Prove $H$ is a proper normal subgroup of $G$ if $H$ is generated by ${[x,y] mid x,y in G} cup {x^p mid x in...
$begingroup$
I am trying to solve the following problem:
Let $G$ be a nontrivial finite $p$-group, where $p$ is a prime, and let $H$ be the subgroup of $G$ generated by the set ${[x,y] mid x,y in G} cup {x^p mid x in G}$.
Prove that $H$ is a proper normal subgroup of $G$.
It is not hard to see that $H$ is a normal subgroup of $G$. Indeed, any element of $g in G$ commutes with any element of the commutator. Moreover, $gx^pg^{-1} = (gxg^{-1})^p$. Since any generator of $H$ commutes with any element of $G$, the same is true for $H$. However, I do not know how to prove $H neq G$.
Any thoughts?
abstract-algebra group-theory finite-groups normal-subgroups nilpotent-groups
edited Dec 30 '18 at 13:17
Shaun
9,759113684
asked Dec 29 '18 at 20:32
PeterPeter
1106
$endgroup$
add a comment |
$begingroup$
I am trying to solve the following problem:
Let $G$ be a nontrivial finite $p$-group, where $p$ is a prime, and let $H$ be the subgroup of $G$ generated by the set ${[x,y] mid x,y in G} cup {x^p mid x in G}$.
Prove that $H$ is a proper normal subgroup of $G$.
It is not hard to see that $H$ is a normal subgroup of $G$. Indeed, any element of $g in G$ commutes with any element of the commutator. Moreover, $gx^pg^{-1} = (gxg^{-1})^p$. Since any generator of $H$ commutes with any element of $G$, the same is true for $H$. However, I do not know how to prove $H neq G$.
Any thoughts?
abstract-algebra group-theory finite-groups normal-subgroups nilpotent-groups
edited Dec 30 '18 at 13:17
Shaun
9,759113684
asked Dec 29 '18 at 20:32
PeterPeter
1106
$endgroup$
add a comment |
$begingroup$
I am trying to solve the following problem:
Let $G$ be a nontrivial finite $p$-group, where $p$ is a prime, and let $H$ be the subgroup of $G$ generated by the set ${[x,y] mid x,y in G} cup {x^p mid x in G}$.
Prove that $H$ is a proper normal subgroup of $G$.
It is not hard to see that $H$ is a normal subgroup of $G$. Indeed, any element of $g in G$ commutes with any element of the commutator. Moreover, $gx^pg^{-1} = (gxg^{-1})^p$. Since any generator of $H$ commutes with any element of $G$, the same is true for $H$. However, I do not know how to prove $H neq G$.
Any thoughts?
abstract-algebra group-theory finite-groups normal-subgroups nilpotent-groups
edited Dec 30 '18 at 13:17
Shaun
9,759113684
asked Dec 29 '18 at 20:32
PeterPeter
1106
$endgroup$
I am trying to solve the following problem:
Let $G$ be a nontrivial finite $p$-group, where $p$ is a prime, and let $H$ be the subgroup of $G$ generated by the set ${[x,y] mid x,y in G} cup {x^p mid x in G}$.
Prove that $H$ is a proper normal subgroup of $G$.
It is not hard to see that $H$ is a normal subgroup of $G$. Indeed, any element of $g in G$ commutes with any element of the commutator. Moreover, $gx^pg^{-1} = (gxg^{-1})^p$. Since any generator of $H$ commutes with any element of $G$, the same is true for $H$. However, I do not know how to prove $H neq G$.
Any thoughts?
abstract-algebra group-theory finite-groups normal-subgroups nilpotent-groups
abstract-algebra group-theory finite-groups normal-subgroups nilpotent-groups
edited Dec 30 '18 at 13:17
Shaun
9,759113684
asked Dec 29 '18 at 20:32
PeterPeter
1106
edited Dec 30 '18 at 13:17
Shaun
9,759113684
asked Dec 29 '18 at 20:32
PeterPeter
1106
edited Dec 30 '18 at 13:17
Shaun
9,759113684
edited Dec 30 '18 at 13:17
Shaun
9,759113684
edited Dec 30 '18 at 13:17
Shaun
9,759113684
9,759113684
asked Dec 29 '18 at 20:32
PeterPeter
1106
asked Dec 29 '18 at 20:32
PeterPeter
1106
asked Dec 29 '18 at 20:32
PeterPeter
1106
1106
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Note first that $[G,G]$ is a proper subgroup of $G$ and that $H/[G,G]$ is just the subgroup of $G/[G,G]$ generated by $p$th powers (why?). So, we may replace $G$ with $G/[G,G]$ and thus assume $G$ is abelian. But if $G$ is abelian then $H$ is just the set of $p$th powers, and so any element of $G$ of maximal order will not be in $H$.
answered Dec 29 '18 at 20:37
Eric WofseyEric Wofsey
190k14216348
$endgroup$
$begingroup$
We can say $H/[G,G]$ is the subgroup of $G/[G,G]$ generated by $p$th powers because $p$ divides $G/[G,G]$; so, Cauchy's theorem ensures there is an element of order $p$ in $G/[G,G]$. Am I right @EricWofsey?
