Nilpotent ring and Nilpotent groups.
$begingroup$
Let $R$ be a ring (associative and with unity) and $B$ be a subring with the property that $B^n = 0$ i.e.
$$ forall; x_1, x_2, dots, x_n in B: ; x_1 cdot x_2 cdots x_n = 0$$
My aim is to prove that the set $G = {1+x mid xin B}$ is a nilpotent group.
It is easy to see that $G$ is a group. For $n=3$, I found
$$[1+x,1+y]_G = 1+[x,y]_R,$$
where $$[1+x,1+y]_G = (1-x+x^2)(1-y+y^2)(1+x)(1+y)$$ is the group commutator
and $[x,y]_R = xy-yx$ is the ring commutator. And $G$ is a nilpotent group of degree no more than $3$.
But if $n=4$, I got
$$[1+x, 1+y]_G = 1 + x^2y+xy+xyx+y^2x-yx-yxy,$$ and I don't know what to do.
abstract-algebra group-theory ring-theory nilpotence nilpotent-groups
edited Feb 13 at 18:25
Yanior Weg
2,37211144
asked Dec 21 '18 at 16:06
NoctusNoctus
575
$endgroup$
add a comment |
$begingroup$
Let $R$ be a ring (associative and with unity) and $B$ be a subring with the property that $B^n = 0$ i.e.
$$ forall; x_1, x_2, dots, x_n in B: ; x_1 cdot x_2 cdots x_n = 0$$
My aim is to prove that the set $G = {1+x mid xin B}$ is a nilpotent group.
It is easy to see that $G$ is a group. For $n=3$, I found
$$[1+x,1+y]_G = 1+[x,y]_R,$$
where $$[1+x,1+y]_G = (1-x+x^2)(1-y+y^2)(1+x)(1+y)$$ is the group commutator
and $[x,y]_R = xy-yx$ is the ring commutator. And $G$ is a nilpotent group of degree no more than $3$.
But if $n=4$, I got
$$[1+x, 1+y]_G = 1 + x^2y+xy+xyx+y^2x-yx-yxy,$$ and I don't know what to do.
abstract-algebra group-theory ring-theory nilpotence nilpotent-groups
edited Feb 13 at 18:25
Yanior Weg
2,37211144
asked Dec 21 '18 at 16:06
NoctusNoctus
575
$endgroup$
add a comment |
$begingroup$
Let $R$ be a ring (associative and with unity) and $B$ be a subring with the property that $B^n = 0$ i.e.
$$ forall; x_1, x_2, dots, x_n in B: ; x_1 cdot x_2 cdots x_n = 0$$
My aim is to prove that the set $G = {1+x mid xin B}$ is a nilpotent group.
It is easy to see that $G$ is a group. For $n=3$, I found
$$[1+x,1+y]_G = 1+[x,y]_R,$$
where $$[1+x,1+y]_G = (1-x+x^2)(1-y+y^2)(1+x)(1+y)$$ is the group commutator
and $[x,y]_R = xy-yx$ is the ring commutator. And $G$ is a nilpotent group of degree no more than $3$.
But if $n=4$, I got
$$[1+x, 1+y]_G = 1 + x^2y+xy+xyx+y^2x-yx-yxy,$$ and I don't know what to do.
abstract-algebra group-theory ring-theory nilpotence nilpotent-groups
edited Feb 13 at 18:25
Yanior Weg
2,37211144
asked Dec 21 '18 at 16:06
NoctusNoctus
575
$endgroup$
Let $R$ be a ring (associative and with unity) and $B$ be a subring with the property that $B^n = 0$ i.e.
$$ forall; x_1, x_2, dots, x_n in B: ; x_1 cdot x_2 cdots x_n = 0$$
My aim is to prove that the set $G = {1+x mid xin B}$ is a nilpotent group.
It is easy to see that $G$ is a group. For $n=3$, I found
$$[1+x,1+y]_G = 1+[x,y]_R,$$
where $$[1+x,1+y]_G = (1-x+x^2)(1-y+y^2)(1+x)(1+y)$$ is the group commutator
and $[x,y]_R = xy-yx$ is the ring commutator. And $G$ is a nilpotent group of degree no more than $3$.
