Mixing
Showing commutator subgroup is a subgroup
$begingroup$
Now, the common introduction question to a commutator subgroup $G'$ is showing that it is normal in the group $G$. However, I'm having a problem with something even more basic than this.
Let $[a,b],[c,d] in G'$,
Then we have:
$[a,b][c,d]=a^{-1}b^{-1}abc^{-1}d^{-1}cd$
I don't see an easy to conclude that this of the form $[g,h]=g^{-1}h^{-1}gh$ for $g,h in G$.
What am i missing here?
$G' neq { a^{-1}b^{-1}ab : a,b in G }$
$G' = {<a^{-1}b^{-1}ab> : a,b in G}$
group-theory
asked Feb 1 at 23:16
Mathematical MushroomMathematical Mushroom
1278
$endgroup$
|
show 3 more comments
$begingroup$
Now, the common introduction question to a commutator subgroup $G'$ is showing that it is normal in the group $G$. However, I'm having a problem with something even more basic than this.
Let $[a,b],[c,d] in G'$,
Then we have:
$[a,b][c,d]=a^{-1}b^{-1}abc^{-1}d^{-1}cd$
I don't see an easy to conclude that this of the form $[g,h]=g^{-1}h^{-1}gh$ for $g,h in G$.
What am i missing here?
$G' neq { a^{-1}b^{-1}ab : a,b in G }$
$G' = {<a^{-1}b^{-1}ab> : a,b in G}$
group-theory
asked Feb 1 at 23:16
Mathematical MushroomMathematical Mushroom
1278
$endgroup$
3
$begingroup$
How are you defining the commutator subgroup? The standard definition says it is the subgroup generated by commutators. Once can also define it to be any word in the elements of $G$ that has a rearrangement that multiplies to the identity, but that's not standard (I believe).
$endgroup$
– lulu
Feb 1 at 23:19
5
$begingroup$
In general, the product of two commutator elements is not a commutator
$endgroup$
– Hagen von Eitzen
Feb 1 at 23:21
$begingroup$
Oh. Then what is the operation that makes $G'$ a group?
$endgroup$
– Mathematical Mushroom
Feb 1 at 23:22
2
$begingroup$
It's the product inherited from $G$, of course.
$endgroup$
– lulu
Feb 1 at 23:22
5
$begingroup$
No. $G'$ is the subgroup generated by commutators, it is not the set of commutators (which, as you remark, is generally not closed under multiplication).
$endgroup$
– lulu
Feb 1 at 23:24
|
show 3 more comments
$begingroup$
Now, the common introduction question to a commutator subgroup $G'$ is showing that it is normal in the group $G$. However, I'm having a problem with something even more basic than this.
Let $[a,b],[c,d] in G'$,
Then we have:
$[a,b][c,d]=a^{-1}b^{-1}abc^{-1}d^{-1}cd$
I don't see an easy to conclude that this of the form $[g,h]=g^{-1}h^{-1}gh$ for $g,h in G$.
What am i missing here?
$G' neq { a^{-1}b^{-1}ab : a,b in G }$
$G' = {<a^{-1}b^{-1}ab> : a,b in G}$
group-theory
asked Feb 1 at 23:16
Mathematical MushroomMathematical Mushroom
1278
$endgroup$
Now, the common introduction question to a commutator subgroup $G'$ is showing that it is normal in the group $G$. However, I'm having a problem with something even more basic than this.
Let $[a,b],[c,d] in G'$,
Then we have:
$[a,b][c,d]=a^{-1}b^{-1}abc^{-1}d^{-1}cd$
I don't see an easy to conclude that this of the form $[g,h]=g^{-1}h^{-1}gh$ for $g,h in G$.
What am i missing here?
