How to prove the closure of Commutator Subgroup?
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I know that a commutator subgroup $C$ of a group $G$ is defined as $C=a,bin G$. When I reading the textbook, it always state that $C$ is a subgroup of $G$ and only prove if it is normal.
To prove a set is a group, we need to prove is it closure, associative, have an identity element and has inverse, but when I check my textbook on how to prove $C$ is a group, it only shows identity element and inverse, although associative is too obvious, but as the closure I cannot find a way to prove that.
The question is, how to prove that $aba^-1b^-1hkh^-1k^-1in C$ for all $a,b,h,kin G$?
abstract-algebra group-theory normal-subgroups
asked Aug 26 at 6:49
kelvin hong 方
3137
 |Â
show 6 more comments
up vote
2
down vote
favorite
I know that a commutator subgroup $C$ of a group $G$ is defined as $C=a,bin G$. When I reading the textbook, it always state that $C$ is a subgroup of $G$ and only prove if it is normal.
To prove a set is a group, we need to prove is it closure, associative, have an identity element and has inverse, but when I check my textbook on how to prove $C$ is a group, it only shows identity element and inverse, although associative is too obvious, but as the closure I cannot find a way to prove that.
The question is, how to prove that $aba^-1b^-1hkh^-1k^-1in C$ for all $a,b,h,kin G$?
abstract-algebra group-theory normal-subgroups
asked Aug 26 at 6:49
kelvin hong 方
3137
5
That's not the definition of the commutator subgroup. The commutator subgroup is defined as the subgroup generated by the commutators.
– Lord Shark the Unknown
Aug 26 at 6:50
i.e. $C = langle aba^-1b^-1 mid a,bin Grangle$
– Vishnu M
Aug 26 at 6:53
@LordSharktheUnknown You mean there are some elements in $C$ cannot express in the form $aba^-1b^-1$ right? It is just 'generated' by it but not all of them have the form?
– kelvin hong 方
Aug 26 at 6:57
@postmortes Sir can you expressed explicitly how $aba^-1cb^-1c^-1$ can be the form $xyx^-1y^-1$? Additionally, it seems that the two elements you choose have common variable $b$, can you write how can I write the product of the two with no common variable?
– kelvin hong 方
Aug 26 at 7:21
@postmortes This sounds stranges, can you show me what other things I should have, to prove I can actually do that?
– kelvin hong 方
Aug 26 at 7:33
 |Â
show 6 more comments
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I know that a commutator subgroup $C$ of a group $G$ is defined as $C=a,bin G$. When I reading the textbook, it always state that $C$ is a subgroup of $G$ and only prove if it is normal.
To prove a set is a group, we need to prove is it closure, associative, have an identity element and has inverse, but when I check my textbook on how to prove $C$ is a group, it only shows identity element and inverse, although associative is too obvious, but as the closure I cannot find a way to prove that.
The question is, how to prove that $aba^-1b^-1hkh^-1k^-1in C$ for all $a,b,h,kin G$?
abstract-algebra group-theory normal-subgroups
asked Aug 26 at 6:49
kelvin hong 方
3137
I know that a commutator subgroup $C$ of a group $G$ is defined as $C=a,bin G$. When I reading the textbook, it always state that $C$ is a subgroup of $G$ and only prove if it is normal.
To prove a set is a group, we need to prove is it closure, associative, have an identity element and has inverse, but when I check my textbook on how to prove $C$ is a group, it only shows identity element and inverse, although associative is too obvious, but as the closure I cannot find a way to prove that.
The question is, how to prove that $aba^-1b^-1hkh^-1k^-1in C$ for all $a,b,h,kin G$?
abstract-algebra group-theory normal-subgroups
asked Aug 26 at 6:49
kelvin hong 方
3137
asked Aug 26 at 6:49
kelvin hong 方
3137
asked Aug 26 at 6:49
kelvin hong 方
3137
asked Aug 26 at 6:49
kelvin hong 方
3137
3137
5
That's not the definition of the commutator subgroup. The commutator subgroup is defined as the subgroup generated by the commutators.
– Lord Shark the Unknown
Aug 26 at 6:50
i.e. $C = langle aba^-1b^-1 mid a,bin Grangle$
– Vishnu M
Aug 26 at 6:53
@LordSharktheUnknown You mean there are some elements in $C$ cannot express in the form $aba^-1b^-1$ right? It is just 'generated' by it but not all of them have the form?
