Why $mathbb{Z}_{lambda(n)}^* le mathbb{Z}_{phi(n)}^*$ does not hold?
$begingroup$
Let $lambda(n) = lambda(mathbb{Z}_n^*)$ be the Carmichael function and $phi(n) = | mathbb{Z}_{n}^{*} |$ be the Euler function. I need to show:
$$mathbb{Z}_{lambda(n)}^* le mathbb{Z}_{phi(n)}^*$$
But why should $mathbb{Z}_{lambda(n)}^* le mathbb{Z}_{phi(n)}^*$ one be a subgroup of the other? This was stated in one course on cryptography but I cannot find a good justification for it.
The claim was wrong
In general this only holds for the subset relation. Still it would be great to find a counterexample. In return, I leave here an argument why the subset relation holds.
abstract-algebra group-theory number-theory
asked Jan 6 at 22:55
JavierJavier
2,02321133
$endgroup$
|
show 1 more comment
$begingroup$
Let $lambda(n) = lambda(mathbb{Z}_n^*)$ be the Carmichael function and $phi(n) = | mathbb{Z}_{n}^{*} |$ be the Euler function. I need to show:
$$mathbb{Z}_{lambda(n)}^* le mathbb{Z}_{phi(n)}^*$$
But why should $mathbb{Z}_{lambda(n)}^* le mathbb{Z}_{phi(n)}^*$ one be a subgroup of the other? This was stated in one course on cryptography but I cannot find a good justification for it.
The claim was wrong
In general this only holds for the subset relation. Still it would be great to find a counterexample. In return, I leave here an argument why the subset relation holds.
abstract-algebra group-theory number-theory
asked Jan 6 at 22:55
JavierJavier
2,02321133
$endgroup$
$begingroup$
If $G$ is abelian then $f_k(x) = x^k$ is an homomorphism $G to G$. If $G = (mathbb{Z}/n mathbb{Z})^times$ with $n$ odd (having $phi(n) =prod_{p^r | n} (p-1)p^{r-1}$ elements) then $ker(f_k)$ has $prod_{p^r | n} frac{(p-1)p^{r-1}}{gcd((p-1)p^{r-1},k)}$ elements. $lambda(n)$ is the exponent of $(mathbb{Z}/n mathbb{Z})^times$ ie. the least integer such that $ker(f_k) = (mathbb{Z}/n mathbb{Z})^times$ ie. $lambda(n) = lcm( ((p-1)p^{r-1})_{p^r |n})$.
$endgroup$
– reuns
Jan 6 at 23:05
3
$begingroup$
We rather have an embedding from the cyclic group $Bbb Z_{lambda(n)}$ (of additive notation) to the multiplicative group $Bbb Z_{mathbb n}^*$ of the ring $Bbb Z_n$ which has $varphi(n)$ elements.
$endgroup$
– Berci
Jan 6 at 23:07
$begingroup$
I am confused because you introduced a group $G$, but the question that you ask does not appear to involve $G$.
$endgroup$
– Derek Holt
Jan 7 at 9:03
$begingroup$
@DerekHolt Thank you, I reformulated my question. $G$ is not relevant for the question itself
$endgroup$
– Javier
Jan 7 at 23:17
$begingroup$
@Berci yes, you're right that is what follows from the facts i listed , my interest is to prove the other fact though
$endgroup$
– Javier
Jan 7 at 23:22
|
show 1 more comment
$begingroup$
Let $lambda(n) = lambda(mathbb{Z}_n^*)$ be the Carmichael function and $phi(n) = | mathbb{Z}_{n}^{*} |$ be the Euler function. I need to show:
$$mathbb{Z}_{lambda(n)}^* le mathbb{Z}_{phi(n)}^*$$
But why should $mathbb{Z}_{lambda(n)}^* le mathbb{Z}_{phi(n)}^*$ one be a subgroup of the other? This was stated in one course on cryptography but I cannot find a good justification for it.
The claim was wrong
In general this only holds for the subset relation. Still it would be great to find a counterexample. In return, I leave here an argument why the subset relation holds.
abstract-algebra group-theory number-theory
asked Jan 6 at 22:55
JavierJavier
2,02321133
$endgroup$
Let $lambda(n) = lambda(mathbb{Z}_n^*)$ be the Carmichael function and $phi(n) = | mathbb{Z}_{n}^{*} |$ be the Euler function. I need to show:
$$mathbb{Z}_{lambda(n)}^* le mathbb{Z}_{phi(n)}^*$$
But why should $mathbb{Z}_{lambda(n)}^* le mathbb{Z}_{phi(n)}^*$ one be a subgroup of the other? This was stated in one course on cryptography but I cannot find a good justification for it.