$G' neq { a^{-1}b^{-1}ab : a,b in G }$
$G' = {<a^{-1}b^{-1}ab> : a,b in G}$
group-theory
group-theory
asked Feb 1 at 23:16
Mathematical MushroomMathematical Mushroom
1278
asked Feb 1 at 23:16
Mathematical MushroomMathematical Mushroom
1278
edited Feb 1 at 23:26
Mathematical Mushroom
asked Feb 1 at 23:16
Mathematical MushroomMathematical Mushroom
1278
asked Feb 1 at 23:16
Mathematical MushroomMathematical Mushroom
1278
asked Feb 1 at 23:16
Mathematical MushroomMathematical Mushroom
1278
1278
3
$begingroup$
How are you defining the commutator subgroup? The standard definition says it is the subgroup generated by commutators. Once can also define it to be any word in the elements of $G$ that has a rearrangement that multiplies to the identity, but that's not standard (I believe).
$endgroup$
– lulu
Feb 1 at 23:19
5
$begingroup$
In general, the product of two commutator elements is not a commutator
$endgroup$
– Hagen von Eitzen
Feb 1 at 23:21
$begingroup$
Oh. Then what is the operation that makes $G'$ a group?
$endgroup$
– Mathematical Mushroom
Feb 1 at 23:22
2
$begingroup$
It's the product inherited from $G$, of course.
$endgroup$
– lulu
Feb 1 at 23:22
5
$begingroup$
No. $G'$ is the subgroup generated by commutators, it is not the set of commutators (which, as you remark, is generally not closed under multiplication).
$endgroup$
– lulu
Feb 1 at 23:24
|
show 3 more comments
3
$begingroup$
How are you defining the commutator subgroup? The standard definition says it is the subgroup generated by commutators. Once can also define it to be any word in the elements of $G$ that has a rearrangement that multiplies to the identity, but that's not standard (I believe).
$endgroup$
– lulu
Feb 1 at 23:19
5
$begingroup$
In general, the product of two commutator elements is not a commutator
$endgroup$
– Hagen von Eitzen
Feb 1 at 23:21
$begingroup$
Oh. Then what is the operation that makes $G'$ a group?
$endgroup$
– Mathematical Mushroom
Feb 1 at 23:22
2
$begingroup$
It's the product inherited from $G$, of course.
$endgroup$
– lulu
Feb 1 at 23:22
5
$begingroup$
No. $G'$ is the subgroup generated by commutators, it is not the set of commutators (which, as you remark, is generally not closed under multiplication).
$endgroup$
– lulu
Feb 1 at 23:24
3
3
$begingroup$
How are you defining the commutator subgroup? The standard definition says it is the subgroup generated by commutators. Once can also define it to be any word in the elements of $G$ that has a rearrangement that multiplies to the identity, but that's not standard (I believe).
$endgroup$
– lulu
Feb 1 at 23:19
$begingroup$
How are you defining the commutator subgroup? The standard definition says it is the subgroup generated by commutators. Once can also define it to be any word in the elements of $G$ that has a rearrangement that multiplies to the identity, but that's not standard (I believe).
$endgroup$
– lulu
Feb 1 at 23:19
5
5
$begingroup$
In general, the product of two commutator elements is not a commutator
$endgroup$
– Hagen von Eitzen
Feb 1 at 23:21
$begingroup$
In general, the product of two commutator elements is not a commutator
$endgroup$
– Hagen von Eitzen
Feb 1 at 23:21
$begingroup$
Oh. Then what is the operation that makes $G'$ a group?
$endgroup$
– Mathematical Mushroom
Feb 1 at 23:22
$begingroup$
Oh. Then what is the operation that makes $G'$ a group?
$endgroup$
– Mathematical Mushroom
Feb 1 at 23:22
2
2
$begingroup$
It's the product inherited from $G$, of course.
$endgroup$
– lulu
Feb 1 at 23:22
$begingroup$
It's the product inherited from $G$, of course.
$endgroup$
– lulu
Feb 1 at 23:22
5
5
$begingroup$
No. $G'$ is the subgroup generated by commutators, it is not the set of commutators (which, as you remark, is generally not closed under multiplication).
