Finite field generator exponentiation properties












0












$begingroup$


For a finite integer field $mathbb{Z}_n$ with a generator $g$ and $xinmathbb{Z}_n$, a property I do not understand arises and I have not come across an explanation for it. If another set of values $x_1=n+x, x_2=2n+xdots$ then $g^{x}neq g^{x_1}neq g^{x_2}dots$. Why is this? I've noticed and conjectured that for an $x_k=kn+x$, $g^{x}=g^{x_k +k}$ at least for several examples I have been toying with to get a feel for the mechanics. Since all these values are members of the same congruence class, I'd expect them to generate identical field elements. Is it because the multiplicative subgroup is of order $n-1 $ due to the removal of the additive identity? I'd appreciate any guidance and clarification on this matter.






abstract-algebra finite-fields







share|cite|improve this question









$endgroup$












  • $begingroup$
    Is $n$ a prime number?
    $endgroup$
    – Bernard
    Dec 3 '18 at 23:19






  • 1




    $begingroup$
    Every element of $mathbb{F}_p$ satisfies $g^p = g$, so $g^{2p} = g^2, g^{3p} = g^3$, etc. In other words, since the multiplicative group has order $p - 1$, the exponents are numbers that make sense $bmod (p - 1)$, not $bmod p$. In general exponentiation takes in two different types of objects as input; the fact that exponentiation seems like it takes in the same types of objects as input over the real numbers is misleading.
    $endgroup$
    – Qiaochu Yuan
    Dec 3 '18 at 23:22












  • $begingroup$
    @Bernard yes it's a prime modulus, sorry for not including that detail explicitly.
    $endgroup$
    – Ken Goss
    Dec 3 '18 at 23:38






  • 1




    $begingroup$
    I have a long rant about this; I wrote a (bad) version of it here: math.stackexchange.com/questions/56663/…
    $endgroup$
    – Qiaochu Yuan
    Dec 3 '18 at 23:45






  • 1




    $begingroup$
    @QiaochuYuan you’re right that is a long rant, but I rather like it.
    $endgroup$
    – Randall
    Dec 4 '18 at 0:47
















0












$begingroup$


For a finite integer field $mathbb{Z}_n$ with a generator $g$ and $xinmathbb{Z}_n$, a property I do not understand arises and I have not come across an explanation for it. If another set of values $x_1=n+x, x_2=2n+xdots$ then $g^{x}neq g^{x_1}neq g^{x_2}dots$. Why is this? I've noticed and conjectured that for an $x_k=kn+x$, $g^{x}=g^{x_k +k}$ at least for several examples I have been toying with to get a feel for the mechanics. Since all these values are members of the same congruence class, I'd expect them to generate identical field elements. Is it because the multiplicative subgroup is of order $n-1 $ due to the removal of the additive identity? I'd appreciate any guidance and clarification on this matter.






abstract-algebra finite-fields







share|cite|improve this question









$endgroup$












  • $begingroup$
    Is $n$ a prime number?
    $endgroup$
    – Bernard
    Dec 3 '18 at 23:19






  • 1




    $begingroup$
    Every element of $mathbb{F}_p$ satisfies $g^p = g$, so $g^{2p} = g^2, g^{3p} = g^3$, etc. In other words, since the multiplicative group has order $p - 1$, the exponents are numbers that make sense $bmod (p - 1)$, not $bmod p$. In general exponentiation takes in two different types of objects as input; the fact that exponentiation seems like it takes in the same types of objects as input over the real numbers is misleading.
    $endgroup$
    – Qiaochu Yuan
    Dec 3 '18 at 23:22












  • $begingroup$
    @Bernard yes it's a prime modulus, sorry for not including that detail explicitly.
    $endgroup$
    – Ken Goss
    Dec 3 '18 at 23:38






  • 1




    $begingroup$
    I have a long rant about this; I wrote a (bad) version of it here: math.stackexchange.com/questions/56663/…
    $endgroup$
    – Qiaochu Yuan
    Dec 3 '18 at 23:45






  • 1




    $begingroup$
    @QiaochuYuan you’re right that is a long rant, but I rather like it.
    $endgroup$
    – Randall
    Dec 4 '18 at 0:47