– kelvin hong 方
Aug 26 at 6:57
@postmortes Sir can you expressed explicitly how $aba^-1cb^-1c^-1$ can be the form $xyx^-1y^-1$? Additionally, it seems that the two elements you choose have common variable $b$, can you write how can I write the product of the two with no common variable?
– kelvin hong 方
Aug 26 at 7:21
@postmortes This sounds stranges, can you show me what other things I should have, to prove I can actually do that?
– kelvin hong 方
Aug 26 at 7:33
 |Â
show 6 more comments
5
That's not the definition of the commutator subgroup. The commutator subgroup is defined as the subgroup generated by the commutators.
– Lord Shark the Unknown
Aug 26 at 6:50
i.e. $C = langle aba^-1b^-1 mid a,bin Grangle$
– Vishnu M
Aug 26 at 6:53
@LordSharktheUnknown You mean there are some elements in $C$ cannot express in the form $aba^-1b^-1$ right? It is just 'generated' by it but not all of them have the form?
– kelvin hong 方
Aug 26 at 6:57
@postmortes Sir can you expressed explicitly how $aba^-1cb^-1c^-1$ can be the form $xyx^-1y^-1$? Additionally, it seems that the two elements you choose have common variable $b$, can you write how can I write the product of the two with no common variable?
– kelvin hong 方
Aug 26 at 7:21
@postmortes This sounds stranges, can you show me what other things I should have, to prove I can actually do that?
– kelvin hong 方
Aug 26 at 7:33
5
5
That's not the definition of the commutator subgroup. The commutator subgroup is defined as the subgroup generated by the commutators.
– Lord Shark the Unknown
Aug 26 at 6:50
That's not the definition of the commutator subgroup. The commutator subgroup is defined as the subgroup generated by the commutators.
– Lord Shark the Unknown
Aug 26 at 6:50
i.e. $C = langle aba^-1b^-1 mid a,bin Grangle$
– Vishnu M
Aug 26 at 6:53
i.e. $C = langle aba^-1b^-1 mid a,bin Grangle$
– Vishnu M
Aug 26 at 6:53
@LordSharktheUnknown You mean there are some elements in $C$ cannot express in the form $aba^-1b^-1$ right? It is just 'generated' by it but not all of them have the form?
– kelvin hong 方
Aug 26 at 6:57
@LordSharktheUnknown You mean there are some elements in $C$ cannot express in the form $aba^-1b^-1$ right? It is just 'generated' by it but not all of them have the form?
– kelvin hong 方
Aug 26 at 6:57
@postmortes Sir can you expressed explicitly how $aba^-1cb^-1c^-1$ can be the form $xyx^-1y^-1$? Additionally, it seems that the two elements you choose have common variable $b$, can you write how can I write the product of the two with no common variable?
– kelvin hong 方
Aug 26 at 7:21
@postmortes Sir can you expressed explicitly how $aba^-1cb^-1c^-1$ can be the form $xyx^-1y^-1$? Additionally, it seems that the two elements you choose have common variable $b$, can you write how can I write the product of the two with no common variable?
– kelvin hong 方
Aug 26 at 7:21
@postmortes This sounds stranges, can you show me what other things I should have, to prove I can actually do that?
– kelvin hong 方
Aug 26 at 7:33
@postmortes This sounds stranges, can you show me what other things I should have, to prove I can actually do that?
– kelvin hong 方
Aug 26 at 7:33
 |Â
show 6 more comments
1 Answer
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In general, the set of commutators does not equal the commutator subgroup, as pointed out above by several remarks. See also this paper of Kappe and Morse, where criteria are discussed for equality.
I. D. MacDonald provides counterexamples in [I. D. MacDonald, Commutators and Their Products, The American Mathematical Monthly Vol. 93, No. 6 (1986), pp. 440-444]. In particular, he proves by a simple counting argument the following theorem.
If $G$ is a finite group and $|G:Z(G)|^2<|G'|$, then $G'$ has elements which are not commutators.
