How do generators of a group work?
$G$ is a group, $H$ is a subgroup of $G$, and $[G:H]$ stands for the index of $H$ in $G$ in the following example:
Let $G=S_3$, $H=left<(1,2)right>$. Then $[G:H]=3$.
I know the definition of group generators:
A set of generators $(g_1,...,g_n)$ is a set of group elements such that possibly repeated application of the generators on themselves and each other is capable of producing all the elements in the group.
What does the individual elements of $H=left<(1,2)right>$ look like? Any help would be greatly aprciated.
P.S. I know how to find the index when the groups don’t involve a group generator, the thing I need help with is understanding the group generator.
abstract-algebra finite-groups normal-subgroups
edited Dec 10 '18 at 5:10
Shaun
8,750113680
asked Dec 10 '18 at 4:16
AMN52
326
|
show 2 more comments
$G$ is a group, $H$ is a subgroup of $G$, and $[G:H]$ stands for the index of $H$ in $G$ in the following example:
Let $G=S_3$, $H=left<(1,2)right>$. Then $[G:H]=3$.
I know the definition of group generators:
A set of generators $(g_1,...,g_n)$ is a set of group elements such that possibly repeated application of the generators on themselves and each other is capable of producing all the elements in the group.
What does the individual elements of $H=left<(1,2)right>$ look like? Any help would be greatly aprciated.
P.S. I know how to find the index when the groups don’t involve a group generator, the thing I need help with is understanding the group generator.
abstract-algebra finite-groups normal-subgroups
edited Dec 10 '18 at 5:10
Shaun
8,750113680
asked Dec 10 '18 at 4:16
AMN52
326
What's the actual question here?
– Lord Shark the Unknown
Dec 10 '18 at 4:18
My question is more or less this: What does the individual elements of $H$ look like?
– AMN52
Dec 10 '18 at 4:21
$H$ must have the identity element. It must also have the element $(1 2)$.
– Lord Shark the Unknown
Dec 10 '18 at 4:22
$(1,2)$ has order 2 so it's the only non identity element in $H$.
– Justin Stevenson
Dec 10 '18 at 4:22
Why does $H$ only contain $(1,2)$ and the identity element?
– AMN52
Dec 10 '18 at 4:25
|
show 2 more comments
$G$ is a group, $H$ is a subgroup of $G$, and $[G:H]$ stands for the index of $H$ in $G$ in the following example:
Let $G=S_3$, $H=left<(1,2)right>$. Then $[G:H]=3$.
I know the definition of group generators:
A set of generators $(g_1,...,g_n)$ is a set of group elements such that possibly repeated application of the generators on themselves and each other is capable of producing all the elements in the group.
What does the individual elements of $H=left<(1,2)right>$ look like? Any help would be greatly aprciated.
P.S. I know how to find the index when the groups don’t involve a group generator, the thing I need help with is understanding the group generator.
abstract-algebra finite-groups normal-subgroups
edited Dec 10 '18 at 5:10
Shaun
8,750113680
asked Dec 10 '18 at 4:16
AMN52
326
$G$ is a group, $H$ is a subgroup of $G$, and $[G:H]$ stands for the index of $H$ in $G$ in the following example:
Let $G=S_3$, $H=left<(1,2)right>$. Then $[G:H]=3$.
I know the definition of group generators:
A set of generators $(g_1,...,g_n)$ is a set of group elements such that possibly repeated application of the generators on themselves and each other is capable of producing all the elements in the group.
What does the individual elements of $H=left<(1,2)right>$ look like? Any help would be greatly aprciated.
P.S. I know how to find the index when the groups don’t involve a group generator, the thing I need help with is understanding the group generator.
abstract-algebra finite-groups normal-subgroups
abstract-algebra finite-groups normal-subgroups
edited Dec 10 '18 at 5:10
Shaun
8,750113680
asked Dec 10 '18 at 4:16
AMN52
326
edited Dec 10 '18 at 5:10
Shaun
8,750113680
asked Dec 10 '18 at 4:16
AMN52
326
edited Dec 10 '18 at 5:10
Shaun
8,750113680
edited Dec 10 '18 at 5:10
Shaun
8,750113680
edited Dec 10 '18 at 5:10
Shaun
8,750113680
8,750113680
asked Dec 10 '18 at 4:16
AMN52
326
asked Dec 10 '18 at 4:16
AMN52
326
asked Dec 10 '18 at 4:16
AMN52
326
326
What's the actual question here?
