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Proof of A Proposition on Dumit and Foote's Abstract Algebra
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I am reading Section 2.3 of Dummit and Foote's Abstract Algebra, 3rd edition, and am having a hard time proving the following proposition on page 57:
Let $H=langle xrangle$ and assume $|x|=n<infty$. Then $H=langle x^arangle$ if and only if $(a,n)=1$, where $(a,n)$ denotes the greatest common devisor of $a$ and $n$.
My attempt;
If $H=langle x^arangle$, then by Prop. 2, $|H|=|x^a|$. Since $H=langle xrangle$, Prop. 2 implies $|H|=|x|=n$. So we have $|x^a|=n$. By Prop. 5, $|x^a|=dfracn(n, a)$. This and $|x^a|=n$ imply $(n,a)=1$.
Conversely, suppose $(n,a)=1$. Then by Prop. 5, $|x^a|=dfracn(n,a)=n=|x|$. Since $H=langle xrangle$, $|x^a|=|x|=|H|$.
From here, how do I proceed to show $H=langle x^arangle$?
Proposition 2. If $H=langle xrangle$, then $|H|=|x|$.
Proposition 5. Let $G$ be a group, let $xin G$ and let $ainmathbbZ-0$. If $|x|=n<infty$, then $|x^a|=dfracn(n,a)$.
abstract-algebra proof-verification
asked Aug 3 at 2:01
kmiyazaki
12110
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I am reading Section 2.3 of Dummit and Foote's Abstract Algebra, 3rd edition, and am having a hard time proving the following proposition on page 57:
Let $H=langle xrangle$ and assume $|x|=n<infty$. Then $H=langle x^arangle$ if and only if $(a,n)=1$, where $(a,n)$ denotes the greatest common devisor of $a$ and $n$.
My attempt;
If $H=langle x^arangle$, then by Prop. 2, $|H|=|x^a|$. Since $H=langle xrangle$, Prop. 2 implies $|H|=|x|=n$. So we have $|x^a|=n$. By Prop. 5, $|x^a|=dfracn(n, a)$. This and $|x^a|=n$ imply $(n,a)=1$.
Conversely, suppose $(n,a)=1$. Then by Prop. 5, $|x^a|=dfracn(n,a)=n=|x|$. Since $H=langle xrangle$, $|x^a|=|x|=|H|$.
From here, how do I proceed to show $H=langle x^arangle$?
Proposition 2. If $H=langle xrangle$, then $|H|=|x|$.
Proposition 5. Let $G$ be a group, let $xin G$ and let $ainmathbbZ-0$. If $|x|=n<infty$, then $|x^a|=dfracn(n,a)$.
abstract-algebra proof-verification
asked Aug 3 at 2:01
kmiyazaki
12110
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am reading Section 2.3 of Dummit and Foote's Abstract Algebra, 3rd edition, and am having a hard time proving the following proposition on page 57:
Let $H=langle xrangle$ and assume $|x|=n<infty$. Then $H=langle x^arangle$ if and only if $(a,n)=1$, where $(a,n)$ denotes the greatest common devisor of $a$ and $n$.
My attempt;
If $H=langle x^arangle$, then by Prop. 2, $|H|=|x^a|$. Since $H=langle xrangle$, Prop. 2 implies $|H|=|x|=n$. So we have $|x^a|=n$. By Prop. 5, $|x^a|=dfracn(n, a)$. This and $|x^a|=n$ imply $(n,a)=1$.
Conversely, suppose $(n,a)=1$. Then by Prop. 5, $|x^a|=dfracn(n,a)=n=|x|$. Since $H=langle xrangle$, $|x^a|=|x|=|H|$.
From here, how do I proceed to show $H=langle x^arangle$?
Proposition 2. If $H=langle xrangle$, then $|H|=|x|$.
Proposition 5. Let $G$ be a group, let $xin G$ and let $ainmathbbZ-0$. If $|x|=n<infty$, then $|x^a|=dfracn(n,a)$.
abstract-algebra proof-verification
asked Aug 3 at 2:01
kmiyazaki
12110
I am reading Section 2.3 of Dummit and Foote's Abstract Algebra, 3rd edition, and am having a hard time proving the following proposition on page 57:
Let $H=langle xrangle$ and assume $|x|=n<infty$. Then $H=langle x^arangle$ if and only if $(a,n)=1$, where $(a,n)$ denotes the greatest common devisor of $a$ and $n$.
My attempt;
If $H=langle x^arangle$, then by Prop. 2, $|H|=|x^a|$. Since $H=langle xrangle$, Prop. 2 implies $|H|=|x|=n$. So we have $|x^a|=n$. By Prop. 5, $|x^a|=dfracn(n, a)$. This and $|x^a|=n$ imply $(n,a)=1$.
Conversely, suppose $(n,a)=1$. Then by Prop. 5, $|x^a|=dfracn(n,a)=n=|x|$. Since $H=langle xrangle$, $|x^a|=|x|=|H|$.
From here, how do I proceed to show $H=langle x^arangle$?
Proposition 2. If $H=langle xrangle$, then $|H|=|x|$.
