Proving exponentiation in $mathbb{Z}^{*}_{pq}$ is one to one
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We saw in our crypto class that in the group $mathbb{Z}^{*}_{pq}$ where $p$ and $q$ are primes, that if for some $a$, $gcd(a, phi(pq) = (p-1)(q-1))$ = 1 (where $phi$ is Euler's totient function) then the function $f(x) = x^a$ is one to one, and he lecturer proved this by calculating the inverse of $a$ (the RSA trapdoor). What I have trouble understanding is why this proves the function is one to one. We computed an inverse function, that is showed that there is an inverse, but how does this show that there is no other inverse for an element?
group-theory number-theory cryptography
edited Dec 28 '18 at 17:21
Shaun
9,620113684
asked Dec 28 '18 at 17:07
ElooEloo
1418
$endgroup$
add a comment |
$begingroup$
We saw in our crypto class that in the group $mathbb{Z}^{*}_{pq}$ where $p$ and $q$ are primes, that if for some $a$, $gcd(a, phi(pq) = (p-1)(q-1))$ = 1 (where $phi$ is Euler's totient function) then the function $f(x) = x^a$ is one to one, and he lecturer proved this by calculating the inverse of $a$ (the RSA trapdoor). What I have trouble understanding is why this proves the function is one to one. We computed an inverse function, that is showed that there is an inverse, but how does this show that there is no other inverse for an element?
group-theory number-theory cryptography
edited Dec 28 '18 at 17:21
Shaun
9,620113684
asked Dec 28 '18 at 17:07
ElooEloo
1418
$endgroup$
$begingroup$
If you can compute an inverse function, then by definition your function is both one-to-one and onto.
$endgroup$
– DudeMan
Dec 28 '18 at 17:12
add a comment |
$begingroup$
We saw in our crypto class that in the group $mathbb{Z}^{*}_{pq}$ where $p$ and $q$ are primes, that if for some $a$, $gcd(a, phi(pq) = (p-1)(q-1))$ = 1 (where $phi$ is Euler's totient function) then the function $f(x) = x^a$ is one to one, and he lecturer proved this by calculating the inverse of $a$ (the RSA trapdoor). What I have trouble understanding is why this proves the function is one to one. We computed an inverse function, that is showed that there is an inverse, but how does this show that there is no other inverse for an element?
group-theory number-theory cryptography
edited Dec 28 '18 at 17:21
Shaun
9,620113684
asked Dec 28 '18 at 17:07
ElooEloo
1418
$endgroup$
We saw in our crypto class that in the group $mathbb{Z}^{*}_{pq}$ where $p$ and $q$ are primes, that if for some $a$, $gcd(a, phi(pq) = (p-1)(q-1))$ = 1 (where $phi$ is Euler's totient function) then the function $f(x) = x^a$ is one to one, and he lecturer proved this by calculating the inverse of $a$ (the RSA trapdoor). What I have trouble understanding is why this proves the function is one to one. We computed an inverse function, that is showed that there is an inverse, but how does this show that there is no other inverse for an element?
group-theory number-theory cryptography
group-theory number-theory cryptography
edited Dec 28 '18 at 17:21
Shaun
9,620113684
asked Dec 28 '18 at 17:07
ElooEloo
1418
edited Dec 28 '18 at 17:21
Shaun
9,620113684
asked Dec 28 '18 at 17:07
ElooEloo
1418
edited Dec 28 '18 at 17:21
Shaun
9,620113684
edited Dec 28 '18 at 17:21
Shaun
9,620113684
edited Dec 28 '18 at 17:21
Shaun
9,620113684
9,620113684
asked Dec 28 '18 at 17:07
ElooEloo
1418
asked Dec 28 '18 at 17:07
ElooEloo
1418
asked Dec 28 '18 at 17:07
ElooEloo
1418
1418
$begingroup$
If you can compute an inverse function, then by definition your function is both one-to-one and onto.
$endgroup$
– DudeMan
Dec 28 '18 at 17:12
add a comment |
$begingroup$
If you can compute an inverse function, then by definition your function is both one-to-one and onto.
$endgroup$
– DudeMan
Dec 28 '18 at 17:12
$begingroup$
If you can compute an inverse function, then by definition your function is both one-to-one and onto.
$endgroup$
– DudeMan
Dec 28 '18 at 17:12
$begingroup$
If you can compute an inverse function, then by definition your function is both one-to-one and onto.
