How to find order of the subgroup?











up vote
0
down vote

favorite












Having two prime numbers $p$, $q$ and a relation $N=pq$.



How could I find the order of group $mathbb{Z}_{N^2}^{*}$?



I know that $ord(mathbb{Z}_{N^2})=N^2$ and that $mathbb{Z}_{N^2}^{*}$ is a subgroup of $mathbb{Z}_{N^2}$.



Also I know that the order of the group is divided by the order of its subgroups, so $ord(mathbb{Z}_{N^2}^{*})|ord(mathbb{Z}_{N^2})$.



So, the possible orders could be either $N^2, N, p , q, 1$, but thats where I get stuck: I don't know how to determine which one would be the right order?



Could you give me any hints how to solve this problem?



Edit:



"What does the ∗ mean in this case?"



$mathbb{Z}_{n}^{*}={a in mathbb{Z}_{n} : gcd(a, n) = 1 }$






group-theory







share|cite|improve this question
























  • What does the $*$ mean in this case?
    – Melody
    2 days ago








  • 3




    $mathbb{Z}^*_{N^2}$ is not a subgroup of $mathbb{Z}_{N^2}$; one is a group under multiplication, the other is a group under addition...
    – Matt B
    2 days ago






  • 2




    The order of $mathbb{Z}_{n}^*$ is $phi(n)$. so $phi(p^2q^2)=p(p-1)*q*(q-1)$ Here is the wiki link for $phi(n)$: en.wikipedia.org/wiki/Euler%27s_totient_function also:math.stackexchange.com/questions/194705/…
    – mathnoob
    2 days ago












  • Added definition in my post. @Melody
    – Andry
    2 days ago










  • Thanks for pointing out that my problem was already based on my wrong assumption @MattB
    – Andry
    2 days ago















up vote
0
down vote

favorite












Having two prime numbers $p$, $q$ and a relation $N=pq$.



How could I find the order of group $mathbb{Z}_{N^2}^{*}$?



I know that $ord(mathbb{Z}_{N^2})=N^2$ and that $mathbb{Z}_{N^2}^{*}$ is a subgroup of $mathbb{Z}_{N^2}$.



Also I know that the order of the group is divided by the order of its subgroups, so $ord(mathbb{Z}_{N^2}^{*})|ord(mathbb{Z}_{N^2})$.



So, the possible orders could be either $N^2, N, p , q, 1$, but thats where I get stuck: I don't know how to determine which one would be the right order?



Could you give me any hints how to solve this problem?



Edit:



"What does the ∗ mean in this case?"



$mathbb{Z}_{n}^{*}={a in mathbb{Z}_{n} : gcd(a, n) = 1 }$






group-theory







share|cite|improve this question
























  • What does the $*$ mean in this case?
    – Melody
    2 days ago








  • 3




    $mathbb{Z}^*_{N^2}$ is not a subgroup of $mathbb{Z}_{N^2}$; one is a group under multiplication, the other is a group under addition...
    – Matt B
    2 days ago






  • 2




    The order of $mathbb{Z}_{n}^*$ is $phi(n)$. so $phi(p^2q^2)=p(p-1)*q*(q-1)$ Here is the wiki link for $phi(n)$: en.wikipedia.org/wiki/Euler%27s_totient_function also:math.stackexchange.com/questions/194705/…
    – mathnoob
    2 days ago












  • Added definition in my post. @Melody
    – Andry
    2 days ago










  • Thanks for pointing out that my problem was already based on my wrong assumption @MattB
    – Andry
    2 days ago













up vote
0
down vote

favorite









up vote
0
down vote

favorite











Having two prime numbers $p$, $q$ and a relation $N=pq$.



How could I find the order of group $mathbb{Z}_{N^2}^{*}$?



I know that $ord(mathbb{Z}_{N^2})=N^2$ and that $mathbb{Z}_{N^2}^{*}$ is a subgroup of $mathbb{Z}_{N^2}$.



Also I know that the order of the group is divided by the order of its subgroups, so $ord(mathbb{Z}_{N^2}^{*})|ord(mathbb{Z}_{N^2})$.



So, the possible orders could be either $N^2, N, p , q, 1$, but thats where I get stuck: I don't know how to determine which one would be the right order?



Could you give me any hints how to solve this problem?



Edit:



"What does the ∗ mean in this case?"



$mathbb{Z}_{n}^{*}={a in mathbb{Z}_{n} : gcd(a, n) = 1 }$






group-theory







share|cite|improve this question















Having two prime numbers $p$, $q$ and a relation $N=pq$.



How could I find the order of group $mathbb{Z}_{N^2}^{*}$?



I know that $ord(mathbb{Z}_{N^2})=N^2$ and that $mathbb{Z}_{N^2}^{*}$ is a subgroup of $mathbb{Z}_{N^2}$.



Also I know that the order of the group is divided by the order of its subgroups, so $ord(mathbb{Z}_{N^2}^{*})|ord(mathbb{Z}_{N^2})$.



So, the possible orders could be either $N^2, N, p , q, 1$, but thats where I get stuck: I don't know how to determine which one would be the right order?