$endgroup$
– lulu
Feb 1 at 23:24
$begingroup$
No. $G'$ is the subgroup generated by commutators, it is not the set of commutators (which, as you remark, is generally not closed under multiplication).
$endgroup$
– lulu
Feb 1 at 23:24
|
show 3 more comments
1 Answer
1
active
oldest
votes
$begingroup$
Note: $$boxed{G'=langle{ [a, b]mid a,bin G}rangle.}$$ Hence for $[f,g], [h, k]in G'$ with $f,g,h,kin G$, we have that $[f,g][h,k]in G'$, since it is a product of some generators of the subgroup.
More generally, for $mathfrak{g}=prod_{iin I}[g_i, g'_i]$ and $mathfrak{h}=prod_{jin J}[h_j, h'_j]$ belonging to $G'$, where $I$ and $J$ are some index sets for some element sequences $(g_i), (g'_i)in G^I$ and $(h_j), (h'_j)in G^J$, we have that
begin{align}
mathfrak{g}mathfrak{h}&=prod_{iin I}[g_i, g'_i]prod_{jin J}[h_j, h'_j] tag{1} \
&=prod_{ellin L}[f_ell, f'_ell],
end{align}
where $L$ and the sequences $(f_ell), (f'_ell)in G^L$ run through the commutators in $(1)$ in order; that is, one notices that it is a product of commutators as it is the product of two products of commutators.
As for showing it's a subgroup, I suggest the "two-step subgroup lemma". However, as pointed out in the comments below, it is much better to prove that the group $langle Xrangle$ generated by a subset $X$ of a group $G$ with respect to the operation of $G$ is a subgroup of $G$ in general.
This question is about how to show the subgroup is normal.
answered Feb 1 at 23:45
ShaunShaun
10.5k113687
$endgroup$
2
$begingroup$
If it is defined as “the subgroup generated by...” then why would you need to show it is a subgroup? Did you mean “As for showing it’s a normal subgroup”?
$endgroup$
– Arturo Magidin
Feb 2 at 1:03
$begingroup$
Well, for the sake of rigour, some of us prefer to check whether such definitions make sense. I think what I have typed in this answer is illustrative, at least, of how one works with $G'$ and what the general element of $G'$ is. I'll add a note giving a hint on how to prove that the subgroup is normal.
$endgroup$
– Shaun
Feb 2 at 10:38
$begingroup$
D'you think that's a fair comment of mine, @ArturoMagidin? You seem more experienced than I am so your feedback would be appreciated $ddotsmile$
$endgroup$
– Shaun
Feb 2 at 10:54
4
$begingroup$
There are two definitions of "the subgroup of group $G$ generated by subset $X$ of $G$". One is the intersection of all subgroups of $G$ containing $X$, and the other is the set of all products of elements of $X$ and their inverses. The first of these clearly does define a subgroup of $G$ and then you need to prove that the two definitions are equivalent. It is preferable to do all of that in a general setting, since then it becomes unnecessary to prove that you are defining a subgroup in specific cases. For normality, it is easier to prove the stronger property that $G'$ is characteristic.
$endgroup$
– Derek Holt
Feb 2 at 14:25
1
$begingroup$
For general comments along the lines of Derek Holt’s comment, see this prior answer.
$endgroup$
– Arturo Magidin
Feb 2 at 23:36
|
show 2 more comments
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1 Answer
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$begingroup$
Note: $$boxed{G'=langle{ [a, b]mid a,bin G}rangle.}$$ Hence for $[f,g], [h, k]in G'$ with $f,g,h,kin G$, we have that $[f,g][h,k]in G'$, since it is a product of some generators of the subgroup.
