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
9,690113684
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
9,690113684
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
9,690113684
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
9,690113684
asked Dec 29 '18 at 20:32
PeterPeter
1106
edited Dec 30 '18 at 13:17
Shaun
9,690113684
asked Dec 29 '18 at 20:32
PeterPeter
1106
edited Dec 30 '18 at 13:17
Shaun
9,690113684
edited Dec 30 '18 at 13:17
Shaun
9,690113684
edited Dec 30 '18 at 13:17
Shaun
9,690113684
9,690113684
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
9,690113684
answered Dec 29 '18 at 20:42
C MonsourC Monsour
6,3141326
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3056229%2fprove-h-is-a-proper-normal-subgroup-of-g-if-h-is-generated-by-x-y-mi%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
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 |
$begingroup$
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
9,690113684
answered Dec 29 '18 at 20:42
C MonsourC Monsour
6,3141326
$endgroup$
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
9,690113684
answered Dec 29 '18 at 20:42
C MonsourC Monsour
6,3141326
$endgroup$
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
9,690113684
answered Dec 29 '18 at 20:42
C MonsourC Monsour
6,3141326
$endgroup$
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
9,690113684
answered Dec 29 '18 at 20:42
C MonsourC Monsour
6,3141326
edited Dec 30 '18 at 13:15
Shaun
9,690113684
edited Dec 30 '18 at 13:15
Shaun
9,690113684
edited Dec 30 '18 at 13:15
Shaun
9,690113684
9,690113684
answered Dec 29 '18 at 20:42
C MonsourC Monsour
6,3141326
answered Dec 29 '18 at 20:42
C MonsourC Monsour
6,3141326
answered Dec 29 '18 at 20:42
C MonsourC Monsour
6,3141326
6,3141326
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3056229%2fprove-h-is-a-proper-normal-subgroup-of-g-if-h-is-generated-by-x-y-mi%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
