Is $[ G^i, G^i]$ always normal in $G$?
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Let $G^1 = [G, G]$ and $G^i = [G^i-1, G^i-1]$, then is it true that $G^i$ is normal in $G$?
My approach:
Since it is true for $i = 1$. Assume by induction it is true for $i leq n-1$. If $alpha in [G^i, G^i] = g^i+1$, then $alpha = aba^-1b^-1 $ where $a, b in G^i$. Therefore $$galpha g^-1 = gaba^-1b^-1g^-1 = gaba^-1(g^-1b^-1bg)b^-1g^-1 = big((ga)b(a^-1g^-1)b^-1big) cdotbig(bgb^-1g^-1big) in G^i+1.$$ Hence $G^i+1$ is normal.
Somehow I am finding this result powerful and yet I am unable to find any reference which talks abouts this, so I am guessing I am making some mistake. Can someone confirm this.
group-theory
asked Sep 8 at 12:41
henceproved
172
add a comment |Â
up vote
0
down vote
favorite
Let $G^1 = [G, G]$ and $G^i = [G^i-1, G^i-1]$, then is it true that $G^i$ is normal in $G$?
My approach:
Since it is true for $i = 1$. Assume by induction it is true for $i leq n-1$. If $alpha in [G^i, G^i] = g^i+1$, then $alpha = aba^-1b^-1 $ where $a, b in G^i$. Therefore $$galpha g^-1 = gaba^-1b^-1g^-1 = gaba^-1(g^-1b^-1bg)b^-1g^-1 = big((ga)b(a^-1g^-1)b^-1big) cdotbig(bgb^-1g^-1big) in G^i+1.$$ Hence $G^i+1$ is normal.
Somehow I am finding this result powerful and yet I am unable to find any reference which talks abouts this, so I am guessing I am making some mistake. Can someone confirm this.
group-theory
asked Sep 8 at 12:41
henceproved
172
1
Yes it's normal, it's called the derived series and any introductory algebra book should talk about it.
– Steve D
Sep 8 at 12:53
1
The usual way to prove this is to show that $[G,G]$ is characteristic in $G$, and since being characteristic is a transitive property, it follows that $G^i$ is characteristic and hence normal.
– Derek Holt
Sep 8 at 13:10
Actually it's even better : $G^i$ is characteristic in $G$ and this follows by induction from the fact that $[G,G]$ is characteristic in $G$ and that characteristicity is transitive
– Max
Sep 8 at 19:59
The other way is to show that $[H,K]$ is normal in $G$ if both $H$ and $K$ are. To see this, note that the conjugate of a commutator of two elements is equal to the commutator of the conjugates.
– C Monsour
Sep 8 at 20:25
Also please note that you mean $G^(i)$ for the derived series, not $G^i$, which is from the lower central series. $G^(i+1)=[G^(i),G^(i)]$, whilst $G^i+1=[G^i,G]$.
– C Monsour
Sep 8 at 20:30
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $G^1 = [G, G]$ and $G^i = [G^i-1, G^i-1]$, then is it true that $G^i$ is normal in $G$?
My approach:
Since it is true for $i = 1$. Assume by induction it is true for $i leq n-1$. If $alpha in [G^i, G^i] = g^i+1$, then $alpha = aba^-1b^-1 $ where $a, b in G^i$. Therefore $$galpha g^-1 = gaba^-1b^-1g^-1 = gaba^-1(g^-1b^-1bg)b^-1g^-1 = big((ga)b(a^-1g^-1)b^-1big) cdotbig(bgb^-1g^-1big) in G^i+1.$$ Hence $G^i+1$ is normal.
Somehow I am finding this result powerful and yet I am unable to find any reference which talks abouts this, so I am guessing I am making some mistake. Can someone confirm this.
group-theory
asked Sep 8 at 12:41
henceproved
172
Let $G^1 = [G, G]$ and $G^i = [G^i-1, G^i-1]$, then is it true that $G^i$ is normal in $G$?
My approach:
Since it is true for $i = 1$. Assume by induction it is true for $i leq n-1$. If $alpha in [G^i, G^i] = g^i+1$, then $alpha = aba^-1b^-1 $ where $a, b in G^i$. Therefore $$galpha g^-1 = gaba^-1b^-1g^-1 = gaba^-1(g^-1b^-1bg)b^-1g^-1 = big((ga)b(a^-1g^-1)b^-1big) cdotbig(bgb^-1g^-1big) in G^i+1.$$ Hence $G^i+1$ is normal.
Somehow I am finding this result powerful and yet I am unable to find any reference which talks abouts this, so I am guessing I am making some mistake. Can someone confirm this.
group-theory
group-theory
asked Sep 8 at 12:41
henceproved
172
asked Sep 8 at 12:41
henceproved
172
asked Sep 8 at 12:41
henceproved
172
asked Sep 8 at 12:41
henceproved
172
asked Sep 8 at 12:41
henceproved
172
172
1
Yes it's normal, it's called the derived series and any introductory algebra book should talk about it.