More generally, for $mathfrak{g}=prod_{iin I}[g_i, g'_i]$ and $mathfrak{h}=prod_{jin J}[h_j, h'_j]$ belonging to $G'$, where $I$ and $J$ are some index sets for some element sequences $(g_i), (g'_i)in G^I$ and $(h_j), (h'_j)in G^J$, we have that
begin{align}
mathfrak{g}mathfrak{h}&=prod_{iin I}[g_i, g'_i]prod_{jin J}[h_j, h'_j] tag{1} \
&=prod_{ellin L}[f_ell, f'_ell],
end{align}
where $L$ and the sequences $(f_ell), (f'_ell)in G^L$ run through the commutators in $(1)$ in order; that is, one notices that it is a product of commutators as it is the product of two products of commutators.
As for showing it's a subgroup, I suggest the "two-step subgroup lemma". However, as pointed out in the comments below, it is much better to prove that the group $langle Xrangle$ generated by a subset $X$ of a group $G$ with respect to the operation of $G$ is a subgroup of $G$ in general.
This question is about how to show the subgroup is normal.
answered Feb 1 at 23:45
ShaunShaun
10.5k113687
$endgroup$
2
$begingroup$
If it is defined as “the subgroup generated by...” then why would you need to show it is a subgroup? Did you mean “As for showing it’s a normal subgroup”?
$endgroup$
– Arturo Magidin
Feb 2 at 1:03
$begingroup$
Well, for the sake of rigour, some of us prefer to check whether such definitions make sense. I think what I have typed in this answer is illustrative, at least, of how one works with $G'$ and what the general element of $G'$ is. I'll add a note giving a hint on how to prove that the subgroup is normal.
$endgroup$
– Shaun
Feb 2 at 10:38
$begingroup$
D'you think that's a fair comment of mine, @ArturoMagidin? You seem more experienced than I am so your feedback would be appreciated $ddotsmile$
$endgroup$
– Shaun
Feb 2 at 10:54
4
$begingroup$
There are two definitions of "the subgroup of group $G$ generated by subset $X$ of $G$". One is the intersection of all subgroups of $G$ containing $X$, and the other is the set of all products of elements of $X$ and their inverses. The first of these clearly does define a subgroup of $G$ and then you need to prove that the two definitions are equivalent. It is preferable to do all of that in a general setting, since then it becomes unnecessary to prove that you are defining a subgroup in specific cases. For normality, it is easier to prove the stronger property that $G'$ is characteristic.
$endgroup$
– Derek Holt
Feb 2 at 14:25
1
$begingroup$
For general comments along the lines of Derek Holt’s comment, see this prior answer.
$endgroup$
– Arturo Magidin
Feb 2 at 23:36
|
show 2 more comments
$begingroup$
Note: $$boxed{G'=langle{ [a, b]mid a,bin G}rangle.}$$ Hence for $[f,g], [h, k]in G'$ with $f,g,h,kin G$, we have that $[f,g][h,k]in G'$, since it is a product of some generators of the subgroup.
More generally, for $mathfrak{g}=prod_{iin I}[g_i, g'_i]$ and $mathfrak{h}=prod_{jin J}[h_j, h'_j]$ belonging to $G'$, where $I$ and $J$ are some index sets for some element sequences $(g_i), (g'_i)in G^I$ and $(h_j), (h'_j)in G^J$, we have that
begin{align}
mathfrak{g}mathfrak{h}&=prod_{iin I}[g_i, g'_i]prod_{jin J}[h_j, h'_j] tag{1} \
&=prod_{ellin L}[f_ell, f'_ell],
end{align}
where $L$ and the sequences $(f_ell), (f'_ell)in G^L$ run through the commutators in $(1)$ in order; that is, one notices that it is a product of commutators as it is the product of two products of commutators.
As for showing it's a subgroup, I suggest the "two-step subgroup lemma". However, as pointed out in the comments below, it is much better to prove that the group $langle Xrangle$ generated by a subset $X$ of a group $G$ with respect to the operation of $G$ is a subgroup of $G$ in general.