Could you give me any hints how to solve this problem?



Edit:



"What does the ∗ mean in this case?"



$mathbb{Z}_{n}^{*}={a in mathbb{Z}_{n} : gcd(a, n) = 1 }$






group-theory




group-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 2 days ago

























asked 2 days ago









Andry

154




154












  • What does the $*$ mean in this case?
    – Melody
    2 days ago








  • 3




    $mathbb{Z}^*_{N^2}$ is not a subgroup of $mathbb{Z}_{N^2}$; one is a group under multiplication, the other is a group under addition...
    – Matt B
    2 days ago






  • 2




    The order of $mathbb{Z}_{n}^*$ is $phi(n)$. so $phi(p^2q^2)=p(p-1)*q*(q-1)$ Here is the wiki link for $phi(n)$: en.wikipedia.org/wiki/Euler%27s_totient_function also:math.stackexchange.com/questions/194705/…
    – mathnoob
    2 days ago












  • Added definition in my post. @Melody
    – Andry
    2 days ago










  • Thanks for pointing out that my problem was already based on my wrong assumption @MattB
    – Andry
    2 days ago


















  • What does the $*$ mean in this case?
    – Melody
    2 days ago








  • 3




    $mathbb{Z}^*_{N^2}$ is not a subgroup of $mathbb{Z}_{N^2}$; one is a group under multiplication, the other is a group under addition...
    – Matt B
    2 days ago






  • 2




    The order of $mathbb{Z}_{n}^*$ is $phi(n)$. so $phi(p^2q^2)=p(p-1)*q*(q-1)$ Here is the wiki link for $phi(n)$: en.wikipedia.org/wiki/Euler%27s_totient_function also:math.stackexchange.com/questions/194705/…
    – mathnoob
    2 days ago












  • Added definition in my post. @Melody
    – Andry
    2 days ago










  • Thanks for pointing out that my problem was already based on my wrong assumption @MattB
    – Andry
    2 days ago
















What does the $*$ mean in this case?
– Melody
2 days ago






What does the $*$ mean in this case?
– Melody
2 days ago






3




3




$mathbb{Z}^*_{N^2}$ is not a subgroup of $mathbb{Z}_{N^2}$; one is a group under multiplication, the other is a group under addition...
– Matt B
2 days ago




$mathbb{Z}^*_{N^2}$ is not a subgroup of $mathbb{Z}_{N^2}$; one is a group under multiplication, the other is a group under addition...
– Matt B
2 days ago




2




2




The order of $mathbb{Z}_{n}^*$ is $phi(n)$. so $phi(p^2q^2)=p(p-1)*q*(q-1)$ Here is the wiki link for $phi(n)$: en.wikipedia.org/wiki/Euler%27s_totient_function also:math.stackexchange.com/questions/194705/…
– mathnoob
2 days ago






The order of $mathbb{Z}_{n}^*$ is $phi(n)$. so $phi(p^2q^2)=p(p-1)*q*(q-1)$ Here is the wiki link for $phi(n)$: en.wikipedia.org/wiki/Euler%27s_totient_function also:math.stackexchange.com/questions/194705/…
– mathnoob
2 days ago














Added definition in my post. @Melody
– Andry
2 days ago




Added definition in my post. @Melody
– Andry
2 days ago












Thanks for pointing out that my problem was already based on my wrong assumption @MattB
– Andry
2 days ago




Thanks for pointing out that my problem was already based on my wrong assumption @MattB
– Andry
2 days ago










1 Answer
1






active

oldest

votes

















up vote
3
down vote



accepted










First of all, $mathbb{Z}^*_n$ is not a subgroup of $mathbb{Z}_n$ since they have different operations, with this in mind; it is well known that for an integer $n$, $ord(mathbb{Z}^*_n)=varphi(n)$, so you just have to find $varphi(p^2q^2)$, first we analyze when $p,q$ are different primes, we have



$$varphi(p^2q^2)=varphi(p^2)varphi(q^2)$$
$$=(p^2-p)(q^2-q)$$
$$=N^2-Np -Nq+N$$



If $p=q$ we have to calculate $varphi(p^4)$



$$varphi(p^4)=p^4-p^3 $$
$$=N^2-Np$$






share|cite|improve this answer























    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',
    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%2f3003557%2fhow-to-find-order-of-the-subgroup%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








    up vote
    3
    down vote



    accepted










    First of all, $mathbb{Z}^*_n$ is not a subgroup of $mathbb{Z}_n$ since they have different operations, with this in mind; it is well known that for an integer $n$, $ord(mathbb{Z}^*_n)=varphi(n)$, so you just have to find $varphi(p^2q^2)$, first we analyze when $p,q$ are different primes, we have



    $$varphi(p^2q^2)=varphi(p^2)varphi(q^2)$$
    $$=(p^2-p)(q^2-q)$$
    $$=N^2-Np -Nq+N$$