Finally see also this StackExchange entry, which mentions another interesting paper of Marty Isaacs (also behind Jstor) with more counterexamples.
answered Aug 26 at 11:49
Nicky Hekster
27.1k53152
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
In general, the set of commutators does not equal the commutator subgroup, as pointed out above by several remarks. See also this paper of Kappe and Morse, where criteria are discussed for equality.
I. D. MacDonald provides counterexamples in [I. D. MacDonald, Commutators and Their Products, The American Mathematical Monthly Vol. 93, No. 6 (1986), pp. 440-444]. In particular, he proves by a simple counting argument the following theorem.
If $G$ is a finite group and $|G:Z(G)|^2<|G'|$, then $G'$ has elements which are not commutators.
Finally see also this StackExchange entry, which mentions another interesting paper of Marty Isaacs (also behind Jstor) with more counterexamples.
answered Aug 26 at 11:49
Nicky Hekster
27.1k53152
add a comment |Â
up vote
3
down vote
accepted
In general, the set of commutators does not equal the commutator subgroup, as pointed out above by several remarks. See also this paper of Kappe and Morse, where criteria are discussed for equality.
I. D. MacDonald provides counterexamples in [I. D. MacDonald, Commutators and Their Products, The American Mathematical Monthly Vol. 93, No. 6 (1986), pp. 440-444]. In particular, he proves by a simple counting argument the following theorem.
If $G$ is a finite group and $|G:Z(G)|^2<|G'|$, then $G'$ has elements which are not commutators.
Finally see also this StackExchange entry, which mentions another interesting paper of Marty Isaacs (also behind Jstor) with more counterexamples.
answered Aug 26 at 11:49
Nicky Hekster
27.1k53152
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
In general, the set of commutators does not equal the commutator subgroup, as pointed out above by several remarks. See also this paper of Kappe and Morse, where criteria are discussed for equality.
I. D. MacDonald provides counterexamples in [I. D. MacDonald, Commutators and Their Products, The American Mathematical Monthly Vol. 93, No. 6 (1986), pp. 440-444]. In particular, he proves by a simple counting argument the following theorem.
If $G$ is a finite group and $|G:Z(G)|^2<|G'|$, then $G'$ has elements which are not commutators.
Finally see also this StackExchange entry, which mentions another interesting paper of Marty Isaacs (also behind Jstor) with more counterexamples.
answered Aug 26 at 11:49
Nicky Hekster
27.1k53152
In general, the set of commutators does not equal the commutator subgroup, as pointed out above by several remarks. See also this paper of Kappe and Morse, where criteria are discussed for equality.
I. D. MacDonald provides counterexamples in [I. D. MacDonald, Commutators and Their Products, The American Mathematical Monthly Vol. 93, No. 6 (1986), pp. 440-444]. In particular, he proves by a simple counting argument the following theorem.
If $G$ is a finite group and $|G:Z(G)|^2<|G'|$, then $G'$ has elements which are not commutators.
Finally see also this StackExchange entry, which mentions another interesting paper of Marty Isaacs (also behind Jstor) with more counterexamples.
answered Aug 26 at 11:49
Nicky Hekster
27.1k53152
answered Aug 26 at 11:49
Nicky Hekster
27.1k53152
answered Aug 26 at 11:49
Nicky Hekster
27.1k53152
answered Aug 26 at 11:49
Nicky Hekster
27.1k53152
27.1k53152
add a comment |Â
add a comment |Â
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5
That's not the definition of the commutator subgroup. The commutator subgroup is defined as the subgroup generated by the commutators.
– Lord Shark the Unknown
Aug 26 at 6:50
i.e. $C = langle aba^-1b^-1 mid a,bin Grangle$
– Vishnu M
Aug 26 at 6:53
@LordSharktheUnknown You mean there are some elements in $C$ cannot express in the form $aba^-1b^-1$ right? It is just 'generated' by it but not all of them have the form?
– kelvin hong 方
Aug 26 at 6:57
@postmortes Sir can you expressed explicitly how $aba^-1cb^-1c^-1$ can be the form $xyx^-1y^-1$? Additionally, it seems that the two elements you choose have common variable $b$, can you write how can I write the product of the two with no common variable?
– kelvin hong 方
Aug 26 at 7:21
@postmortes This sounds stranges, can you show me what other things I should have, to prove I can actually do that?
– kelvin hong 方
Aug 26 at 7:33