But if $n=4$, I got
$$[1+x, 1+y]_G = 1 + x^2y+xy+xyx+y^2x-yx-yxy,$$ and I don't know what to do.
abstract-algebra group-theory ring-theory nilpotence nilpotent-groups
abstract-algebra group-theory ring-theory nilpotence nilpotent-groups
edited Feb 13 at 18:25
Yanior Weg
2,37211144
asked Dec 21 '18 at 16:06
NoctusNoctus
575
edited Feb 13 at 18:25
Yanior Weg
2,37211144
asked Dec 21 '18 at 16:06
NoctusNoctus
575
edited Feb 13 at 18:25
Yanior Weg
2,37211144
edited Feb 13 at 18:25
Yanior Weg
2,37211144
edited Feb 13 at 18:25
Yanior Weg
2,37211144
2,37211144
asked Dec 21 '18 at 16:06
NoctusNoctus
575
asked Dec 21 '18 at 16:06
NoctusNoctus
575
asked Dec 21 '18 at 16:06
NoctusNoctus
575
575
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
First, note that since every element of $B$ is nilpotent, we have the following explicit formula for the inverse of $1-x$:
$$(1-x)^{-1}=1+x+x^2+...+x^{n-1}$$
So let's define our descending series of normal subgroups as:
$$G_k=1+B_k$$ Where $B_k$ is the additive subgroup of $B$ generated by elements of the form $x_1 x_2 ...x_m$ for $x_iin B$, $mgeq k$.
One may check that $G_k$ is a normal subgroup in $G$, and we have $G=G_1$ and $G_n=1$.
So to show $G$ is nilpotent, it suffices to show that $[g,G_k]subset G_{k+1}$ for all $gin G$, where $[,]$ is the group commutator.
Doing this in the case of the element of $G_k$ being $1-x_1 x_2 ....x_m$:
$(1-y)(1-x_1 x_2 .... x_m)(1+y+y^2...+y^{n-1})(1+x_1 x_2 ..x_m + (x_1 x_2...x_m)^2+...)$=
$big(1-(1-y)x_1 x_2 ...x_m(1+y+y^2+...y^{n-1})big)(1+x_1 x_2 ..x_m + (x_1 x_2...x_m)^2+...)$=
$big(1-x_1 x_2 ...x_m(1+y+y^2+...y^{n-1})+y(x_1 x_2 ...x_m(1+y+y^2+...y^{n-1})big)(1+x_1 x_2 ..x_m + (x_1 x_2...x_m)^2+...)$
$alpha=-x_1 x_2 ..x_m(y+y^2+...y^{n-1})$
$beta= y x_1 x_2 ...x_m(1+y+y^2+...y^{n-1})$
=$(1-x_1 x_2 ..x_m +alpha +beta)(1+x_1 x_2 ..x_m + (x_1 x_2...x_m)^2+...)$
=$1-(x_1...x_m)^2+(alpha+beta)big((x_1 x_2...x_m)^2+...(x_1...x_m)^{n-1}big)+(1-x_1 x_2 ..x_m)(x_1 x_2...x_m)^2+...)+(alpha+beta)(1+x_1 ...x_m)$
Which is in $G_{k+1}$, since all the terms after the $1$ in this expression are of length at least $m+1$.
The general element case is the exact same, its just a bit more of a mess.
edited Feb 9 at 22:59
darij grinberg
11k33167
answered Dec 22 '18 at 5:22
user277182user277182
456212
$endgroup$
$begingroup$
You can simplify the argument by just showing $G_k/G_{k+1}$ is abelian. If $y, zin B_k$ then $$ (1+y)(1+z)=1 +y+z+yz $$ which is just $1+y+zpmod{G_{k+1}}$. This is the same as $(1+z)(1+y)$.
$endgroup$
– Hempelicious
Dec 22 '18 at 17:20
$begingroup$
@Hempelicious's computation (which, by the way, proves $left[G, G_kright] subseteq G_{k+1}$ as well, because we can set $y in B$ instead of $y in B_k$ and get the same final congruence) relies on the following lemma: Given any $k geq 0$ and any two elements $a, b in G$. If $a-b in B_{k+1}$, then $b^{-1}a in G_{k+1}$. (To prove this lemma, assume $a-b in B_{k+1}$. Then, $b^{-1}a - 1 = b^{-1}left(a-bright) in B_{k+1}$ as well, because $b^{-1} in G$ and $G B_{k+1} subseteq B_{k+1}$. In other words, $b^{-1}a in 1 + B_{k+1} = G_{k+1}$.)