– Lord Shark the Unknown
Dec 10 '18 at 4:18
My question is more or less this: What does the individual elements of $H$ look like?
– AMN52
Dec 10 '18 at 4:21
$H$ must have the identity element. It must also have the element $(1 2)$.
– Lord Shark the Unknown
Dec 10 '18 at 4:22
$(1,2)$ has order 2 so it's the only non identity element in $H$.
– Justin Stevenson
Dec 10 '18 at 4:22
Why does $H$ only contain $(1,2)$ and the identity element?
– AMN52
Dec 10 '18 at 4:25
|
show 2 more comments
What's the actual question here?
– Lord Shark the Unknown
Dec 10 '18 at 4:18
My question is more or less this: What does the individual elements of $H$ look like?
– AMN52
Dec 10 '18 at 4:21
$H$ must have the identity element. It must also have the element $(1 2)$.
– Lord Shark the Unknown
Dec 10 '18 at 4:22
$(1,2)$ has order 2 so it's the only non identity element in $H$.
– Justin Stevenson
Dec 10 '18 at 4:22
Why does $H$ only contain $(1,2)$ and the identity element?
– AMN52
Dec 10 '18 at 4:25
What's the actual question here?
– Lord Shark the Unknown
Dec 10 '18 at 4:18
What's the actual question here?
– Lord Shark the Unknown
Dec 10 '18 at 4:18
My question is more or less this: What does the individual elements of $H$ look like?
– AMN52
Dec 10 '18 at 4:21
My question is more or less this: What does the individual elements of $H$ look like?
– AMN52
Dec 10 '18 at 4:21
$H$ must have the identity element. It must also have the element $(1 2)$.
– Lord Shark the Unknown
Dec 10 '18 at 4:22
$H$ must have the identity element. It must also have the element $(1 2)$.
– Lord Shark the Unknown
Dec 10 '18 at 4:22
$(1,2)$ has order 2 so it's the only non identity element in $H$.
– Justin Stevenson
Dec 10 '18 at 4:22
$(1,2)$ has order 2 so it's the only non identity element in $H$.
– Justin Stevenson
Dec 10 '18 at 4:22
Why does $H$ only contain $(1,2)$ and the identity element?
– AMN52
Dec 10 '18 at 4:25
Why does $H$ only contain $(1,2)$ and the identity element?
– AMN52
Dec 10 '18 at 4:25
|
show 2 more comments
2 Answers
2
active
oldest
votes
$(12)$ is a transposition. $(12)^{-1}=(12)$, that is, it's its own inverse. All that can be gotten by taking powers of $(12)$ is $(12)$ and $e$, the identity. (Note: In general, $langle arangle ={a^n:ninmathbb Z}$). Thus $langle (12)rangle ={(12),e}$, a two element group.
Since $mid S_3mid=6$, we get $[S_3:langle (12)rangle] =3$.
answered Dec 10 '18 at 4:33
Chris Custer
10.8k3824
add a comment |
Because $H$ is the set of all elements of the form $(1,2)^n$ and $(1,2)^2=e$.
By Justin Stevenson.
answered Dec 10 '18 at 4:33
AMN52
326
technically this is plagiarism...lol :)
– maridia
Dec 10 '18 at 4:36
Sorry, I will add the guy who gave me the answer, I was just to trying end the question so no else kept answering.
– AMN52
Dec 10 '18 at 4:50
add a comment |
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2 Answers
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2 Answers
2
active
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$(12)$ is a transposition. $(12)^{-1}=(12)$, that is, it's its own inverse. All that can be gotten by taking powers of $(12)$ is $(12)$ and $e$, the identity. (Note: In general, $langle arangle ={a^n:ninmathbb Z}$). Thus $langle (12)rangle ={(12),e}$, a two element group.
Since $mid S_3mid=6$, we get $[S_3:langle (12)rangle] =3$.
answered Dec 10 '18 at 4:33
Chris Custer
10.8k3824
add a comment |
$(12)$ is a transposition. $(12)^{-1}=(12)$, that is, it's its own inverse. All that can be gotten by taking powers of $(12)$ is $(12)$ and $e$, the identity. (Note: In general, $langle arangle ={a^n:ninmathbb Z}$). Thus $langle (12)rangle ={(12),e}$, a two element group.
