Mixing
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
asked 2 days ago
Andry
154
|
show 1 more comment
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
asked 2 days ago
Andry
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
|
show 1 more comment
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
asked 2 days ago
Andry
154
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
asked 2 days ago
Andry
154
asked 2 days ago
Andry
154
edited 2 days ago
asked 2 days ago
Andry
154
asked 2 days ago
Andry
154
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
|
show 1 more comment
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
|
show 1 more comment
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 $varphi(p^4)$
$$varphi(p^4)=p^4-p^3 $$
$$=N^2-Np$$
answered 2 days ago
Bruno Andrades
1515
add a comment |
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 $varphi(p^4)$
$$varphi(p^4)=p^4-p^3 $$
$$=N^2-Np$$
answered 2 days ago
Bruno Andrades
1515
add a comment |
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 $varphi(p^4)$
$$varphi(p^4)=p^4-p^3 $$
$$=N^2-Np$$
answered 2 days ago
Bruno Andrades
1515
add a comment |
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 $varphi(p^4)$
$$varphi(p^4)=p^4-p^3 $$
$$=N^2-Np$$
answered 2 days ago
Bruno Andrades
1515
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 $varphi(p^4)$
$$varphi(p^4)=p^4-p^3 $$
$$=N^2-Np$$
answered 2 days ago
Bruno Andrades
1515
edited yesterday
answered 2 days ago
Bruno Andrades
1515
answered 2 days ago
Bruno Andrades
1515
answered 2 days ago
Bruno Andrades
1515
1515
add a comment |
add a comment |
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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