Field of discrete logarithms of a generator of a finite group
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Suppose that $mathbbG$ is finite group of order $p$ ($p$ can be a prime number if necessary) and that $g$ is a generator of $mathbbG$. What is the field of discrete logarithms of $g$?
I can see that the set $g^mathbbZsimeq 1,ldots p-1$ is finite and contains $p$ elements, it also has a commutative group structure with $0$ as neutral element, so it is (isomorphic to) $(mathbbZ_p-1,+)$. Addition is determined by
$c:=a+b$ with $c$ determined by $g^c:= g^ag^b$. This is the discrete logarithm problem
But how can we deduces that there is also a multiplication? Can we derive it from the group law and the generator?
Because $(g^a)^b neq g^a cdot_mathbbZ_pb$ in general.
finite-groups finite-fields elliptic-curves cryptography
asked Sep 2 at 8:49
Mark Neuhaus
5671312
add a comment |Â
up vote
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down vote
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Suppose that $mathbbG$ is finite group of order $p$ ($p$ can be a prime number if necessary) and that $g$ is a generator of $mathbbG$. What is the field of discrete logarithms of $g$?
I can see that the set $g^mathbbZsimeq 1,ldots p-1$ is finite and contains $p$ elements, it also has a commutative group structure with $0$ as neutral element, so it is (isomorphic to) $(mathbbZ_p-1,+)$. Addition is determined by
$c:=a+b$ with $c$ determined by $g^c:= g^ag^b$. This is the discrete logarithm problem
But how can we deduces that there is also a multiplication? Can we derive it from the group law and the generator?
Because $(g^a)^b neq g^a cdot_mathbbZ_pb$ in general.
finite-groups finite-fields elliptic-curves cryptography
asked Sep 2 at 8:49
Mark Neuhaus
5671312
$g$ is a generator of $G$. So $G$ is cyclic. Now consider the exponents.
– Wuestenfux
Sep 2 at 9:23
I see how addition work, because $g^acdot g^b= g^a+b$, but I don't see how multiplication works.
– Mark Neuhaus
Sep 4 at 18:41
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Suppose that $mathbbG$ is finite group of order $p$ ($p$ can be a prime number if necessary) and that $g$ is a generator of $mathbbG$. What is the field of discrete logarithms of $g$?
I can see that the set $g^mathbbZsimeq 1,ldots p-1$ is finite and contains $p$ elements, it also has a commutative group structure with $0$ as neutral element, so it is (isomorphic to) $(mathbbZ_p-1,+)$. Addition is determined by
$c:=a+b$ with $c$ determined by $g^c:= g^ag^b$. This is the discrete logarithm problem
But how can we deduces that there is also a multiplication? Can we derive it from the group law and the generator?
Because $(g^a)^b neq g^a cdot_mathbbZ_pb$ in general.
finite-groups finite-fields elliptic-curves cryptography
asked Sep 2 at 8:49
Mark Neuhaus
5671312
Suppose that $mathbbG$ is finite group of order $p$ ($p$ can be a prime number if necessary) and that $g$ is a generator of $mathbbG$. What is the field of discrete logarithms of $g$?
I can see that the set $g^mathbbZsimeq 1,ldots p-1$ is finite and contains $p$ elements, it also has a commutative group structure with $0$ as neutral element, so it is (isomorphic to) $(mathbbZ_p-1,+)$. Addition is determined by
$c:=a+b$ with $c$ determined by $g^c:= g^ag^b$. This is the discrete logarithm problem
But how can we deduces that there is also a multiplication? Can we derive it from the group law and the generator?
Because $(g^a)^b neq g^a cdot_mathbbZ_pb$ in general.
finite-groups finite-fields elliptic-curves cryptography
finite-groups finite-fields elliptic-curves cryptography
asked Sep 2 at 8:49
Mark Neuhaus
5671312
asked Sep 2 at 8:49
Mark Neuhaus
5671312
edited Sep 4 at 19:17
asked Sep 2 at 8:49
Mark Neuhaus
5671312
asked Sep 2 at 8:49
Mark Neuhaus
5671312
asked Sep 2 at 8:49
Mark Neuhaus
5671312
5671312
$g$ is a generator of $G$. So $G$ is cyclic. Now consider the exponents.
– Wuestenfux
Sep 2 at 9:23
I see how addition work, because $g^acdot g^b= g^a+b$, but I don't see how multiplication works.
– Mark Neuhaus
Sep 4 at 18:41
add a comment |Â
$g$ is a generator of $G$. So $G$ is cyclic. Now consider the exponents.
– Wuestenfux
Sep 2 at 9:23
I see how addition work, because $g^acdot g^b= g^a+b$, but I don't see how multiplication works.
– Mark Neuhaus
Sep 4 at 18:41
$g$ is a generator of $G$. So $G$ is cyclic. Now consider the exponents.
– Wuestenfux
Sep 2 at 9:23
$g$ is a generator of $G$. So $G$ is cyclic. Now consider the exponents.
– Wuestenfux
Sep 2 at 9:23
I see how addition work, because $g^acdot g^b= g^a+b$, but I don't see how multiplication works.
– Mark Neuhaus
Sep 4 at 18:41
I see how addition work, because $g^acdot g^b= g^a+b$, but I don't see how multiplication works.
– Mark Neuhaus
Sep 4 at 18:41
add a comment |Â
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$g$ is a generator of $G$. So $G$ is cyclic. Now consider the exponents.
– Wuestenfux
Sep 2 at 9:23
I see how addition work, because $g^acdot g^b= g^a+b$, but I don't see how multiplication works.
– Mark Neuhaus
Sep 4 at 18:41