    If $p=q$ we have to calculate $varphi(p^4)$



    $$varphi(p^4)=p^4-p^3 $$
    $$=N^2-Np$$






    share|cite|improve this answer



























      up vote
      3
      down vote



      accepted










      First of all, $mathbb{Z}^*_n$ is not a subgroup of $mathbb{Z}_n$ since they have different operations, with this in mind; it is well known that for an integer $n$, $ord(mathbb{Z}^*_n)=varphi(n)$, so you just have to find $varphi(p^2q^2)$, first we analyze when $p,q$ are different primes, we have



      $$varphi(p^2q^2)=varphi(p^2)varphi(q^2)$$
      $$=(p^2-p)(q^2-q)$$
      $$=N^2-Np -Nq+N$$



      If $p=q$ we have to calculate $varphi(p^4)$



      $$varphi(p^4)=p^4-p^3 $$
      $$=N^2-Np$$






      share|cite|improve this answer

























        up vote
        3
        down vote



        accepted







        up vote
        3
        down vote



        accepted






        First of all, $mathbb{Z}^*_n$ is not a subgroup of $mathbb{Z}_n$ since they have different operations, with this in mind; it is well known that for an integer $n$, $ord(mathbb{Z}^*_n)=varphi(n)$, so you just have to find $varphi(p^2q^2)$, first we analyze when $p,q$ are different primes, we have



        $$varphi(p^2q^2)=varphi(p^2)varphi(q^2)$$
        $$=(p^2-p)(q^2-q)$$
        $$=N^2-Np -Nq+N$$



        If $p=q$ we have to calculate $varphi(p^4)$



        $$varphi(p^4)=p^4-p^3 $$
        $$=N^2-Np$$






        share|cite|improve this answer














        First of all, $mathbb{Z}^*_n$ is not a subgroup of $mathbb{Z}_n$ since they have different operations, with this in mind; it is well known that for an integer $n$, $ord(mathbb{Z}^*_n)=varphi(n)$, so you just have to find $varphi(p^2q^2)$, first we analyze when $p,q$ are different primes, we have



        $$varphi(p^2q^2)=varphi(p^2)varphi(q^2)$$
        $$=(p^2-p)(q^2-q)$$
        $$=N^2-Np -Nq+N$$



        If $p=q$ we have to calculate $varphi(p^4)$



        $$varphi(p^4)=p^4-p^3 $$
        $$=N^2-Np$$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited yesterday

























        answered 2 days ago









        Bruno Andrades

        1515




        1515






























             

            draft saved


            draft discarded



















































             


            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3003557%2fhow-to-find-order-of-the-subgroup%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

            Can a sorcerer learn a 5th-level spell early by creating spell slots using the Font of Magic feature?

            Image
            28 6 $begingroup$ Per the Font of Magic feature, sorcerer can use Flexible Casting to create 5th-level spell slots at level 7, even though they are not typically available until level 9. You can transform unexpended sorcery points into one spell slot as a bonus action on your turn. The Creating Spell Slots table shows the cost of creating a spell slot of [5th level is 7] If such a sorcerer levels up while still having this 5th-level spell slot, can they choose a 5th-level spell as the spell they gain upon levelling up? The Spellcasting feature states: Additionally, when you gain a level in this class, you can choose one of the sorcerer spells you know and replace it with another spell from the sorcerer spell list, which also must be of a level for which you have spell slots.
            Read more

            Does disintegrating a polymorphed enemy still kill it after the 2018 errata?

            Image
            up vote 13 down vote favorite 1 After the 2018 errata, the disintegrate spell description now reads: A creature targeted by this spell must make a Dexterity saving throw. On a failed save, the target takes 10d6 + 40 force damage. The target is disintegrated if this damage leaves it with 0 hit points. If you were to polymorph an enemy into a rat and then disintegrate it, would the enemy be disintegrated or would it just return to its original form? I know that for Druids, it’s not an instant kill anymore, but is this the case for polymorph as well? dnd-5e spells polymorph errata share | improve this question edited 18 hours ago
            Read more

            A Topological Invariant for $pi_3(U(n))$

            Image
            3 3 $begingroup$ Recently, I saw a construction of topological invariant for $pi_3(U(n))$ with $ngeq 2$ : $$ N=frac{1}{24pi^2}int_{S^3} d^3x epsilon^{ijk} Tr[(U^{-1}partial_{x_i}U)(U^{-1}partial_{x_j}U)(U^{-1}partial_{x_k}U)] , $$ where $Uin U(n)$ depends on $boldsymbol{x}=(x_1,x_2,x_3)in S^3$ , $epsilon^{ijk}$ is the Levi-Civita symbol, $i,j,k=1,2,3$ , and the duplicated indexes are summed over. It is claimed that $N$ is an integer, but why? Update 02/02/2019 I think I got an argument for $n=2$ . In this case, $U=e^{i varphi} q$ with $qin SU(2)$ . Due to the trace and the Levi-Civita symbol in $N$ , $varphi$ does not contribute to $N$ . As $Tr[q^{dagger}partial_i q]=0$ and $(q^{dagger}partial_i q)^{dagger}=-q^{dagger}partial_i q$ , $q^{dagger}partial_i q$ in geneeral has the form $q^{dagger}partia
            Read more