$endgroup$
– darij grinberg
Feb 9 at 23:11
add a comment |
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1 Answer
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$begingroup$
First, note that since every element of $B$ is nilpotent, we have the following explicit formula for the inverse of $1-x$:
$$(1-x)^{-1}=1+x+x^2+...+x^{n-1}$$
So let's define our descending series of normal subgroups as:
$$G_k=1+B_k$$ Where $B_k$ is the additive subgroup of $B$ generated by elements of the form $x_1 x_2 ...x_m$ for $x_iin B$, $mgeq k$.
One may check that $G_k$ is a normal subgroup in $G$, and we have $G=G_1$ and $G_n=1$.
So to show $G$ is nilpotent, it suffices to show that $[g,G_k]subset G_{k+1}$ for all $gin G$, where $[,]$ is the group commutator.
Doing this in the case of the element of $G_k$ being $1-x_1 x_2 ....x_m$:
$(1-y)(1-x_1 x_2 .... x_m)(1+y+y^2...+y^{n-1})(1+x_1 x_2 ..x_m + (x_1 x_2...x_m)^2+...)$=
$big(1-(1-y)x_1 x_2 ...x_m(1+y+y^2+...y^{n-1})big)(1+x_1 x_2 ..x_m + (x_1 x_2...x_m)^2+...)$=
$big(1-x_1 x_2 ...x_m(1+y+y^2+...y^{n-1})+y(x_1 x_2 ...x_m(1+y+y^2+...y^{n-1})big)(1+x_1 x_2 ..x_m + (x_1 x_2...x_m)^2+...)$
$alpha=-x_1 x_2 ..x_m(y+y^2+...y^{n-1})$
$beta= y x_1 x_2 ...x_m(1+y+y^2+...y^{n-1})$
=$(1-x_1 x_2 ..x_m +alpha +beta)(1+x_1 x_2 ..x_m + (x_1 x_2...x_m)^2+...)$
=$1-(x_1...x_m)^2+(alpha+beta)big((x_1 x_2...x_m)^2+...(x_1...x_m)^{n-1}big)+(1-x_1 x_2 ..x_m)(x_1 x_2...x_m)^2+...)+(alpha+beta)(1+x_1 ...x_m)$
Which is in $G_{k+1}$, since all the terms after the $1$ in this expression are of length at least $m+1$.
The general element case is the exact same, its just a bit more of a mess.
edited Feb 9 at 22:59
darij grinberg
11k33167
answered Dec 22 '18 at 5:22
user277182user277182
456212
$endgroup$
$begingroup$
You can simplify the argument by just showing $G_k/G_{k+1}$ is abelian. If $y, zin B_k$ then $$ (1+y)(1+z)=1 +y+z+yz $$ which is just $1+y+zpmod{G_{k+1}}$. This is the same as $(1+z)(1+y)$.
$endgroup$
– Hempelicious
Dec 22 '18 at 17:20
$begingroup$
@Hempelicious's computation (which, by the way, proves $left[G, G_kright] subseteq G_{k+1}$ as well, because we can set $y in B$ instead of $y in B_k$ and get the same final congruence) relies on the following lemma: Given any $k geq 0$ and any two elements $a, b in G$. If $a-b in B_{k+1}$, then $b^{-1}a in G_{k+1}$. (To prove this lemma, assume $a-b in B_{k+1}$. Then, $b^{-1}a - 1 = b^{-1}left(a-bright) in B_{k+1}$ as well, because $b^{-1} in G$ and $G B_{k+1} subseteq B_{k+1}$. In other words, $b^{-1}a in 1 + B_{k+1} = G_{k+1}$.)
$endgroup$
– darij grinberg
Feb 9 at 23:11
add a comment |
$begingroup$
First, note that since every element of $B$ is nilpotent, we have the following explicit formula for the inverse of $1-x$:
$$(1-x)^{-1}=1+x+x^2+...+x^{n-1}$$
So let's define our descending series of normal subgroups as:
$$G_k=1+B_k$$ Where $B_k$ is the additive subgroup of $B$ generated by elements of the form $x_1 x_2 ...x_m$ for $x_iin B$, $mgeq k$.
One may check that $G_k$ is a normal subgroup in $G$, and we have $G=G_1$ and $G_n=1$.