– Steve D
Sep 8 at 12:53
1
The usual way to prove this is to show that $[G,G]$ is characteristic in $G$, and since being characteristic is a transitive property, it follows that $G^i$ is characteristic and hence normal.
– Derek Holt
Sep 8 at 13:10
Actually it's even better : $G^i$ is characteristic in $G$ and this follows by induction from the fact that $[G,G]$ is characteristic in $G$ and that characteristicity is transitive
– Max
Sep 8 at 19:59
The other way is to show that $[H,K]$ is normal in $G$ if both $H$ and $K$ are. To see this, note that the conjugate of a commutator of two elements is equal to the commutator of the conjugates.
– C Monsour
Sep 8 at 20:25
Also please note that you mean $G^(i)$ for the derived series, not $G^i$, which is from the lower central series. $G^(i+1)=[G^(i),G^(i)]$, whilst $G^i+1=[G^i,G]$.
– C Monsour
Sep 8 at 20:30
add a comment |Â
1
Yes it's normal, it's called the derived series and any introductory algebra book should talk about it.
– Steve D
Sep 8 at 12:53
1
The usual way to prove this is to show that $[G,G]$ is characteristic in $G$, and since being characteristic is a transitive property, it follows that $G^i$ is characteristic and hence normal.
– Derek Holt
Sep 8 at 13:10
Actually it's even better : $G^i$ is characteristic in $G$ and this follows by induction from the fact that $[G,G]$ is characteristic in $G$ and that characteristicity is transitive
– Max
Sep 8 at 19:59
The other way is to show that $[H,K]$ is normal in $G$ if both $H$ and $K$ are. To see this, note that the conjugate of a commutator of two elements is equal to the commutator of the conjugates.
– C Monsour
Sep 8 at 20:25
Also please note that you mean $G^(i)$ for the derived series, not $G^i$, which is from the lower central series. $G^(i+1)=[G^(i),G^(i)]$, whilst $G^i+1=[G^i,G]$.
– C Monsour
Sep 8 at 20:30
1
1
Yes it's normal, it's called the derived series and any introductory algebra book should talk about it.
– Steve D
Sep 8 at 12:53
Yes it's normal, it's called the derived series and any introductory algebra book should talk about it.
– Steve D
Sep 8 at 12:53
1
1
The usual way to prove this is to show that $[G,G]$ is characteristic in $G$, and since being characteristic is a transitive property, it follows that $G^i$ is characteristic and hence normal.
– Derek Holt
Sep 8 at 13:10
The usual way to prove this is to show that $[G,G]$ is characteristic in $G$, and since being characteristic is a transitive property, it follows that $G^i$ is characteristic and hence normal.
– Derek Holt
Sep 8 at 13:10
Actually it's even better : $G^i$ is characteristic in $G$ and this follows by induction from the fact that $[G,G]$ is characteristic in $G$ and that characteristicity is transitive
– Max
Sep 8 at 19:59
Actually it's even better : $G^i$ is characteristic in $G$ and this follows by induction from the fact that $[G,G]$ is characteristic in $G$ and that characteristicity is transitive
– Max
Sep 8 at 19:59
The other way is to show that $[H,K]$ is normal in $G$ if both $H$ and $K$ are. To see this, note that the conjugate of a commutator of two elements is equal to the commutator of the conjugates.
– C Monsour
Sep 8 at 20:25
The other way is to show that $[H,K]$ is normal in $G$ if both $H$ and $K$ are. To see this, note that the conjugate of a commutator of two elements is equal to the commutator of the conjugates.
– C Monsour
Sep 8 at 20:25
Also please note that you mean $G^(i)$ for the derived series, not $G^i$, which is from the lower central series. $G^(i+1)=[G^(i),G^(i)]$, whilst $G^i+1=[G^i,G]$.
– C Monsour
Sep 8 at 20:30
Also please note that you mean $G^(i)$ for the derived series, not $G^i$, which is from the lower central series. $G^(i+1)=[G^(i),G^(i)]$, whilst $G^i+1=[G^i,G]$.
– C Monsour
Sep 8 at 20:30
add a comment |Â
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1
Yes it's normal, it's called the derived series and any introductory algebra book should talk about it.
– Steve D
Sep 8 at 12:53
1
The usual way to prove this is to show that $[G,G]$ is characteristic in $G$, and since being characteristic is a transitive property, it follows that $G^i$ is characteristic and hence normal.
– Derek Holt
Sep 8 at 13:10
Actually it's even better : $G^i$ is characteristic in $G$ and this follows by induction from the fact that $[G,G]$ is characteristic in $G$ and that characteristicity is transitive
– Max
Sep 8 at 19:59
The other way is to show that $[H,K]$ is normal in $G$ if both $H$ and $K$ are. To see this, note that the conjugate of a commutator of two elements is equal to the commutator of the conjugates.
– C Monsour
Sep 8 at 20:25
Also please note that you mean $G^(i)$ for the derived series, not $G^i$, which is from the lower central series. $G^(i+1)=[G^(i),G^(i)]$, whilst $G^i+1=[G^i,G]$.
– C Monsour
Sep 8 at 20:30