Show that $y=x^k$ with $gcd(k,n)=1$ is a generator of $G$. [duplicate]

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How to find a generator of a cyclic group?

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Could someone please verify whether my solution is okay?

Let $G$ be a finite cyclic group with $|G|=n$ and generator $x$. If $y=x^k$ and $gcd(k,n)=1$, then show that $y$ is a generator of $G$.

Let $y=x^k$ with $gcd(k,n)=1$. Then $|langle yrangle|=|y|=fracngcd(k,n)=fracn1=n$. Then $G=langle yrangle$.

edited Aug 13 at 3:53

Alan Wang

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asked Aug 13 at 3:51

numericalorange

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marked as duplicate by Jyrki LahtonenÂ group-theory
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Aug 17 at 6:44

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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Yes you are correct.
â€“Â Alan Wang
Aug 13 at 3:52

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up vote
3
down vote

favorite

This question already has an answer here:

How to find a generator of a cyclic group?

8 answers

Could someone please verify whether my solution is okay?

Let $G$ be a finite cyclic group with $|G|=n$ and generator $x$. If $y=x^k$ and $gcd(k,n)=1$, then show that $y$ is a generator of $G$.

Let $y=x^k$ with $gcd(k,n)=1$. Then $|langle yrangle|=|y|=fracngcd(k,n)=fracn1=n$. Then $G=langle yrangle$.

edited Aug 13 at 3:53

Alan Wang

4,486932

asked Aug 13 at 3:51

numericalorange

1,266110

marked as duplicate by Jyrki LahtonenÂ group-theory
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Aug 17 at 6:44

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

2

Yes you are correct.
â€“Â Alan Wang
Aug 13 at 3:52

add a commentÂ |Â

up vote
3
down vote

favorite

This question already has an answer here:

How to find a generator of a cyclic group?

8 answers

Could someone please verify whether my solution is okay?

Let $G$ be a finite cyclic group with $|G|=n$ and generator $x$. If $y=x^k$ and $gcd(k,n)=1$, then show that $y$ is a generator of $G$.

Let $y=x^k$ with $gcd(k,n)=1$. Then $|langle yrangle|=|y|=fracngcd(k,n)=fracn1=n$. Then $G=langle yrangle$.

edited Aug 13 at 3:53

Alan Wang

4,486932

asked Aug 13 at 3:51

numericalorange

1,266110

This question already has an answer here:

How to find a generator of a cyclic group?

8 answers

Could someone please verify whether my solution is okay?

Let $G$ be a finite cyclic group with $|G|=n$ and generator $x$. If $y=x^k$ and $gcd(k,n)=1$, then show that $y$ is a generator of $G$.

Let $y=x^k$ with $gcd(k,n)=1$. Then $|langle yrangle|=|y|=fracngcd(k,n)=fracn1=n$. Then $G=langle yrangle$.

This question already has an answer here:

How to find a generator of a cyclic group?

8 answers

edited Aug 13 at 3:53

Alan Wang

4,486932

asked Aug 13 at 3:51

numericalorange

1,266110

edited Aug 13 at 3:53

Alan Wang

4,486932

edited Aug 13 at 3:53

Alan Wang

4,486932

edited Aug 13 at 3:53

Alan Wang

4,486932

asked Aug 13 at 3:51

numericalorange

1,266110

asked Aug 13 at 3:51

numericalorange

1,266110

asked Aug 13 at 3:51

numericalorange

1,266110

marked as duplicate by Jyrki LahtonenÂ group-theory
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Aug 17 at 6:44

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

2

Yes you are correct.
â€“Â Alan Wang
Aug 13 at 3:52

add a commentÂ |Â

2

Yes you are correct.
â€“Â Alan Wang
Aug 13 at 3:52

Yes you are correct.
â€“Â Alan Wang
Aug 13 at 3:52

add a commentÂ |Â

1 Answer
1

active

oldest

votes

up vote
2
down vote

accepted

Here's another way:

If we can show that some power of $y$ is $x$, that is, if we can show that for some $a in Bbb Z$,

$y^a = x, tag 1$

then we are done, since then

$x^l = y^al, tag 2$

so that

$langle x rangle = langle y rangle. tag 3$

Well, since

$gcd(n, k) = 1, tag 4$

we have $a, b in Bbb Z$ with

$ak + bn = 1; tag 5$

then

$x = x^1 = x^ak + bn = x^akx^bn = x^ak = (x^k)^a = y^a, tag 6$

establishing (1), and we are done!