$endgroup$
– DudeMan
Dec 28 '18 at 17:12
add a comment |
2 Answers
2
active
oldest
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If $f$ is an function and $g$ and $h$ are inverse functions for $f,$ then $g=h$ since $$g=g(fh)=(gf)h=h.$$
answered Dec 28 '18 at 17:15
D. BroganD. Brogan
643513
$endgroup$
add a comment |
$begingroup$
In general: if $G$ is a finite group and $k$ a positive integer. Then the map $f: G rightarrow G$ defined by $f(g)=g^k$ is a bijection if and only if $gcd(k,|G|)=1$. (The proof of the "if" uses Bézout's Lemma, the "only if" leverages Cauchy's Theorem)
answered Dec 28 '18 at 20:39
Nicky HeksterNicky Hekster
29k63456
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add a comment |
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2 Answers
2
active
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2 Answers
2
active
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active
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active
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$begingroup$
If $f$ is an function and $g$ and $h$ are inverse functions for $f,$ then $g=h$ since $$g=g(fh)=(gf)h=h.$$
answered Dec 28 '18 at 17:15
D. BroganD. Brogan
643513
$endgroup$
add a comment |
$begingroup$
If $f$ is an function and $g$ and $h$ are inverse functions for $f,$ then $g=h$ since $$g=g(fh)=(gf)h=h.$$
answered Dec 28 '18 at 17:15
D. BroganD. Brogan
643513
$endgroup$
add a comment |
$begingroup$
If $f$ is an function and $g$ and $h$ are inverse functions for $f,$ then $g=h$ since $$g=g(fh)=(gf)h=h.$$
answered Dec 28 '18 at 17:15
D. BroganD. Brogan
643513
$endgroup$
If $f$ is an function and $g$ and $h$ are inverse functions for $f,$ then $g=h$ since $$g=g(fh)=(gf)h=h.$$
answered Dec 28 '18 at 17:15
D. BroganD. Brogan
643513
answered Dec 28 '18 at 17:15
D. BroganD. Brogan
643513
answered Dec 28 '18 at 17:15
D. BroganD. Brogan
643513
answered Dec 28 '18 at 17:15
D. BroganD. Brogan
643513
643513
add a comment |
add a comment |
$begingroup$
In general: if $G$ is a finite group and $k$ a positive integer. Then the map $f: G rightarrow G$ defined by $f(g)=g^k$ is a bijection if and only if $gcd(k,|G|)=1$. (The proof of the "if" uses Bézout's Lemma, the "only if" leverages Cauchy's Theorem)
answered Dec 28 '18 at 20:39
Nicky HeksterNicky Hekster
29k63456
$endgroup$
add a comment |
$begingroup$
In general: if $G$ is a finite group and $k$ a positive integer. Then the map $f: G rightarrow G$ defined by $f(g)=g^k$ is a bijection if and only if $gcd(k,|G|)=1$. (The proof of the "if" uses Bézout's Lemma, the "only if" leverages Cauchy's Theorem)
answered Dec 28 '18 at 20:39
Nicky HeksterNicky Hekster
29k63456
$endgroup$
add a comment |
$begingroup$
In general: if $G$ is a finite group and $k$ a positive integer. Then the map $f: G rightarrow G$ defined by $f(g)=g^k$ is a bijection if and only if $gcd(k,|G|)=1$. (The proof of the "if" uses Bézout's Lemma, the "only if" leverages Cauchy's Theorem)
answered Dec 28 '18 at 20:39
Nicky HeksterNicky Hekster
29k63456
$endgroup$
In general: if $G$ is a finite group and $k$ a positive integer. Then the map $f: G rightarrow G$ defined by $f(g)=g^k$ is a bijection if and only if $gcd(k,|G|)=1$. (The proof of the "if" uses Bézout's Lemma, the "only if" leverages Cauchy's Theorem)
answered Dec 28 '18 at 20:39
Nicky HeksterNicky Hekster
29k63456
answered Dec 28 '18 at 20:39
Nicky HeksterNicky Hekster
29k63456
answered Dec 28 '18 at 20:39
Nicky HeksterNicky Hekster
29k63456
answered Dec 28 '18 at 20:39
Nicky HeksterNicky Hekster
29k63456
29k63456
add a comment |
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$begingroup$
If you can compute an inverse function, then by definition your function is both one-to-one and onto.
$endgroup$
– DudeMan
Dec 28 '18 at 17:12