So to show $G$ is nilpotent, it suffices to show that $[g,G_k]subset G_{k+1}$ for all $gin G$, where $[,]$ is the group commutator.
Doing this in the case of the element of $G_k$ being $1-x_1 x_2 ....x_m$:
$(1-y)(1-x_1 x_2 .... x_m)(1+y+y^2...+y^{n-1})(1+x_1 x_2 ..x_m + (x_1 x_2...x_m)^2+...)$=
$big(1-(1-y)x_1 x_2 ...x_m(1+y+y^2+...y^{n-1})big)(1+x_1 x_2 ..x_m + (x_1 x_2...x_m)^2+...)$=
$big(1-x_1 x_2 ...x_m(1+y+y^2+...y^{n-1})+y(x_1 x_2 ...x_m(1+y+y^2+...y^{n-1})big)(1+x_1 x_2 ..x_m + (x_1 x_2...x_m)^2+...)$
$alpha=-x_1 x_2 ..x_m(y+y^2+...y^{n-1})$
$beta= y x_1 x_2 ...x_m(1+y+y^2+...y^{n-1})$
=$(1-x_1 x_2 ..x_m +alpha +beta)(1+x_1 x_2 ..x_m + (x_1 x_2...x_m)^2+...)$
=$1-(x_1...x_m)^2+(alpha+beta)big((x_1 x_2...x_m)^2+...(x_1...x_m)^{n-1}big)+(1-x_1 x_2 ..x_m)(x_1 x_2...x_m)^2+...)+(alpha+beta)(1+x_1 ...x_m)$
Which is in $G_{k+1}$, since all the terms after the $1$ in this expression are of length at least $m+1$.
The general element case is the exact same, its just a bit more of a mess.
edited Feb 9 at 22:59
darij grinberg
11k33167
answered Dec 22 '18 at 5:22
user277182user277182
456212
$endgroup$
$begingroup$
You can simplify the argument by just showing $G_k/G_{k+1}$ is abelian. If $y, zin B_k$ then $$ (1+y)(1+z)=1 +y+z+yz $$ which is just $1+y+zpmod{G_{k+1}}$. This is the same as $(1+z)(1+y)$.
$endgroup$
– Hempelicious
Dec 22 '18 at 17:20
$begingroup$
@Hempelicious's computation (which, by the way, proves $left[G, G_kright] subseteq G_{k+1}$ as well, because we can set $y in B$ instead of $y in B_k$ and get the same final congruence) relies on the following lemma: Given any $k geq 0$ and any two elements $a, b in G$. If $a-b in B_{k+1}$, then $b^{-1}a in G_{k+1}$. (To prove this lemma, assume $a-b in B_{k+1}$. Then, $b^{-1}a - 1 = b^{-1}left(a-bright) in B_{k+1}$ as well, because $b^{-1} in G$ and $G B_{k+1} subseteq B_{k+1}$. In other words, $b^{-1}a in 1 + B_{k+1} = G_{k+1}$.)
$endgroup$
– darij grinberg
Feb 9 at 23:11
add a comment |
$begingroup$
First, note that since every element of $B$ is nilpotent, we have the following explicit formula for the inverse of $1-x$:
$$(1-x)^{-1}=1+x+x^2+...+x^{n-1}$$
So let's define our descending series of normal subgroups as:
$$G_k=1+B_k$$ Where $B_k$ is the additive subgroup of $B$ generated by elements of the form $x_1 x_2 ...x_m$ for $x_iin B$, $mgeq k$.
One may check that $G_k$ is a normal subgroup in $G$, and we have $G=G_1$ and $G_n=1$.
So to show $G$ is nilpotent, it suffices to show that $[g,G_k]subset G_{k+1}$ for all $gin G$, where $[,]$ is the group commutator.