Since $mid S_3mid=6$, we get $[S_3:langle (12)rangle] =3$.
answered Dec 10 '18 at 4:33
Chris Custer
10.8k3824
add a comment |
$(12)$ is a transposition. $(12)^{-1}=(12)$, that is, it's its own inverse. All that can be gotten by taking powers of $(12)$ is $(12)$ and $e$, the identity. (Note: In general, $langle arangle ={a^n:ninmathbb Z}$). Thus $langle (12)rangle ={(12),e}$, a two element group.
Since $mid S_3mid=6$, we get $[S_3:langle (12)rangle] =3$.
answered Dec 10 '18 at 4:33
Chris Custer
10.8k3824
$(12)$ is a transposition. $(12)^{-1}=(12)$, that is, it's its own inverse. All that can be gotten by taking powers of $(12)$ is $(12)$ and $e$, the identity. (Note: In general, $langle arangle ={a^n:ninmathbb Z}$). Thus $langle (12)rangle ={(12),e}$, a two element group.
Since $mid S_3mid=6$, we get $[S_3:langle (12)rangle] =3$.
answered Dec 10 '18 at 4:33
Chris Custer
10.8k3824
edited Dec 10 '18 at 4:41
answered Dec 10 '18 at 4:33
Chris Custer
10.8k3824
answered Dec 10 '18 at 4:33
Chris Custer
10.8k3824
answered Dec 10 '18 at 4:33
Chris Custer
10.8k3824
10.8k3824
add a comment |
add a comment |
Because $H$ is the set of all elements of the form $(1,2)^n$ and $(1,2)^2=e$.
By Justin Stevenson.
answered Dec 10 '18 at 4:33
AMN52
326
technically this is plagiarism...lol :)
– maridia
Dec 10 '18 at 4:36
Sorry, I will add the guy who gave me the answer, I was just to trying end the question so no else kept answering.
– AMN52
Dec 10 '18 at 4:50
add a comment |
Because $H$ is the set of all elements of the form $(1,2)^n$ and $(1,2)^2=e$.
By Justin Stevenson.
answered Dec 10 '18 at 4:33
AMN52
326
technically this is plagiarism...lol :)
– maridia
Dec 10 '18 at 4:36
Sorry, I will add the guy who gave me the answer, I was just to trying end the question so no else kept answering.
– AMN52
Dec 10 '18 at 4:50
add a comment |
Because $H$ is the set of all elements of the form $(1,2)^n$ and $(1,2)^2=e$.
By Justin Stevenson.
answered Dec 10 '18 at 4:33
AMN52
326
Because $H$ is the set of all elements of the form $(1,2)^n$ and $(1,2)^2=e$.
By Justin Stevenson.
answered Dec 10 '18 at 4:33
AMN52
326
edited Dec 10 '18 at 4:51
answered Dec 10 '18 at 4:33
AMN52
326
answered Dec 10 '18 at 4:33
AMN52
326
answered Dec 10 '18 at 4:33
AMN52
326
326
technically this is plagiarism...lol :)
– maridia
Dec 10 '18 at 4:36
Sorry, I will add the guy who gave me the answer, I was just to trying end the question so no else kept answering.
– AMN52
Dec 10 '18 at 4:50
add a comment |
technically this is plagiarism...lol :)
– maridia
Dec 10 '18 at 4:36
Sorry, I will add the guy who gave me the answer, I was just to trying end the question so no else kept answering.
– AMN52
Dec 10 '18 at 4:50
technically this is plagiarism...lol :)
– maridia
Dec 10 '18 at 4:36
technically this is plagiarism...lol :)
– maridia
Dec 10 '18 at 4:36
Sorry, I will add the guy who gave me the answer, I was just to trying end the question so no else kept answering.
– AMN52
Dec 10 '18 at 4:50
Sorry, I will add the guy who gave me the answer, I was just to trying end the question so no else kept answering.
– AMN52
Dec 10 '18 at 4:50
add a comment |
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What's the actual question here?
– Lord Shark the Unknown
Dec 10 '18 at 4:18
My question is more or less this: What does the individual elements of $H$ look like?
– AMN52
Dec 10 '18 at 4:21
$H$ must have the identity element. It must also have the element $(1 2)$.
– Lord Shark the Unknown
Dec 10 '18 at 4:22
$(1,2)$ has order 2 so it's the only non identity element in $H$.
– Justin Stevenson
Dec 10 '18 at 4:22
Why does $H$ only contain $(1,2)$ and the identity element?
– AMN52
Dec 10 '18 at 4:25