Why $mathbb{Z}_{lambda(n)}^* le mathbb{Z}_{phi(n)}^*$?












6












$begingroup$


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:



$$mathbb{Z}_{lambda(n)}^* le mathbb{Z}_{phi(n)}^*$$



Some facts I know (maybe not relevant)




  1. $lambda(n)|phi(n)$.


  2. Any finite abelian group $G$ contains a cyclic subgroup of order $lambda(G)$.


  3. Any two finite cyclic groups of the same order are isomorphic.



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.






abstract-algebra group-theory number-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    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








  • 2




    $begingroup$
    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
















6












$begingroup$


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:



$$mathbb{Z}_{lambda(n)}^* le mathbb{Z}_{phi(n)}^*$$



Some facts I know (maybe not relevant)




  1. $lambda(n)|phi(n)$.


  2. Any finite abelian group $G$ contains a cyclic subgroup of order $lambda(G)$.


  3. Any two finite cyclic groups of the same order are isomorphic.



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.






abstract-algebra group-theory number-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    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








  • 2




    $begingroup$
    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














6












6








6





$begingroup$


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:



$$mathbb{Z}_{lambda(n)}^* le mathbb{Z}_{phi(n)}^*$$



Some facts I know (maybe not relevant)




  1. $lambda(n)|phi(n)$.


  2. Any finite abelian group $G$ contains a cyclic subgroup of order $lambda(G)$.


  3. Any two finite cyclic groups of the same order are isomorphic.



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.






abstract-algebra group-theory number-theory







share|cite|improve this question











$endgroup$




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:



$$mathbb{Z}_{lambda(n)}^* le mathbb{Z}_{phi(n)}^*$$



Some facts I know (maybe not relevant)




  1. $lambda(n)|phi(n)$.


  2. Any finite abelian group $G$ contains a cyclic subgroup of order $lambda(G)$.


  3. Any two finite cyclic groups of the same order are isomorphic.



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.






abstract-algebra group-theory number-theory




abstract-algebra group-theory number-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 7 at 23:16







Javier

















asked Jan 6 at 22:55









JavierJavier

2,01621133




2,01621133












  • $begingroup$
    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








  • 2




    $begingroup$
    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








  • 2




    $begingroup$
    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






2




2




$begingroup$
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










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