This question is about how to show the subgroup is normal.
answered Feb 1 at 23:45
ShaunShaun
10.5k113687
$endgroup$
2
$begingroup$
If it is defined as “the subgroup generated by...” then why would you need to show it is a subgroup? Did you mean “As for showing it’s a normal subgroup”?
$endgroup$
– Arturo Magidin
Feb 2 at 1:03
$begingroup$
Well, for the sake of rigour, some of us prefer to check whether such definitions make sense. I think what I have typed in this answer is illustrative, at least, of how one works with $G'$ and what the general element of $G'$ is. I'll add a note giving a hint on how to prove that the subgroup is normal.
$endgroup$
– Shaun
Feb 2 at 10:38
$begingroup$
D'you think that's a fair comment of mine, @ArturoMagidin? You seem more experienced than I am so your feedback would be appreciated $ddotsmile$
$endgroup$
– Shaun
Feb 2 at 10:54
4
$begingroup$
There are two definitions of "the subgroup of group $G$ generated by subset $X$ of $G$". One is the intersection of all subgroups of $G$ containing $X$, and the other is the set of all products of elements of $X$ and their inverses. The first of these clearly does define a subgroup of $G$ and then you need to prove that the two definitions are equivalent. It is preferable to do all of that in a general setting, since then it becomes unnecessary to prove that you are defining a subgroup in specific cases. For normality, it is easier to prove the stronger property that $G'$ is characteristic.
$endgroup$
– Derek Holt
Feb 2 at 14:25
1
$begingroup$
For general comments along the lines of Derek Holt’s comment, see this prior answer.
$endgroup$
– Arturo Magidin
Feb 2 at 23:36
|
show 2 more comments
$begingroup$
Note: $$boxed{G'=langle{ [a, b]mid a,bin G}rangle.}$$ Hence for $[f,g], [h, k]in G'$ with $f,g,h,kin G$, we have that $[f,g][h,k]in G'$, since it is a product of some generators of the subgroup.
More generally, for $mathfrak{g}=prod_{iin I}[g_i, g'_i]$ and $mathfrak{h}=prod_{jin J}[h_j, h'_j]$ belonging to $G'$, where $I$ and $J$ are some index sets for some element sequences $(g_i), (g'_i)in G^I$ and $(h_j), (h'_j)in G^J$, we have that
begin{align}
mathfrak{g}mathfrak{h}&=prod_{iin I}[g_i, g'_i]prod_{jin J}[h_j, h'_j] tag{1} \
&=prod_{ellin L}[f_ell, f'_ell],
end{align}
where $L$ and the sequences $(f_ell), (f'_ell)in G^L$ run through the commutators in $(1)$ in order; that is, one notices that it is a product of commutators as it is the product of two products of commutators.
As for showing it's a subgroup, I suggest the "two-step subgroup lemma". However, as pointed out in the comments below, it is much better to prove that the group $langle Xrangle$ generated by a subset $X$ of a group $G$ with respect to the operation of $G$ is a subgroup of $G$ in general.
This question is about how to show the subgroup is normal.
answered Feb 1 at 23:45
ShaunShaun
10.5k113687
$endgroup$
Note: $$boxed{G'=langle{ [a, b]mid a,bin G}rangle.}$$ Hence for $[f,g], [h, k]in G'$ with $f,g,h,kin G$, we have that $[f,g][h,k]in G'$, since it is a product of some generators of the subgroup.
More generally, for $mathfrak{g}=prod_{iin I}[g_i, g'_i]$ and $mathfrak{h}=prod_{jin J}[h_j, h'_j]$ belonging to $G'$, where $I$ and $J$ are some index sets for some element sequences $(g_i), (g'_i)in G^I$ and $(h_j), (h'_j)in G^J$, we have that
begin{align}
mathfrak{g}mathfrak{h}&=prod_{iin I}[g_i, g'_i]prod_{jin J}[h_j, h'_j] tag{1} \
&=prod_{ellin L}[f_ell, f'_ell],
end{align}
where $L$ and the sequences $(f_ell), (f'_ell)in G^L$ run through the commutators in $(1)$ in order; that is, one notices that it is a product of commutators as it is the product of two products of commutators.