The claim was wrong
In general this only holds for the subset relation. Still it would be great to find a counterexample. In return, I leave here an argument why the subset relation holds.
abstract-algebra group-theory number-theory
abstract-algebra group-theory number-theory
asked Jan 6 at 22:55
JavierJavier
2,02321133
asked Jan 6 at 22:55
JavierJavier
2,02321133
edited Jan 28 at 22:39
Javier
asked Jan 6 at 22:55
JavierJavier
2,02321133
asked Jan 6 at 22:55
JavierJavier
2,02321133
asked Jan 6 at 22:55
JavierJavier
2,02321133
2,02321133
$begingroup$
If $G$ is abelian then $f_k(x) = x^k$ is an homomorphism $G to G$. If $G = (mathbb{Z}/n mathbb{Z})^times$ with $n$ odd (having $phi(n) =prod_{p^r | n} (p-1)p^{r-1}$ elements) then $ker(f_k)$ has $prod_{p^r | n} frac{(p-1)p^{r-1}}{gcd((p-1)p^{r-1},k)}$ elements. $lambda(n)$ is the exponent of $(mathbb{Z}/n mathbb{Z})^times$ ie. the least integer such that $ker(f_k) = (mathbb{Z}/n mathbb{Z})^times$ ie. $lambda(n) = lcm( ((p-1)p^{r-1})_{p^r |n})$.
$endgroup$
– reuns
Jan 6 at 23:05
3
$begingroup$
We rather have an embedding from the cyclic group $Bbb Z_{lambda(n)}$ (of additive notation) to the multiplicative group $Bbb Z_{mathbb n}^*$ of the ring $Bbb Z_n$ which has $varphi(n)$ elements.
$endgroup$
– Berci
Jan 6 at 23:07
$begingroup$
I am confused because you introduced a group $G$, but the question that you ask does not appear to involve $G$.
$endgroup$
– Derek Holt
Jan 7 at 9:03
$begingroup$
@DerekHolt Thank you, I reformulated my question. $G$ is not relevant for the question itself
$endgroup$
– Javier
Jan 7 at 23:17
$begingroup$
@Berci yes, you're right that is what follows from the facts i listed , my interest is to prove the other fact though
$endgroup$
– Javier
Jan 7 at 23:22
|
show 1 more comment
$begingroup$
If $G$ is abelian then $f_k(x) = x^k$ is an homomorphism $G to G$. If $G = (mathbb{Z}/n mathbb{Z})^times$ with $n$ odd (having $phi(n) =prod_{p^r | n} (p-1)p^{r-1}$ elements) then $ker(f_k)$ has $prod_{p^r | n} frac{(p-1)p^{r-1}}{gcd((p-1)p^{r-1},k)}$ elements. $lambda(n)$ is the exponent of $(mathbb{Z}/n mathbb{Z})^times$ ie. the least integer such that $ker(f_k) = (mathbb{Z}/n mathbb{Z})^times$ ie. $lambda(n) = lcm( ((p-1)p^{r-1})_{p^r |n})$.
$endgroup$
– reuns
Jan 6 at 23:05
3
$begingroup$
We rather have an embedding from the cyclic group $Bbb Z_{lambda(n)}$ (of additive notation) to the multiplicative group $Bbb Z_{mathbb n}^*$ of the ring $Bbb Z_n$ which has $varphi(n)$ elements.
$endgroup$
– Berci
Jan 6 at 23:07
$begingroup$
I am confused because you introduced a group $G$, but the question that you ask does not appear to involve $G$.
$endgroup$
– Derek Holt
Jan 7 at 9:03
$begingroup$
@DerekHolt Thank you, I reformulated my question. $G$ is not relevant for the question itself
$endgroup$
– Javier
Jan 7 at 23:17
$begingroup$
@Berci yes, you're right that is what follows from the facts i listed , my interest is to prove the other fact though
$endgroup$
– Javier
Jan 7 at 23:22
$begingroup$
If $G$ is abelian then $f_k(x) = x^k$ is an homomorphism $G to G$. If $G = (mathbb{Z}/n mathbb{Z})^times$ with $n$ odd (having $phi(n) =prod_{p^r | n} (p-1)p^{r-1}$ elements) then $ker(f_k)$ has $prod_{p^r | n} frac{(p-1)p^{r-1}}{gcd((p-1)p^{r-1},k)}$ elements. $lambda(n)$ is the exponent of $(mathbb{Z}/n mathbb{Z})^times$ ie. the least integer such that $ker(f_k) = (mathbb{Z}/n mathbb{Z})^times$ ie. $lambda(n) = lcm( ((p-1)p^{r-1})_{p^r |n})$.
