Suppose $G$ has a lower central series and $H$ is a proper subgroup of $G$. Show an index $k$ such that...
Suppose $G$ has a lower central series and $H$ is a proper subgroup of $G$. Show that there is an index $k$ such that $G^{k+1} subset H$ but $G^k notsubset H$.
If $G$ has a lower central series then,
$$G=G^{0} triangleright G^{1} triangleright cdots triangleright G^{n}={e}$$
is a subnormal series where $G^i=[G^{i-1},G]=langle [x,y]=xyx^{-1}y^{-1} mid x in G^{i-1}, y in G rangle$
I'm not sure as to why should the claim be true. Any ideas?
abstract-algebra proof-explanation
asked Dec 12 '18 at 22:49
Al JebrAl Jebr
4,16243176
add a comment |
Suppose $G$ has a lower central series and $H$ is a proper subgroup of $G$. Show that there is an index $k$ such that $G^{k+1} subset H$ but $G^k notsubset H$.
If $G$ has a lower central series then,
$$G=G^{0} triangleright G^{1} triangleright cdots triangleright G^{n}={e}$$
is a subnormal series where $G^i=[G^{i-1},G]=langle [x,y]=xyx^{-1}y^{-1} mid x in G^{i-1}, y in G rangle$
I'm not sure as to why should the claim be true. Any ideas?
abstract-algebra proof-explanation
asked Dec 12 '18 at 22:49
Al JebrAl Jebr
4,16243176
add a comment |
Suppose $G$ has a lower central series and $H$ is a proper subgroup of $G$. Show that there is an index $k$ such that $G^{k+1} subset H$ but $G^k notsubset H$.
If $G$ has a lower central series then,
$$G=G^{0} triangleright G^{1} triangleright cdots triangleright G^{n}={e}$$
is a subnormal series where $G^i=[G^{i-1},G]=langle [x,y]=xyx^{-1}y^{-1} mid x in G^{i-1}, y in G rangle$
I'm not sure as to why should the claim be true. Any ideas?
abstract-algebra proof-explanation
asked Dec 12 '18 at 22:49
Al JebrAl Jebr
4,16243176
Suppose $G$ has a lower central series and $H$ is a proper subgroup of $G$. Show that there is an index $k$ such that $G^{k+1} subset H$ but $G^k notsubset H$.
If $G$ has a lower central series then,
$$G=G^{0} triangleright G^{1} triangleright cdots triangleright G^{n}={e}$$
is a subnormal series where $G^i=[G^{i-1},G]=langle [x,y]=xyx^{-1}y^{-1} mid x in G^{i-1}, y in G rangle$
I'm not sure as to why should the claim be true. Any ideas?
abstract-algebra proof-explanation
abstract-algebra proof-explanation
asked Dec 12 '18 at 22:49
Al JebrAl Jebr
4,16243176
asked Dec 12 '18 at 22:49
Al JebrAl Jebr
4,16243176
asked Dec 12 '18 at 22:49
Al JebrAl Jebr
4,16243176
asked Dec 12 '18 at 22:49
Al JebrAl Jebr
4,16243176
asked Dec 12 '18 at 22:49
Al JebrAl Jebr
4,16243176
4,16243176
add a comment |
add a comment |
1 Answer
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$G^n={e}subset H$. And $G^0=Gnotsubset H$, since $H$ is a proper subgroup. So let $k=operatorname{min}{jmid G^jsubset H}-1$.
answered Dec 12 '18 at 23:04
Chris CusterChris Custer
11k3824
I'm not seeing why such a $k$, besides $k=n$, is guaranteed to exist. How does the definition of lower central series allow such a $k$?
– Al Jebr
Dec 12 '18 at 23:12
@AlJebr I didn't actually use anything except that you have a finite set of nested sets starting at $e$ and going up to $G$. At some point the transition must occur (it seems to me).
– Chris Custer
Dec 12 '18 at 23:24
add a comment |
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1 Answer
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1 Answer
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$G^n={e}subset H$. And $G^0=Gnotsubset H$, since $H$ is a proper subgroup. So let $k=operatorname{min}{jmid G^jsubset H}-1$.
answered Dec 12 '18 at 23:04
Chris CusterChris Custer
11k3824
I'm not seeing why such a $k$, besides $k=n$, is guaranteed to exist. How does the definition of lower central series allow such a $k$?
– Al Jebr
Dec 12 '18 at 23:12
@AlJebr I didn't actually use anything except that you have a finite set of nested sets starting at $e$ and going up to $G$. At some point the transition must occur (it seems to me).
– Chris Custer
Dec 12 '18 at 23:24
add a comment |
$G^n={e}subset H$. And $G^0=Gnotsubset H$, since $H$ is a proper subgroup. So let $k=operatorname{min}{jmid G^jsubset H}-1$.
answered Dec 12 '18 at 23:04
Chris CusterChris Custer
11k3824
I'm not seeing why such a $k$, besides $k=n$, is guaranteed to exist. How does the definition of lower central series allow such a $k$?
– Al Jebr
Dec 12 '18 at 23:12
@AlJebr I didn't actually use anything except that you have a finite set of nested sets starting at $e$ and going up to $G$. At some point the transition must occur (it seems to me).
– Chris Custer
Dec 12 '18 at 23:24
add a comment |
$G^n={e}subset H$. And $G^0=Gnotsubset H$, since $H$ is a proper subgroup. So let $k=operatorname{min}{jmid G^jsubset H}-1$.
answered Dec 12 '18 at 23:04
Chris CusterChris Custer
11k3824
$G^n={e}subset H$. And $G^0=Gnotsubset H$, since $H$ is a proper subgroup. So let $k=operatorname{min}{jmid G^jsubset H}-1$.
answered Dec 12 '18 at 23:04
Chris CusterChris Custer
11k3824
answered Dec 12 '18 at 23:04
Chris CusterChris Custer
11k3824
answered Dec 12 '18 at 23:04
Chris CusterChris Custer
11k3824
answered Dec 12 '18 at 23:04
Chris CusterChris Custer
11k3824
11k3824
I'm not seeing why such a $k$, besides $k=n$, is guaranteed to exist. How does the definition of lower central series allow such a $k$?
– Al Jebr
Dec 12 '18 at 23:12
@AlJebr I didn't actually use anything except that you have a finite set of nested sets starting at $e$ and going up to $G$. At some point the transition must occur (it seems to me).
– Chris Custer
Dec 12 '18 at 23:24
add a comment |
I'm not seeing why such a $k$, besides $k=n$, is guaranteed to exist. How does the definition of lower central series allow such a $k$?
– Al Jebr
Dec 12 '18 at 23:12
@AlJebr I didn't actually use anything except that you have a finite set of nested sets starting at $e$ and going up to $G$. At some point the transition must occur (it seems to me).
– Chris Custer
Dec 12 '18 at 23:24
I'm not seeing why such a $k$, besides $k=n$, is guaranteed to exist. How does the definition of lower central series allow such a $k$?
– Al Jebr
Dec 12 '18 at 23:12
I'm not seeing why such a $k$, besides $k=n$, is guaranteed to exist. How does the definition of lower central series allow such a $k$?
– Al Jebr
Dec 12 '18 at 23:12
@AlJebr I didn't actually use anything except that you have a finite set of nested sets starting at $e$ and going up to $G$. At some point the transition must occur (it seems to me).
– Chris Custer
Dec 12 '18 at 23:24
@AlJebr I didn't actually use anything except that you have a finite set of nested sets starting at $e$ and going up to $G$. At some point the transition must occur (it seems to me).
– Chris Custer
Dec 12 '18 at 23:24
add a comment |
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