Doing this in the case of the element of $G_k$ being $1-x_1 x_2 ....x_m$:
$(1-y)(1-x_1 x_2 .... x_m)(1+y+y^2...+y^{n-1})(1+x_1 x_2 ..x_m + (x_1 x_2...x_m)^2+...)$=
$big(1-(1-y)x_1 x_2 ...x_m(1+y+y^2+...y^{n-1})big)(1+x_1 x_2 ..x_m + (x_1 x_2...x_m)^2+...)$=
$big(1-x_1 x_2 ...x_m(1+y+y^2+...y^{n-1})+y(x_1 x_2 ...x_m(1+y+y^2+...y^{n-1})big)(1+x_1 x_2 ..x_m + (x_1 x_2...x_m)^2+...)$
$alpha=-x_1 x_2 ..x_m(y+y^2+...y^{n-1})$
$beta= y x_1 x_2 ...x_m(1+y+y^2+...y^{n-1})$
=$(1-x_1 x_2 ..x_m +alpha +beta)(1+x_1 x_2 ..x_m + (x_1 x_2...x_m)^2+...)$
=$1-(x_1...x_m)^2+(alpha+beta)big((x_1 x_2...x_m)^2+...(x_1...x_m)^{n-1}big)+(1-x_1 x_2 ..x_m)(x_1 x_2...x_m)^2+...)+(alpha+beta)(1+x_1 ...x_m)$
Which is in $G_{k+1}$, since all the terms after the $1$ in this expression are of length at least $m+1$.
The general element case is the exact same, its just a bit more of a mess.
edited Feb 9 at 22:59
darij grinberg
11k33167
answered Dec 22 '18 at 5:22
user277182user277182
456212
$endgroup$
First, note that since every element of $B$ is nilpotent, we have the following explicit formula for the inverse of $1-x$:
$$(1-x)^{-1}=1+x+x^2+...+x^{n-1}$$
So let's define our descending series of normal subgroups as:
$$G_k=1+B_k$$ Where $B_k$ is the additive subgroup of $B$ generated by elements of the form $x_1 x_2 ...x_m$ for $x_iin B$, $mgeq k$.
One may check that $G_k$ is a normal subgroup in $G$, and we have $G=G_1$ and $G_n=1$.
So to show $G$ is nilpotent, it suffices to show that $[g,G_k]subset G_{k+1}$ for all $gin G$, where $[,]$ is the group commutator.
Doing this in the case of the element of $G_k$ being $1-x_1 x_2 ....x_m$:
$(1-y)(1-x_1 x_2 .... x_m)(1+y+y^2...+y^{n-1})(1+x_1 x_2 ..x_m + (x_1 x_2...x_m)^2+...)$=
$big(1-(1-y)x_1 x_2 ...x_m(1+y+y^2+...y^{n-1})big)(1+x_1 x_2 ..x_m + (x_1 x_2...x_m)^2+...)$=
$big(1-x_1 x_2 ...x_m(1+y+y^2+...y^{n-1})+y(x_1 x_2 ...x_m(1+y+y^2+...y^{n-1})big)(1+x_1 x_2 ..x_m + (x_1 x_2...x_m)^2+...)$
$alpha=-x_1 x_2 ..x_m(y+y^2+...y^{n-1})$
$beta= y x_1 x_2 ...x_m(1+y+y^2+...y^{n-1})$
=$(1-x_1 x_2 ..x_m +alpha +beta)(1+x_1 x_2 ..x_m + (x_1 x_2...x_m)^2+...)$
=$1-(x_1...x_m)^2+(alpha+beta)big((x_1 x_2...x_m)^2+...(x_1...x_m)^{n-1}big)+(1-x_1 x_2 ..x_m)(x_1 x_2...x_m)^2+...)+(alpha+beta)(1+x_1 ...x_m)$
Which is in $G_{k+1}$, since all the terms after the $1$ in this expression are of length at least $m+1$.
The general element case is the exact same, its just a bit more of a mess.
edited Feb 9 at 22:59
darij grinberg
11k33167
answered Dec 22 '18 at 5:22
user277182user277182
456212
edited Feb 9 at 22:59
darij grinberg
11k33167
edited Feb 9 at 22:59
darij grinberg
11k33167
edited Feb 9 at 22:59
darij grinberg
11k33167
11k33167
answered Dec 22 '18 at 5:22
user277182user277182
456212
answered Dec 22 '18 at 5:22
user277182user277182
456212
answered Dec 22 '18 at 5:22
user277182user277182
456212
456212
$begingroup$
You can simplify the argument by just showing $G_k/G_{k+1}$ is abelian. If $y, zin B_k$ then $$ (1+y)(1+z)=1 +y+z+yz $$ which is just $1+y+zpmod{G_{k+1}}$. This is the same as $(1+z)(1+y)$.