P.S. to our OP numericalorange:. Your solution looks fine to me!

answered Aug 13 at 5:06

Robert Lewis

37.3k22357

add a commentÂ |Â

1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

up vote
2
down vote

accepted

Here's another way:

If we can show that some power of $y$ is $x$, that is, if we can show that for some $a in Bbb Z$,

$y^a = x, tag 1$

then we are done, since then

$x^l = y^al, tag 2$

so that

$langle x rangle = langle y rangle. tag 3$

Well, since

$gcd(n, k) = 1, tag 4$

we have $a, b in Bbb Z$ with

$ak + bn = 1; tag 5$

then

$x = x^1 = x^ak + bn = x^akx^bn = x^ak = (x^k)^a = y^a, tag 6$

establishing (1), and we are done!

P.S. to our OP numericalorange:. Your solution looks fine to me!

answered Aug 13 at 5:06

Robert Lewis

37.3k22357

add a commentÂ |Â

up vote
2
down vote

accepted

Here's another way:

If we can show that some power of $y$ is $x$, that is, if we can show that for some $a in Bbb Z$,

$y^a = x, tag 1$

then we are done, since then

$x^l = y^al, tag 2$

so that

$langle x rangle = langle y rangle. tag 3$

Well, since

$gcd(n, k) = 1, tag 4$

we have $a, b in Bbb Z$ with

$ak + bn = 1; tag 5$

then

$x = x^1 = x^ak + bn = x^akx^bn = x^ak = (x^k)^a = y^a, tag 6$

establishing (1), and we are done!

P.S. to our OP numericalorange:. Your solution looks fine to me!

answered Aug 13 at 5:06

Robert Lewis

37.3k22357

add a commentÂ |Â

up vote
2
down vote

accepted

Here's another way:

If we can show that some power of $y$ is $x$, that is, if we can show that for some $a in Bbb Z$,

$y^a = x, tag 1$

then we are done, since then

$x^l = y^al, tag 2$

so that

$langle x rangle = langle y rangle. tag 3$

Well, since

$gcd(n, k) = 1, tag 4$

we have $a, b in Bbb Z$ with

$ak + bn = 1; tag 5$

then

$x = x^1 = x^ak + bn = x^akx^bn = x^ak = (x^k)^a = y^a, tag 6$

establishing (1), and we are done!

P.S. to our OP numericalorange:. Your solution looks fine to me!

answered Aug 13 at 5:06

Robert Lewis

37.3k22357

Here's another way:

If we can show that some power of $y$ is $x$, that is, if we can show that for some $a in Bbb Z$,

$y^a = x, tag 1$

then we are done, since then

$x^l = y^al, tag 2$

so that

$langle x rangle = langle y rangle. tag 3$

Well, since

$gcd(n, k) = 1, tag 4$

we have $a, b in Bbb Z$ with

$ak + bn = 1; tag 5$

then

$x = x^1 = x^ak + bn = x^akx^bn = x^ak = (x^k)^a = y^a, tag 6$

establishing (1), and we are done!

P.S. to our OP numericalorange:. Your solution looks fine to me!

answered Aug 13 at 5:06

Robert Lewis

37.3k22357

answered Aug 13 at 5:06

Robert Lewis

37.3k22357

answered Aug 13 at 5:06

Robert Lewis

37.3k22357

answered Aug 13 at 5:06

Robert Lewis

37.3k22357

add a commentÂ |Â

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