Proposition 5. Let $G$ be a group, let $xin G$ and let $ainmathbbZ-0$. If $|x|=n<infty$, then $|x^a|=dfracn(n,a)$.
abstract-algebra proof-verification
asked Aug 3 at 2:01
kmiyazaki
12110
asked Aug 3 at 2:01
kmiyazaki
12110
asked Aug 3 at 2:01
kmiyazaki
12110
asked Aug 3 at 2:01
kmiyazaki
12110
12110
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
3
down vote
accepted
Have you heard of Bezout's identity? It states that there exists $r,s$ integers so that $ncdot r + acdot s=1$. Now try to use this to show $x^as=x^1=x$, and from that conclude $H=langle x^arangle$.
answered Aug 3 at 2:23
ThomasGrubb
9,71511335
Thank you so much.
– kmiyazaki
Aug 3 at 3:53
How do you conclude from $x^as=x$ that $H=langle x^arangle$?
– kmiyazaki
2 days ago
1
@kmiyazaki the definition of $H$ is $H=1,x,x^2,...$; as sets, clearly $Hsupset langle x^arangle$, so you just have to show the reverse inclusion
– ThomasGrubb
2 days ago
I figure it out. Thank you so much for your help.
– kmiyazaki
2 days ago
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Have you heard of Bezout's identity? It states that there exists $r,s$ integers so that $ncdot r + acdot s=1$. Now try to use this to show $x^as=x^1=x$, and from that conclude $H=langle x^arangle$.
answered Aug 3 at 2:23
ThomasGrubb
9,71511335
Thank you so much.
– kmiyazaki
Aug 3 at 3:53
How do you conclude from $x^as=x$ that $H=langle x^arangle$?
– kmiyazaki
2 days ago
1
@kmiyazaki the definition of $H$ is $H=1,x,x^2,...$; as sets, clearly $Hsupset langle x^arangle$, so you just have to show the reverse inclusion
– ThomasGrubb
2 days ago
I figure it out. Thank you so much for your help.
– kmiyazaki
2 days ago
add a comment |Â
up vote
3
down vote
accepted
Have you heard of Bezout's identity? It states that there exists $r,s$ integers so that $ncdot r + acdot s=1$. Now try to use this to show $x^as=x^1=x$, and from that conclude $H=langle x^arangle$.
answered Aug 3 at 2:23
ThomasGrubb
9,71511335
Thank you so much.
– kmiyazaki
Aug 3 at 3:53
How do you conclude from $x^as=x$ that $H=langle x^arangle$?
– kmiyazaki
2 days ago
1
@kmiyazaki the definition of $H$ is $H=1,x,x^2,...$; as sets, clearly $Hsupset langle x^arangle$, so you just have to show the reverse inclusion
– ThomasGrubb
2 days ago
I figure it out. Thank you so much for your help.
– kmiyazaki
2 days ago
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Have you heard of Bezout's identity? It states that there exists $r,s$ integers so that $ncdot r + acdot s=1$. Now try to use this to show $x^as=x^1=x$, and from that conclude $H=langle x^arangle$.
answered Aug 3 at 2:23
ThomasGrubb
9,71511335
Have you heard of Bezout's identity? It states that there exists $r,s$ integers so that $ncdot r + acdot s=1$. Now try to use this to show $x^as=x^1=x$, and from that conclude $H=langle x^arangle$.
answered Aug 3 at 2:23
ThomasGrubb
9,71511335
answered Aug 3 at 2:23
ThomasGrubb
9,71511335
answered Aug 3 at 2:23
ThomasGrubb
9,71511335
answered Aug 3 at 2:23
ThomasGrubb
9,71511335
9,71511335
Thank you so much.
– kmiyazaki
Aug 3 at 3:53
How do you conclude from $x^as=x$ that $H=langle x^arangle$?
– kmiyazaki
2 days ago
1
@kmiyazaki the definition of $H$ is $H=1,x,x^2,...$; as sets, clearly $Hsupset langle x^arangle$, so you just have to show the reverse inclusion
– ThomasGrubb
2 days ago
I figure it out. Thank you so much for your help.
– kmiyazaki
2 days ago
add a comment |Â
Thank you so much.
– kmiyazaki
Aug 3 at 3:53
How do you conclude from $x^as=x$ that $H=langle x^arangle$?
– kmiyazaki
2 days ago
1
@kmiyazaki the definition of $H$ is $H=1,x,x^2,...$; as sets, clearly $Hsupset langle x^arangle$, so you just have to show the reverse inclusion
– ThomasGrubb
2 days ago
I figure it out. Thank you so much for your help.
– kmiyazaki
2 days ago
Thank you so much.
– kmiyazaki
Aug 3 at 3:53
Thank you so much.
– kmiyazaki
Aug 3 at 3:53
How do you conclude from $x^as=x$ that $H=langle x^arangle$?
– kmiyazaki
2 days ago
How do you conclude from $x^as=x$ that $H=langle x^arangle$?
– kmiyazaki
2 days ago
1
1
@kmiyazaki the definition of $H$ is $H=1,x,x^2,...$; as sets, clearly $Hsupset langle x^arangle$, so you just have to show the reverse inclusion
– ThomasGrubb
2 days ago
@kmiyazaki the definition of $H$ is $H=1,x,x^2,...$; as sets, clearly $Hsupset langle x^arangle$, so you just have to show the reverse inclusion
– ThomasGrubb
2 days ago
I figure it out. Thank you so much for your help.
– kmiyazaki
2 days ago
I figure it out. Thank you so much for your help.
– kmiyazaki
2 days ago
add a comment |Â
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