As for showing it's a subgroup, I suggest the "two-step subgroup lemma". However, as pointed out in the comments below, it is much better to prove that the group $langle Xrangle$ generated by a subset $X$ of a group $G$ with respect to the operation of $G$ is a subgroup of $G$ in general.
This question is about how to show the subgroup is normal.
answered Feb 1 at 23:45
ShaunShaun
10.5k113687
edited Feb 2 at 17:17
answered Feb 1 at 23:45
ShaunShaun
10.5k113687
answered Feb 1 at 23:45
ShaunShaun
10.5k113687
answered Feb 1 at 23:45
ShaunShaun
10.5k113687
10.5k113687
2
$begingroup$
If it is defined as “the subgroup generated by...” then why would you need to show it is a subgroup? Did you mean “As for showing it’s a normal subgroup”?
$endgroup$
– Arturo Magidin
Feb 2 at 1:03
$begingroup$
Well, for the sake of rigour, some of us prefer to check whether such definitions make sense. I think what I have typed in this answer is illustrative, at least, of how one works with $G'$ and what the general element of $G'$ is. I'll add a note giving a hint on how to prove that the subgroup is normal.
$endgroup$
– Shaun
Feb 2 at 10:38
$begingroup$
D'you think that's a fair comment of mine, @ArturoMagidin? You seem more experienced than I am so your feedback would be appreciated $ddotsmile$
$endgroup$
– Shaun
Feb 2 at 10:54
4
$begingroup$
There are two definitions of "the subgroup of group $G$ generated by subset $X$ of $G$". One is the intersection of all subgroups of $G$ containing $X$, and the other is the set of all products of elements of $X$ and their inverses. The first of these clearly does define a subgroup of $G$ and then you need to prove that the two definitions are equivalent. It is preferable to do all of that in a general setting, since then it becomes unnecessary to prove that you are defining a subgroup in specific cases. For normality, it is easier to prove the stronger property that $G'$ is characteristic.
$endgroup$
– Derek Holt
Feb 2 at 14:25
1
$begingroup$
For general comments along the lines of Derek Holt’s comment, see this prior answer.
$endgroup$
– Arturo Magidin
Feb 2 at 23:36
|
show 2 more comments
2
$begingroup$
If it is defined as “the subgroup generated by...” then why would you need to show it is a subgroup? Did you mean “As for showing it’s a normal subgroup”?
$endgroup$
– Arturo Magidin
Feb 2 at 1:03
$begingroup$
Well, for the sake of rigour, some of us prefer to check whether such definitions make sense. I think what I have typed in this answer is illustrative, at least, of how one works with $G'$ and what the general element of $G'$ is. I'll add a note giving a hint on how to prove that the subgroup is normal.
$endgroup$
– Shaun
Feb 2 at 10:38
$begingroup$
D'you think that's a fair comment of mine, @ArturoMagidin? You seem more experienced than I am so your feedback would be appreciated $ddotsmile$
$endgroup$
– Shaun
Feb 2 at 10:54
4
$begingroup$
There are two definitions of "the subgroup of group $G$ generated by subset $X$ of $G$". One is the intersection of all subgroups of $G$ containing $X$, and the other is the set of all products of elements of $X$ and their inverses. The first of these clearly does define a subgroup of $G$ and then you need to prove that the two definitions are equivalent. It is preferable to do all of that in a general setting, since then it becomes unnecessary to prove that you are defining a subgroup in specific cases. For normality, it is easier to prove the stronger property that $G'$ is characteristic.
$endgroup$
– Derek Holt
Feb 2 at 14:25
1
$begingroup$
For general comments along the lines of Derek Holt’s comment, see this prior answer.