$endgroup$
– Peter
Dec 30 '18 at 0:01
2
$begingroup$
@Peter: No, that's totally wrong. Why would it be relevant whether there is an element of order $p$? To prove the claim about $H/[G,G]$, you'll need to use the definition of $H$.
$endgroup$
– Eric Wofsey
Dec 30 '18 at 0:02
$begingroup$
you are totally right!, but I don't see why $H neq [G,G]$. This is also irrelevant as I was asked to prove $H$ is proper; so, it does not matter whether $H neq [G,G]$ or not but I am just curious.
$endgroup$
– Peter
Dec 30 '18 at 0:06
2
$begingroup$
It is entirely possible that $H=[G,G]$; I never said it wasn't.
$endgroup$
– Eric Wofsey
Dec 30 '18 at 0:07
add a comment |
$begingroup$
It's not true that any generator of $H$ commutes with any element of $G$; rather, an element of $G$ conjugates any generator of $H$ to another generator of $H$.
As for why $H<G$, note that $G'le H$. Now in $G/G'$, which is abelian, the subgroup generated by $p$th powers on one hand is proper. In fact, it's just the $p$th powers and so excludes all elements of maximal order. On the other hand, it's clearly equal to $H/G'$. Thus, since $H/G'<G/G'$ we have $H<G$.
edited Dec 30 '18 at 13:15
Shaun
9,759113684
answered Dec 29 '18 at 20:42
C MonsourC Monsour
6,3391326
$endgroup$
add a comment |
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2 Answers
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active
oldest
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2 Answers
2
active
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votes
$begingroup$
Note first that $[G,G]$ is a proper subgroup of $G$ and that $H/[G,G]$ is just the subgroup of $G/[G,G]$ generated by $p$th powers (why?). So, we may replace $G$ with $G/[G,G]$ and thus assume $G$ is abelian. But if $G$ is abelian then $H$ is just the set of $p$th powers, and so any element of $G$ of maximal order will not be in $H$.
answered Dec 29 '18 at 20:37
Eric WofseyEric Wofsey
190k14216348
$endgroup$
$begingroup$
We can say $H/[G,G]$ is the subgroup of $G/[G,G]$ generated by $p$th powers because $p$ divides $G/[G,G]$; so, Cauchy's theorem ensures there is an element of order $p$ in $G/[G,G]$. Am I right @EricWofsey?
$endgroup$
– Peter
Dec 30 '18 at 0:01
2
$begingroup$
@Peter: No, that's totally wrong. Why would it be relevant whether there is an element of order $p$? To prove the claim about $H/[G,G]$, you'll need to use the definition of $H$.
$endgroup$
– Eric Wofsey
Dec 30 '18 at 0:02
$begingroup$
you are totally right!, but I don't see why $H neq [G,G]$. This is also irrelevant as I was asked to prove $H$ is proper; so, it does not matter whether $H neq [G,G]$ or not but I am just curious.
$endgroup$
– Peter
Dec 30 '18 at 0:06
2
$begingroup$
It is entirely possible that $H=[G,G]$; I never said it wasn't.
$endgroup$
– Eric Wofsey
Dec 30 '18 at 0:07
add a comment |
$begingroup$
Note first that $[G,G]$ is a proper subgroup of $G$ and that $H/[G,G]$ is just the subgroup of $G/[G,G]$ generated by $p$th powers (why?). So, we may replace $G$ with $G/[G,G]$ and thus assume $G$ is abelian. But if $G$ is abelian then $H$ is just the set of $p$th powers, and so any element of $G$ of maximal order will not be in $H$.
answered Dec 29 '18 at 20:37
Eric WofseyEric Wofsey
190k14216348
$endgroup$
$begingroup$
We can say $H/[G,G]$ is the subgroup of $G/[G,G]$ generated by $p$th powers because $p$ divides $G/[G,G]$; so, Cauchy's theorem ensures there is an element of order $p$ in $G/[G,G]$. Am I right @EricWofsey?
$endgroup$
– Peter
Dec 30 '18 at 0:01
2
$begingroup$
@Peter: No, that's totally wrong. Why would it be relevant whether there is an element of order $p$? To prove the claim about $H/[G,G]$, you'll need to use the definition of $H$.
