l,n)$.






share|cite|improve this answer









$endgroup$














    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
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3150182%2fhow-to-derive-the-relation-between-k-and-l-given-langle-gk-rangle-lan%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    If $dmid n$, then the order of $langle g^drangle$ is $n/d$. Thus, the order of $langle g^krangle=langle g^{(n,k)}rangle$ is $n/(n,k)$, and similarly, the order of $langle g^lrangle=langle g^{(n,l)}rangle$ is $n/(n,l)$. However, a cyclic group has just one subgroup of every possible order. Therefore, $langle g^krangle=langle g^lrangle$ if and only if $n/(k,n)=n/(l,n)$; that is, $(k,n)=(l,n)$.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      If $dmid n$, then the order of $langle g^drangle$ is $n/d$. Thus, the order of $langle g^krangle=langle g^{(n,k)}rangle$ is $n/(n,k)$, and similarly, the order of $langle g^lrangle=langle g^{(n,l)}rangle$ is $n/(n,l)$. However, a cyclic group has just one subgroup of every possible order. Therefore, $langle g^krangle=langle g^lrangle$ if and only if $n/(k,n)=n/(l,n)$; that is, $(k,n)=(l,n)$.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        If $dmid n$, then the order of $langle g^drangle$ is $n/d$. Thus, the order of $langle g^krangle=langle g^{(n,k)}rangle$ is $n/(n,k)$, and similarly, the order of $langle g^lrangle=langle g^{(n,l)}rangle$ is $n/(n,l)$. However, a cyclic group has just one subgroup of every possible order. Therefore, $langle g^krangle=langle g^lrangle$ if and only if $n/(k,n)=n/(l,n)$; that is, $(k,n)=(l,n)$.






        share|cite|improve this answer









        $endgroup$



        If $dmid n$, then the order of $langle g^drangle$ is $n/d$. Thus, the order of $langle g^krangle=langle g^{(n,k)}rangle$ is $n/(n,k)$, and similarly, the order of $langle g^lrangle=langle g^{(n,l)}rangle$ is $n/(n,l)$. However, a cyclic group has just one subgroup of every possible order. Therefore, $langle g^krangle=langle g^lrangle$ if and only if $n/(k,n)=n/(l,n)$; that is, $(k,n)=(l,n)$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Mar 16 at 17:16









        W-t-PW-t-P

        1,579612




        1,579612






























            draft saved

            draft discarded




















































            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.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3150182%2fhow-to-derive-the-relation-between-k-and-l-given-langle-gk-rangle-lan%23new-answer', 'question_page');
            }
            );

            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







            -

            Popular posts from this blog

            Nidaros erkebispedøme

            Image
            For det lutherske bispedømet, sjå Nidaros bispedøme. Nidaros erkebispedøme var eit katolsk erkebispedøme med sete i Nidaros. Det vart oppretta av kong Olav Kyrre mellom 1070 og 1080 som bispedøme. Bispedømet vart opphøgd til erkebispedøme i 1153. Erkebispedømet vart innstifta av pave Anastasius IV i 1154, og var i funksjon fram til reformasjonen i 1537. Erkebiskopen sitt hovudsete var Erkebispegarden, rett ved domkyrkja Nidarosdomen. Frå 1152 til 1154 var Nicholas Breakspear i Skandinavia som paveleg utsending, og organiserte opprettinga av erkesetet. Innhaldsliste 1 Utbreiing og jurisdiksjon 2 Overhyrdar 2.1 Lydbiskopar 3 Biskopar 3.1 Biskopar 3.2 Erkebiskopar Utbreiing og jurisdiksjon | Kart over den nordiske kyrkjeprovinsen (1153-1387) Erkebispedømet omfatta i tillegg til Noreg øyane i vest i Noregsveldet. Nidaros var eit metropolitanerkebispedøme med fylgjande underliggjande bispedøme suffraganar: Bj...
            Read more

            Birsay

            Image
            Tettstaden Birsay på Orknøyane. Ruinane av jarleplasset om lag midt i biletet. (Fotografi frå 2012) Utsikt mot Birsay og Brough of Brisay (fotografi frå 2009) Birsay er ein tettstad i eit sokn (parish) med same namn, nordvest på øya Mainland [Note 1] på Orknøyane, om lag 31 km frå Kirkwall. Staden er omtala fleire stader i Orknøyingasoga [1] , der han er kalla «Byrgesherad» (norrønt: Birgisherað ). Namnet skriv seg truleg frå elvenamnet Birgisá [2] . Innhaldsliste 1 Den gamle hovudstaden 2 Bispesetet 3 Jarlepalasset 4 Kjelder og utfyllande opplysningar 4.1 Notar 4.2 Fotnotar 4.3 Bakgrunnsstoff Den gamle hovudstaden | Hovudartikkelen for dette emnet er Torfinn Sigurdsson . Soknet er i dag i hovudsak ei jordbrukskommune men er mest kjend for dei mange kulturminna i området. Tettstaden Birsay blir gjerne omtala som «den gamle hovudstaden på Orknøyane» [3] , og ein viser d...
            Read more

            Was Woodrow Wilson really a Liberal?Was World War I a war of liberals against authoritarians?Founding Fathers...

            Image
            Do VLANs within a subnet need to have their own subnet for router on a stick? What are the differences between the usage of 'it' and 'they'? Dragon forelimb placement How do we improve the relationship with a client software team that performs poorly and is becoming less collaborative? What typically incentivizes a professor to change jobs to a lower ranking university? How to write a macro that is braces sensitive? Modeling an IPv4 Address Writing rule stating superpower from different root cause is bad writing Can I ask the recruiters in my resume to put the reason why I am rejected? Approximately how much travel t...
            Read more