$endgroup$
– Arturo Magidin
Feb 2 at 23:36
2
2
$begingroup$
If it is defined as “the subgroup generated by...” then why would you need to show it is a subgroup? Did you mean “As for showing it’s a normal subgroup”?
$endgroup$
– Arturo Magidin
Feb 2 at 1:03
$begingroup$
If it is defined as “the subgroup generated by...” then why would you need to show it is a subgroup? Did you mean “As for showing it’s a normal subgroup”?
$endgroup$
– Arturo Magidin
Feb 2 at 1:03
$begingroup$
Well, for the sake of rigour, some of us prefer to check whether such definitions make sense. I think what I have typed in this answer is illustrative, at least, of how one works with $G'$ and what the general element of $G'$ is. I'll add a note giving a hint on how to prove that the subgroup is normal.
$endgroup$
– Shaun
Feb 2 at 10:38
$begingroup$
Well, for the sake of rigour, some of us prefer to check whether such definitions make sense. I think what I have typed in this answer is illustrative, at least, of how one works with $G'$ and what the general element of $G'$ is. I'll add a note giving a hint on how to prove that the subgroup is normal.
$endgroup$
– Shaun
Feb 2 at 10:38
$begingroup$
D'you think that's a fair comment of mine, @ArturoMagidin? You seem more experienced than I am so your feedback would be appreciated $ddotsmile$
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– Shaun
Feb 2 at 10:54
$begingroup$
D'you think that's a fair comment of mine, @ArturoMagidin? You seem more experienced than I am so your feedback would be appreciated $ddotsmile$
$endgroup$
– Shaun
Feb 2 at 10:54
4
4
$begingroup$
There are two definitions of "the subgroup of group $G$ generated by subset $X$ of $G$". One is the intersection of all subgroups of $G$ containing $X$, and the other is the set of all products of elements of $X$ and their inverses. The first of these clearly does define a subgroup of $G$ and then you need to prove that the two definitions are equivalent. It is preferable to do all of that in a general setting, since then it becomes unnecessary to prove that you are defining a subgroup in specific cases. For normality, it is easier to prove the stronger property that $G'$ is characteristic.
$endgroup$
– Derek Holt
Feb 2 at 14:25
$begingroup$
There are two definitions of "the subgroup of group $G$ generated by subset $X$ of $G$". One is the intersection of all subgroups of $G$ containing $X$, and the other is the set of all products of elements of $X$ and their inverses. The first of these clearly does define a subgroup of $G$ and then you need to prove that the two definitions are equivalent. It is preferable to do all of that in a general setting, since then it becomes unnecessary to prove that you are defining a subgroup in specific cases. For normality, it is easier to prove the stronger property that $G'$ is characteristic.
$endgroup$
– Derek Holt
Feb 2 at 14:25
1
1
$begingroup$
For general comments along the lines of Derek Holt’s comment, see this prior answer.
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– Arturo Magidin
Feb 2 at 23:36
$begingroup$
For general comments along the lines of Derek Holt’s comment, see this prior answer.
$endgroup$
– Arturo Magidin
Feb 2 at 23:36
|
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3
$begingroup$
How are you defining the commutator subgroup? The standard definition says it is the subgroup generated by commutators. Once can also define it to be any word in the elements of $G$ that has a rearrangement that multiplies to the identity, but that's not standard (I believe).
$endgroup$
– lulu
Feb 1 at 23:19
5
$begingroup$
In general, the product of two commutator elements is not a commutator
$endgroup$
– Hagen von Eitzen
Feb 1 at 23:21
$begingroup$
Oh. Then what is the operation that makes $G'$ a group?
$endgroup$
– Mathematical Mushroom
Feb 1 at 23:22
2
$begingroup$
It's the product inherited from $G$, of course.
$endgroup$
– lulu
Feb 1 at 23:22
5
$begingroup$
No. $G'$ is the subgroup generated by commutators, it is not the set of commutators (which, as you remark, is generally not closed under multiplication).
$endgroup$
– lulu
Feb 1 at 23:24