$endgroup$
– Hempelicious
Dec 22 '18 at 17:20
$begingroup$
@Hempelicious's computation (which, by the way, proves $left[G, G_kright] subseteq G_{k+1}$ as well, because we can set $y in B$ instead of $y in B_k$ and get the same final congruence) relies on the following lemma: Given any $k geq 0$ and any two elements $a, b in G$. If $a-b in B_{k+1}$, then $b^{-1}a in G_{k+1}$. (To prove this lemma, assume $a-b in B_{k+1}$. Then, $b^{-1}a - 1 = b^{-1}left(a-bright) in B_{k+1}$ as well, because $b^{-1} in G$ and $G B_{k+1} subseteq B_{k+1}$. In other words, $b^{-1}a in 1 + B_{k+1} = G_{k+1}$.)
$endgroup$
– darij grinberg
Feb 9 at 23:11
add a comment |
$begingroup$
You can simplify the argument by just showing $G_k/G_{k+1}$ is abelian. If $y, zin B_k$ then $$ (1+y)(1+z)=1 +y+z+yz $$ which is just $1+y+zpmod{G_{k+1}}$. This is the same as $(1+z)(1+y)$.
$endgroup$
– Hempelicious
Dec 22 '18 at 17:20
$begingroup$
@Hempelicious's computation (which, by the way, proves $left[G, G_kright] subseteq G_{k+1}$ as well, because we can set $y in B$ instead of $y in B_k$ and get the same final congruence) relies on the following lemma: Given any $k geq 0$ and any two elements $a, b in G$. If $a-b in B_{k+1}$, then $b^{-1}a in G_{k+1}$. (To prove this lemma, assume $a-b in B_{k+1}$. Then, $b^{-1}a - 1 = b^{-1}left(a-bright) in B_{k+1}$ as well, because $b^{-1} in G$ and $G B_{k+1} subseteq B_{k+1}$. In other words, $b^{-1}a in 1 + B_{k+1} = G_{k+1}$.)
$endgroup$
– darij grinberg
Feb 9 at 23:11
$begingroup$
You can simplify the argument by just showing $G_k/G_{k+1}$ is abelian. If $y, zin B_k$ then $$ (1+y)(1+z)=1 +y+z+yz $$ which is just $1+y+zpmod{G_{k+1}}$. This is the same as $(1+z)(1+y)$.
$endgroup$
– Hempelicious
Dec 22 '18 at 17:20
$begingroup$
You can simplify the argument by just showing $G_k/G_{k+1}$ is abelian. If $y, zin B_k$ then $$ (1+y)(1+z)=1 +y+z+yz $$ which is just $1+y+zpmod{G_{k+1}}$. This is the same as $(1+z)(1+y)$.
$endgroup$
– Hempelicious
Dec 22 '18 at 17:20
$begingroup$
@Hempelicious's computation (which, by the way, proves $left[G, G_kright] subseteq G_{k+1}$ as well, because we can set $y in B$ instead of $y in B_k$ and get the same final congruence) relies on the following lemma: Given any $k geq 0$ and any two elements $a, b in G$. If $a-b in B_{k+1}$, then $b^{-1}a in G_{k+1}$. (To prove this lemma, assume $a-b in B_{k+1}$. Then, $b^{-1}a - 1 = b^{-1}left(a-bright) in B_{k+1}$ as well, because $b^{-1} in G$ and $G B_{k+1} subseteq B_{k+1}$. In other words, $b^{-1}a in 1 + B_{k+1} = G_{k+1}$.)
$endgroup$
– darij grinberg
Feb 9 at 23:11
$begingroup$
@Hempelicious's computation (which, by the way, proves $left[G, G_kright] subseteq G_{k+1}$ as well, because we can set $y in B$ instead of $y in B_k$ and get the same final congruence) relies on the following lemma: Given any $k geq 0$ and any two elements $a, b in G$. If $a-b in B_{k+1}$, then $b^{-1}a in G_{k+1}$. (To prove this lemma, assume $a-b in B_{k+1}$. Then, $b^{-1}a - 1 = b^{-1}left(a-bright) in B_{k+1}$ as well, because $b^{-1} in G$ and $G B_{k+1} subseteq B_{k+1}$. In other words, $b^{-1}a in 1 + B_{k+1} = G_{k+1}$.)
$endgroup$
– darij grinberg
Feb 9 at 23:11
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