$endgroup$
– Eric Wofsey
Dec 30 '18 at 0:02
$begingroup$
you are totally right!, but I don't see why $H neq [G,G]$. This is also irrelevant as I was asked to prove $H$ is proper; so, it does not matter whether $H neq [G,G]$ or not but I am just curious.
$endgroup$
– Peter
Dec 30 '18 at 0:06
2
$begingroup$
It is entirely possible that $H=[G,G]$; I never said it wasn't.
$endgroup$
– Eric Wofsey
Dec 30 '18 at 0:07
add a comment |
$begingroup$
Note first that $[G,G]$ is a proper subgroup of $G$ and that $H/[G,G]$ is just the subgroup of $G/[G,G]$ generated by $p$th powers (why?). So, we may replace $G$ with $G/[G,G]$ and thus assume $G$ is abelian. But if $G$ is abelian then $H$ is just the set of $p$th powers, and so any element of $G$ of maximal order will not be in $H$.
answered Dec 29 '18 at 20:37
Eric WofseyEric Wofsey
190k14216348
$endgroup$
Note first that $[G,G]$ is a proper subgroup of $G$ and that $H/[G,G]$ is just the subgroup of $G/[G,G]$ generated by $p$th powers (why?). So, we may replace $G$ with $G/[G,G]$ and thus assume $G$ is abelian. But if $G$ is abelian then $H$ is just the set of $p$th powers, and so any element of $G$ of maximal order will not be in $H$.
answered Dec 29 '18 at 20:37
Eric WofseyEric Wofsey
190k14216348
answered Dec 29 '18 at 20:37
Eric WofseyEric Wofsey
190k14216348
answered Dec 29 '18 at 20:37
Eric WofseyEric Wofsey
190k14216348
answered Dec 29 '18 at 20:37
Eric WofseyEric Wofsey
190k14216348
190k14216348
$begingroup$
We can say $H/[G,G]$ is the subgroup of $G/[G,G]$ generated by $p$th powers because $p$ divides $G/[G,G]$; so, Cauchy's theorem ensures there is an element of order $p$ in $G/[G,G]$. Am I right @EricWofsey?
$endgroup$
– Peter
Dec 30 '18 at 0:01
2
$begingroup$
@Peter: No, that's totally wrong. Why would it be relevant whether there is an element of order $p$? To prove the claim about $H/[G,G]$, you'll need to use the definition of $H$.
$endgroup$
– Eric Wofsey
Dec 30 '18 at 0:02
$begingroup$
you are totally right!, but I don't see why $H neq [G,G]$. This is also irrelevant as I was asked to prove $H$ is proper; so, it does not matter whether $H neq [G,G]$ or not but I am just curious.
$endgroup$
– Peter
Dec 30 '18 at 0:06
2
$begingroup$
It is entirely possible that $H=[G,G]$; I never said it wasn't.
$endgroup$
– Eric Wofsey
Dec 30 '18 at 0:07
add a comment |
$begingroup$
We can say $H/[G,G]$ is the subgroup of $G/[G,G]$ generated by $p$th powers because $p$ divides $G/[G,G]$; so, Cauchy's theorem ensures there is an element of order $p$ in $G/[G,G]$. Am I right @EricWofsey?
$endgroup$
– Peter
Dec 30 '18 at 0:01
2
$begingroup$
@Peter: No, that's totally wrong. Why would it be relevant whether there is an element of order $p$? To prove the claim about $H/[G,G]$, you'll need to use the definition of $H$.
$endgroup$
– Eric Wofsey
Dec 30 '18 at 0:02
$begingroup$
you are totally right!, but I don't see why $H neq [G,G]$. This is also irrelevant as I was asked to prove $H$ is proper; so, it does not matter whether $H neq [G,G]$ or not but I am just curious.
$endgroup$
– Peter
Dec 30 '18 at 0:06
2
$begingroup$
It is entirely possible that $H=[G,G]$; I never said it wasn't.
$endgroup$
– Eric Wofsey
Dec 30 '18 at 0:07
$begingroup$
We can say $H/[G,G]$ is the subgroup of $G/[G,G]$ generated by $p$th powers because $p$ divides $G/[G,G]$; so, Cauchy's theorem ensures there is an element of order $p$ in $G/[G,G]$. Am I right @EricWofsey?
$endgroup$
– Peter
Dec 30 '18 at 0:01
$begingroup$
We can say $H/[G,G]$ is the subgroup of $G/[G,G]$ generated by $p$th powers because $p$ divides $G/[G,G]$; so, Cauchy's theorem ensures there is an element of order $p$ in $G/[G,G]$. Am I right @EricWofsey?