0












0








0





$begingroup$


For a finite integer field $mathbb{Z}_n$ with a generator $g$ and $xinmathbb{Z}_n$, a property I do not understand arises and I have not come across an explanation for it. If another set of values $x_1=n+x, x_2=2n+xdots$ then $g^{x}neq g^{x_1}neq g^{x_2}dots$. Why is this? I've noticed and conjectured that for an $x_k=kn+x$, $g^{x}=g^{x_k +k}$ at least for several examples I have been toying with to get a feel for the mechanics. Since all these values are members of the same congruence class, I'd expect them to generate identical field elements. Is it because the multiplicative subgroup is of order $n-1 $ due to the removal of the additive identity? I'd appreciate any guidance and clarification on this matter.






abstract-algebra finite-fields







share|cite|improve this question









$endgroup$




For a finite integer field $mathbb{Z}_n$ with a generator $g$ and $xinmathbb{Z}_n$, a property I do not understand arises and I have not come across an explanation for it. If another set of values $x_1=n+x, x_2=2n+xdots$ then $g^{x}neq g^{x_1}neq g^{x_2}dots$. Why is this? I've noticed and conjectured that for an $x_k=kn+x$, $g^{x}=g^{x_k +k}$ at least for several examples I have been toying with to get a feel for the mechanics. Since all these values are members of the same congruence class, I'd expect them to generate identical field elements. Is it because the multiplicative subgroup is of order $n-1 $ due to the removal of the additive identity? I'd appreciate any guidance and clarification on this matter.






abstract-algebra finite-fields




abstract-algebra finite-fields






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 3 '18 at 23:08









Ken GossKen Goss

1035




1035












  • $begingroup$
    Is $n$ a prime number?
    $endgroup$
    – Bernard
    Dec 3 '18 at 23:19






  • 1




    $begingroup$
    Every element of $mathbb{F}_p$ satisfies $g^p = g$, so $g^{2p} = g^2, g^{3p} = g^3$, etc. In other words, since the multiplicative group has order $p - 1$, the exponents are numbers that make sense $bmod (p - 1)$, not $bmod p$. In general exponentiation takes in two different types of objects as input; the fact that exponentiation seems like it takes in the same types of objects as input over the real numbers is misleading.
    $endgroup$
    – Qiaochu Yuan
    Dec 3 '18 at 23:22












  • $begingroup$
    @Bernard yes it's a prime modulus, sorry for not including that detail explicitly.
    $endgroup$
    – Ken Goss
    Dec 3 '18 at 23:38






  • 1




    $begingroup$
    I have a long rant about this; I wrote a (bad) version of it here: math.stackexchange.com/questions/56663/…
    $endgroup$
    – Qiaochu Yuan
    Dec 3 '18 at 23:45






  • 1




    $begingroup$
    @QiaochuYuan you’re right that is a long rant, but I rather like it.
    $endgroup$
    – Randall
    Dec 4 '18 at 0:47


















  • $begingroup$
    Is $n$ a prime number?
    $endgroup$
    – Bernard
    Dec 3 '18 at 23:19






  • 1




    $begingroup$
    Every element of $mathbb{F}_p$ satisfies $g^p = g$, so $g^{2p} = g^2, g^{3p} = g^3$, etc. In other words, since the multiplicative group has order $p - 1$, the exponents are numbers that make sense $bmod (p - 1)$, not $bmod p$. In general exponentiation takes in two different types of objects as input; the fact that exponentiation seems like it takes in the same types of objects as input over the real numbers is misleading.
    $endgroup$
    – Qiaochu Yuan
    Dec 3 '18 at 23:22












  • $begingroup$
    @Bernard yes it's a prime modulus, sorry for not including that detail explicitly.
    $endgroup$
    – Ken Goss
    Dec 3 '18 at 23:38






  • 1




    $begingroup$
    I have a long rant about this; I wrote a (bad) version of it here: math.stackexchange.com/questions/56663/…
    $endgroup$
    – Qiaochu Yuan
    Dec 3 '18 at 23:45






  • 1




    $begingroup$
    @QiaochuYuan you’re right that is a long rant, but I rather like it.
    $endgroup$
    – Randall
    Dec 4 '18 at 0:47
















$begingroup$
Is $n$ a prime number?
$endgroup$
– Bernard
Dec 3 '18 at 23:19




$begingroup$
Is $n$ a prime number?
$endgroup$
– Bernard
Dec 3 '18 at 23:19




1




1




$begingroup$
Every element of $mathbb{F}_p$ satisfies $g^p = g$, so $g^{2p} = g^2, g^{3p} = g^3$, etc. In other words, since the multiplicative group has order $p - 1$, the exponents are numbers that make sense $bmod (p - 1)$, not $bmod p$. In general exponentiation takes in two different types of objects as input; the fact that exponentiation seems like it takes in the same types of objects as input over the real numbers is misleading.
$endgroup$
– Qiaochu Yuan
Dec 3 '18 at 23:22






