Exercise 4.11 Allufi
This exercise is on page 69 of Algebra chapter 0.
Problem:
In due time we will prove the easy fact that if $p$ is a prime integer, then the equation $x^d = 1$ can have at most $d$ solutions in $mathbb{Z}/pmathbb{Z}$. Assume this fact, and prove that the multiplicative group $G =(mathbb{Z}/pmathbb{Z})^{ast}$ is cyclic.
Attempt:
Using the this fact along with another exercise one can show that $h^{|g|} = 1$ for $g in G$ element of maximal order. This shows that for all $|h| leq |g|$ for all $h in G$. We can also show that $|g| leq |G| = p - 1$.
I am not sure how can we show that $|G| leq |g|$ though ? I don't want to use Lagrange theorem as we don't know this result yet in the book.
abstract-algebra group-theory polynomials cyclic-groups
edited Dec 27 '18 at 0:56
the_fox
2,43411431
asked Dec 26 '18 at 20:37
Newbie
426211
add a comment |
This exercise is on page 69 of Algebra chapter 0.
Problem:
In due time we will prove the easy fact that if $p$ is a prime integer, then the equation $x^d = 1$ can have at most $d$ solutions in $mathbb{Z}/pmathbb{Z}$. Assume this fact, and prove that the multiplicative group $G =(mathbb{Z}/pmathbb{Z})^{ast}$ is cyclic.
Attempt:
Using the this fact along with another exercise one can show that $h^{|g|} = 1$ for $g in G$ element of maximal order. This shows that for all $|h| leq |g|$ for all $h in G$. We can also show that $|g| leq |G| = p - 1$.
I am not sure how can we show that $|G| leq |g|$ though ? I don't want to use Lagrange theorem as we don't know this result yet in the book.
abstract-algebra group-theory polynomials cyclic-groups
edited Dec 27 '18 at 0:56
the_fox
2,43411431
asked Dec 26 '18 at 20:37
Newbie
426211
See this question.
– Dietrich Burde
Dec 26 '18 at 20:51
See this theorem.
– Bill Dubuque
Dec 27 '18 at 2:23
add a comment |
This exercise is on page 69 of Algebra chapter 0.
Problem:
In due time we will prove the easy fact that if $p$ is a prime integer, then the equation $x^d = 1$ can have at most $d$ solutions in $mathbb{Z}/pmathbb{Z}$. Assume this fact, and prove that the multiplicative group $G =(mathbb{Z}/pmathbb{Z})^{ast}$ is cyclic.
Attempt:
Using the this fact along with another exercise one can show that $h^{|g|} = 1$ for $g in G$ element of maximal order. This shows that for all $|h| leq |g|$ for all $h in G$. We can also show that $|g| leq |G| = p - 1$.
I am not sure how can we show that $|G| leq |g|$ though ? I don't want to use Lagrange theorem as we don't know this result yet in the book.
abstract-algebra group-theory polynomials cyclic-groups
edited Dec 27 '18 at 0:56
the_fox
2,43411431
asked Dec 26 '18 at 20:37
Newbie
426211
This exercise is on page 69 of Algebra chapter 0.
Problem:
In due time we will prove the easy fact that if $p$ is a prime integer, then the equation $x^d = 1$ can have at most $d$ solutions in $mathbb{Z}/pmathbb{Z}$. Assume this fact, and prove that the multiplicative group $G =(mathbb{Z}/pmathbb{Z})^{ast}$ is cyclic.
Attempt:
Using the this fact along with another exercise one can show that $h^{|g|} = 1$ for $g in G$ element of maximal order. This shows that for all $|h| leq |g|$ for all $h in G$. We can also show that $|g| leq |G| = p - 1$.
I am not sure how can we show that $|G| leq |g|$ though ? I don't want to use Lagrange theorem as we don't know this result yet in the book.
abstract-algebra group-theory polynomials cyclic-groups
abstract-algebra group-theory polynomials cyclic-groups
edited Dec 27 '18 at 0:56
the_fox
2,43411431
asked Dec 26 '18 at 20:37
Newbie
426211
edited Dec 27 '18 at 0:56
the_fox
2,43411431
asked Dec 26 '18 at 20:37
Newbie
426211
edited Dec 27 '18 at 0:56
the_fox
2,43411431
edited Dec 27 '18 at 0:56
the_fox
2,43411431
edited Dec 27 '18 at 0:56
the_fox
2,43411431
2,43411431
asked Dec 26 '18 at 20:37
Newbie
426211
asked Dec 26 '18 at 20:37
Newbie
426211
asked Dec 26 '18 at 20:37
Newbie
426211
426211
See this question.
– Dietrich Burde
Dec 26 '18 at 20:51
See this theorem.
– Bill Dubuque
Dec 27 '18 at 2:23
add a comment |
See this question.
– Dietrich Burde
Dec 26 '18 at 20:51
See this theorem.
– Bill Dubuque
Dec 27 '18 at 2:23
See this question.
– Dietrich Burde
Dec 26 '18 at 20:51
See this question.
– Dietrich Burde
Dec 26 '18 at 20:51
See this theorem.
– Bill Dubuque
Dec 27 '18 at 2:23
See this theorem.
– Bill Dubuque
Dec 27 '18 at 2:23
add a comment |
1 Answer
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If $mid gmidlt mid Gmid=p-1$, then $x^{mid gmid}=1$ has $p-1gtmid gmid$ solutions. This contradicts the fact you were given about $x^d=1$.
answered Dec 27 '18 at 4:10
Chris Custer
10.8k3824
add a comment |
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If $mid gmidlt mid Gmid=p-1$, then $x^{mid gmid}=1$ has $p-1gtmid gmid$ solutions. This contradicts the fact you were given about $x^d=1$.
answered Dec 27 '18 at 4:10
Chris Custer
10.8k3824
add a comment |
If $mid gmidlt mid Gmid=p-1$, then $x^{mid gmid}=1$ has $p-1gtmid gmid$ solutions. This contradicts the fact you were given about $x^d=1$.
answered Dec 27 '18 at 4:10
Chris Custer
10.8k3824
add a comment |
If $mid gmidlt mid Gmid=p-1$, then $x^{mid gmid}=1$ has $p-1gtmid gmid$ solutions. This contradicts the fact you were given about $x^d=1$.
answered Dec 27 '18 at 4:10
Chris Custer
10.8k3824
If $mid gmidlt mid Gmid=p-1$, then $x^{mid gmid}=1$ has $p-1gtmid gmid$ solutions. This contradicts the fact you were given about $x^d=1$.
answered Dec 27 '18 at 4:10
Chris Custer
10.8k3824
answered Dec 27 '18 at 4:10
Chris Custer
10.8k3824
answered Dec 27 '18 at 4:10
Chris Custer
10.8k3824
answered Dec 27 '18 at 4:10
Chris Custer
10.8k3824
10.8k3824
add a comment |
add a comment |
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See this question.
– Dietrich Burde
Dec 26 '18 at 20:51
See this theorem.
– Bill Dubuque
Dec 27 '18 at 2:23