How to find order of the subgroup?











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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
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3
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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























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    up vote
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    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

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