$begingroup$
Every element of $mathbb{F}_p$ satisfies $g^p = g$, so $g^{2p} = g^2, g^{3p} = g^3$, etc. In other words, since the multiplicative group has order $p - 1$, the exponents are numbers that make sense $bmod (p - 1)$, not $bmod p$. In general exponentiation takes in two different types of objects as input; the fact that exponentiation seems like it takes in the same types of objects as input over the real numbers is misleading.
$endgroup$
– Qiaochu Yuan
Dec 3 '18 at 23:22














$begingroup$
@Bernard yes it's a prime modulus, sorry for not including that detail explicitly.
$endgroup$
– Ken Goss
Dec 3 '18 at 23:38




$begingroup$
@Bernard yes it's a prime modulus, sorry for not including that detail explicitly.
$endgroup$
– Ken Goss
Dec 3 '18 at 23:38




1




1




$begingroup$
I have a long rant about this; I wrote a (bad) version of it here: math.stackexchange.com/questions/56663/…
$endgroup$
– Qiaochu Yuan
Dec 3 '18 at 23:45




$begingroup$
I have a long rant about this; I wrote a (bad) version of it here: math.stackexchange.com/questions/56663/…
$endgroup$
– Qiaochu Yuan
Dec 3 '18 at 23:45




1




1




$begingroup$
@QiaochuYuan you’re right that is a long rant, but I rather like it.
$endgroup$
– Randall
Dec 4 '18 at 0:47




$begingroup$
@QiaochuYuan you’re right that is a long rant, but I rather like it.
$endgroup$
– Randall
Dec 4 '18 at 0:47










1 Answer
1






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$begingroup$

If your $n$ is a prime number, by lil' Fermat, we have $g^nequiv gmod n$, hence
$$g^{x_k}=g^{x+k}enspaceforall kinmathbf Z.$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Ah, so straight-forward. I just didn't make the connection. Thank you! Between your answer and Qiaochu Yuan's comments it is clear to me now.
    $endgroup$
    – Ken Goss
    Dec 3 '18 at 23:53











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1 Answer
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active

oldest

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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









1












$begingroup$

If your $n$ is a prime number, by lil' Fermat, we have $g^nequiv gmod n$, hence
$$g^{x_k}=g^{x+k}enspaceforall kinmathbf Z.$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Ah, so straight-forward. I just didn't make the connection. Thank you! Between your answer and Qiaochu Yuan's comments it is clear to me now.
    $endgroup$
    – Ken Goss
    Dec 3 '18 at 23:53
















1












$begingroup$

If your $n$ is a prime number, by lil' Fermat, we have $g^nequiv gmod n$, hence
$$g^{x_k}=g^{x+k}enspaceforall kinmathbf Z.$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Ah, so straight-forward. I just didn't make the connection. Thank you! Between your answer and Qiaochu Yuan's comments it is clear to me now.
    $endgroup$
    – Ken Goss
    Dec 3 '18 at 23:53














1












1








1





$begingroup$

If your $n$ is a prime number, by lil' Fermat, we have $g^nequiv gmod n$, hence
$$g^{x_k}=g^{x+k}enspaceforall kinmathbf Z.$$






share|cite|improve this answer











$endgroup$



If your $n$ is a prime number, by lil' Fermat, we have $g^nequiv gmod n$, hence
$$g^{x_k}=g^{x+k}enspaceforall kinmathbf Z.$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 3 '18 at 23:41

























answered Dec 3 '18 at 23:33









BernardBernard

119k740113




119k740113












  • $begingroup$
    Ah, so straight-forward. I just didn't make the connection. Thank you! Between your answer and Qiaochu Yuan's comments it is clear to me now.
    $endgroup$
    – Ken Goss
    Dec 3 '18 at 23:53


















  • $begingroup$
    Ah, so straight-forward. I just didn't make the connection. Thank you! Between your answer and Qiaochu Yuan's comments it is clear to me now.
    $endgroup$
    – Ken Goss
    Dec 3 '18 at 23:53
















$begingroup$
Ah, so straight-forward. I just didn't make the connection. Thank you! Between your answer and Qiaochu Yuan's comments it is clear to me now.
$endgroup$
– Ken Goss
Dec 3 '18 at 23:53




$begingroup$
Ah, so straight-forward. I just didn't make the connection. Thank you! Between your answer and Qiaochu Yuan's comments it is clear to me now.
$endgroup$
– Ken Goss
Dec 3 '18 at 23:53


















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