Euler totient function and groups.

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Say $phi(n)=k$ is the number of integers less than n relatively prime to n. Then prove that any integer a relatively prime to n $a^phi(n)=1 quad modquad n$.
My proof
U(n) be the group of integers coprime to n less than n. Then a belongs to this group and order of group is k. Hence $a^k=1$.
This proof is correct right.
finite-groups totient-function
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up vote
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Say $phi(n)=k$ is the number of integers less than n relatively prime to n. Then prove that any integer a relatively prime to n $a^phi(n)=1 quad modquad n$.
My proof
U(n) be the group of integers coprime to n less than n. Then a belongs to this group and order of group is k. Hence $a^k=1$.
This proof is correct right.
finite-groups totient-function
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Say $phi(n)=k$ is the number of integers less than n relatively prime to n. Then prove that any integer a relatively prime to n $a^phi(n)=1 quad modquad n$.
My proof
U(n) be the group of integers coprime to n less than n. Then a belongs to this group and order of group is k. Hence $a^k=1$.
This proof is correct right.
finite-groups totient-function
Say $phi(n)=k$ is the number of integers less than n relatively prime to n. Then prove that any integer a relatively prime to n $a^phi(n)=1 quad modquad n$.
My proof
U(n) be the group of integers coprime to n less than n. Then a belongs to this group and order of group is k. Hence $a^k=1$.
This proof is correct right.
finite-groups totient-function
finite-groups totient-function
asked Sep 10 at 8:45
Piyush Divyanakar
3,315222
3,315222
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1 Answer
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If a (finite) group has order $k$ then of course every element $x$ of the group satisfies $x^k=1$; this is pure group theory.
However to apply this theorem one has to first show that U(n) is a group in the first place.
SO one has to do some number-theoretic proof, there is no way to avoid it.
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
If a (finite) group has order $k$ then of course every element $x$ of the group satisfies $x^k=1$; this is pure group theory.
However to apply this theorem one has to first show that U(n) is a group in the first place.
SO one has to do some number-theoretic proof, there is no way to avoid it.
add a comment |Â
up vote
1
down vote
If a (finite) group has order $k$ then of course every element $x$ of the group satisfies $x^k=1$; this is pure group theory.
However to apply this theorem one has to first show that U(n) is a group in the first place.
SO one has to do some number-theoretic proof, there is no way to avoid it.
add a comment |Â
up vote
1
down vote
up vote
1
down vote
If a (finite) group has order $k$ then of course every element $x$ of the group satisfies $x^k=1$; this is pure group theory.
However to apply this theorem one has to first show that U(n) is a group in the first place.
SO one has to do some number-theoretic proof, there is no way to avoid it.
If a (finite) group has order $k$ then of course every element $x$ of the group satisfies $x^k=1$; this is pure group theory.
However to apply this theorem one has to first show that U(n) is a group in the first place.
SO one has to do some number-theoretic proof, there is no way to avoid it.
answered Sep 10 at 9:05
P Vanchinathan
14.2k12036
14.2k12036
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