$endgroup$
– Peter
Dec 30 '18 at 0:01
2
2
$begingroup$
@Peter: No, that's totally wrong. Why would it be relevant whether there is an element of order $p$? To prove the claim about $H/[G,G]$, you'll need to use the definition of $H$.
$endgroup$
– Eric Wofsey
Dec 30 '18 at 0:02
$begingroup$
@Peter: No, that's totally wrong. Why would it be relevant whether there is an element of order $p$? To prove the claim about $H/[G,G]$, you'll need to use the definition of $H$.
$endgroup$
– Eric Wofsey
Dec 30 '18 at 0:02
$begingroup$
you are totally right!, but I don't see why $H neq [G,G]$. This is also irrelevant as I was asked to prove $H$ is proper; so, it does not matter whether $H neq [G,G]$ or not but I am just curious.
$endgroup$
– Peter
Dec 30 '18 at 0:06
$begingroup$
you are totally right!, but I don't see why $H neq [G,G]$. This is also irrelevant as I was asked to prove $H$ is proper; so, it does not matter whether $H neq [G,G]$ or not but I am just curious.
$endgroup$
– Peter
Dec 30 '18 at 0:06
2
2
$begingroup$
It is entirely possible that $H=[G,G]$; I never said it wasn't.
$endgroup$
– Eric Wofsey
Dec 30 '18 at 0:07
$begingroup$
It is entirely possible that $H=[G,G]$; I never said it wasn't.
$endgroup$
– Eric Wofsey
Dec 30 '18 at 0:07
add a comment |
$begingroup$
It's not true that any generator of $H$ commutes with any element of $G$; rather, an element of $G$ conjugates any generator of $H$ to another generator of $H$.
As for why $H<G$, note that $G'le H$. Now in $G/G'$, which is abelian, the subgroup generated by $p$th powers on one hand is proper. In fact, it's just the $p$th powers and so excludes all elements of maximal order. On the other hand, it's clearly equal to $H/G'$. Thus, since $H/G'<G/G'$ we have $H<G$.
edited Dec 30 '18 at 13:15
Shaun
9,759113684
answered Dec 29 '18 at 20:42
C MonsourC Monsour
6,3391326
$endgroup$
add a comment |
$begingroup$
It's not true that any generator of $H$ commutes with any element of $G$; rather, an element of $G$ conjugates any generator of $H$ to another generator of $H$.
As for why $H<G$, note that $G'le H$. Now in $G/G'$, which is abelian, the subgroup generated by $p$th powers on one hand is proper. In fact, it's just the $p$th powers and so excludes all elements of maximal order. On the other hand, it's clearly equal to $H/G'$. Thus, since $H/G'<G/G'$ we have $H<G$.
edited Dec 30 '18 at 13:15
Shaun
9,759113684
answered Dec 29 '18 at 20:42
C MonsourC Monsour
6,3391326
$endgroup$
add a comment |
$begingroup$
It's not true that any generator of $H$ commutes with any element of $G$; rather, an element of $G$ conjugates any generator of $H$ to another generator of $H$.
As for why $H<G$, note that $G'le H$. Now in $G/G'$, which is abelian, the subgroup generated by $p$th powers on one hand is proper. In fact, it's just the $p$th powers and so excludes all elements of maximal order. On the other hand, it's clearly equal to $H/G'$. Thus, since $H/G'<G/G'$ we have $H<G$.
edited Dec 30 '18 at 13:15
Shaun
9,759113684
answered Dec 29 '18 at 20:42
C MonsourC Monsour
6,3391326
$endgroup$
It's not true that any generator of $H$ commutes with any element of $G$; rather, an element of $G$ conjugates any generator of $H$ to another generator of $H$.
As for why $H<G$, note that $G'le H$. Now in $G/G'$, which is abelian, the subgroup generated by $p$th powers on one hand is proper. In fact, it's just the $p$th powers and so excludes all elements of maximal order. On the other hand, it's clearly equal to $H/G'$. Thus, since $H/G'<G/G'$ we have $H<G$.
edited Dec 30 '18 at 13:15
Shaun
9,759113684
answered Dec 29 '18 at 20:42
C MonsourC Monsour
6,3391326
edited Dec 30 '18 at 13:15
Shaun
9,759113684
edited Dec 30 '18 at 13:15
Shaun
9,759113684
edited Dec 30 '18 at 13:15
Shaun
9,759113684
9,759113684
answered Dec 29 '18 at 20:42
C MonsourC Monsour
6,3391326
answered Dec 29 '18 at 20:42
C MonsourC Monsour
6,3391326
answered Dec 29 '18 at 20:42
C MonsourC Monsour
6,3391326
6,3391326
add a comment |
add a comment |
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