$endgroup$
– reuns
Jan 6 at 23:05
$begingroup$
If $G$ is abelian then $f_k(x) = x^k$ is an homomorphism $G to G$. If $G = (mathbb{Z}/n mathbb{Z})^times$ with $n$ odd (having $phi(n) =prod_{p^r | n} (p-1)p^{r-1}$ elements) then $ker(f_k)$ has $prod_{p^r | n} frac{(p-1)p^{r-1}}{gcd((p-1)p^{r-1},k)}$ elements. $lambda(n)$ is the exponent of $(mathbb{Z}/n mathbb{Z})^times$ ie. the least integer such that $ker(f_k) = (mathbb{Z}/n mathbb{Z})^times$ ie. $lambda(n) = lcm( ((p-1)p^{r-1})_{p^r |n})$.
$endgroup$
– reuns
Jan 6 at 23:05
3
3
$begingroup$
We rather have an embedding from the cyclic group $Bbb Z_{lambda(n)}$ (of additive notation) to the multiplicative group $Bbb Z_{mathbb n}^*$ of the ring $Bbb Z_n$ which has $varphi(n)$ elements.
$endgroup$
– Berci
Jan 6 at 23:07
$begingroup$
We rather have an embedding from the cyclic group $Bbb Z_{lambda(n)}$ (of additive notation) to the multiplicative group $Bbb Z_{mathbb n}^*$ of the ring $Bbb Z_n$ which has $varphi(n)$ elements.
$endgroup$
– Berci
Jan 6 at 23:07
$begingroup$
I am confused because you introduced a group $G$, but the question that you ask does not appear to involve $G$.
$endgroup$
– Derek Holt
Jan 7 at 9:03
$begingroup$
I am confused because you introduced a group $G$, but the question that you ask does not appear to involve $G$.
$endgroup$
– Derek Holt
Jan 7 at 9:03
$begingroup$
@DerekHolt Thank you, I reformulated my question. $G$ is not relevant for the question itself
$endgroup$
– Javier
Jan 7 at 23:17
$begingroup$
@DerekHolt Thank you, I reformulated my question. $G$ is not relevant for the question itself
$endgroup$
– Javier
Jan 7 at 23:17
$begingroup$
@Berci yes, you're right that is what follows from the facts i listed , my interest is to prove the other fact though
$endgroup$
– Javier
Jan 7 at 23:22
$begingroup$
@Berci yes, you're right that is what follows from the facts i listed , my interest is to prove the other fact though
$endgroup$
– Javier
Jan 7 at 23:22
|
show 1 more comment
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$begingroup$
If $G$ is abelian then $f_k(x) = x^k$ is an homomorphism $G to G$. If $G = (mathbb{Z}/n mathbb{Z})^times$ with $n$ odd (having $phi(n) =prod_{p^r | n} (p-1)p^{r-1}$ elements) then $ker(f_k)$ has $prod_{p^r | n} frac{(p-1)p^{r-1}}{gcd((p-1)p^{r-1},k)}$ elements. $lambda(n)$ is the exponent of $(mathbb{Z}/n mathbb{Z})^times$ ie. the least integer such that $ker(f_k) = (mathbb{Z}/n mathbb{Z})^times$ ie. $lambda(n) = lcm( ((p-1)p^{r-1})_{p^r |n})$.
$endgroup$
– reuns
Jan 6 at 23:05
3
$begingroup$
We rather have an embedding from the cyclic group $Bbb Z_{lambda(n)}$ (of additive notation) to the multiplicative group $Bbb Z_{mathbb n}^*$ of the ring $Bbb Z_n$ which has $varphi(n)$ elements.
$endgroup$
– Berci
Jan 6 at 23:07
$begingroup$
I am confused because you introduced a group $G$, but the question that you ask does not appear to involve $G$.
$endgroup$
– Derek Holt
Jan 7 at 9:03
$begingroup$
@DerekHolt Thank you, I reformulated my question. $G$ is not relevant for the question itself
$endgroup$
– Javier
Jan 7 at 23:17
$begingroup$
@Berci yes, you're right that is what follows from the facts i listed , my interest is to prove the other fact though
$endgroup$
– Javier
Jan 7 at 23:22