Let $Z_i(G)$ be the terms of the upper central series of $G$. Let $H trianglelefteq G$. Show $Z_i(G)H/H...
$begingroup$
Let $Z_0(G) = { 1 }$ and:
$$Z_{i + 1}(G)/Z_i(G) = Z(G/Z_i(G))$$
(as defined in dummit and foote).
I want to show that $Z_i(G)H/H subseteq Z_i(G/H)$.
I can see it's true for $i = 0$ and $i = 1$, but I'm stuck in the inductive case because I have to work with quotients of quotients (and not in the form $(G/H)/(K/H)$ so no 3rd isomorphism theorem).
In other words, I have an element $zH in Z_{i + 1}(G)H/H$.
I want to show that it is in $Z_{i + 1}(G/H)$, which is defined recursively as
$$Z_{i + 1}(G/H)/Z_i(G/H) = Z((G/H)/Z_i(G/H)).$$
group-theory nilpotent-groups
edited Dec 7 '18 at 2:27
darij grinberg
10.5k33062
asked Dec 7 '18 at 2:25
RaekyeRaekye
24539
$endgroup$
add a comment |
$begingroup$
Let $Z_0(G) = { 1 }$ and:
$$Z_{i + 1}(G)/Z_i(G) = Z(G/Z_i(G))$$
(as defined in dummit and foote).
I want to show that $Z_i(G)H/H subseteq Z_i(G/H)$.
I can see it's true for $i = 0$ and $i = 1$, but I'm stuck in the inductive case because I have to work with quotients of quotients (and not in the form $(G/H)/(K/H)$ so no 3rd isomorphism theorem).
In other words, I have an element $zH in Z_{i + 1}(G)H/H$.
I want to show that it is in $Z_{i + 1}(G/H)$, which is defined recursively as
$$Z_{i + 1}(G/H)/Z_i(G/H) = Z((G/H)/Z_i(G/H)).$$
group-theory nilpotent-groups
edited Dec 7 '18 at 2:27
darij grinberg
10.5k33062
asked Dec 7 '18 at 2:25
RaekyeRaekye
24539
$endgroup$
$begingroup$
Rewrite the recursive definition as $Z_{i+1}left(Gright) = left{g in G mid ghg^{-1}h^{-1} in Z_ileft(Gright) text{ for all } h in G right}$. Do you now see why each $Z_i$ is functorial wrt surjective homomorphisms (i.e., each surjective group homomorphism $f : G to H$ sends $Z_ileft(Gright)$ into $Z_ileft(Hright)$ for each $i$)?
$endgroup$
– darij grinberg
Dec 7 '18 at 2:44
add a comment |
$begingroup$
Let $Z_0(G) = { 1 }$ and:
$$Z_{i + 1}(G)/Z_i(G) = Z(G/Z_i(G))$$
(as defined in dummit and foote).
I want to show that $Z_i(G)H/H subseteq Z_i(G/H)$.
I can see it's true for $i = 0$ and $i = 1$, but I'm stuck in the inductive case because I have to work with quotients of quotients (and not in the form $(G/H)/(K/H)$ so no 3rd isomorphism theorem).
In other words, I have an element $zH in Z_{i + 1}(G)H/H$.
I want to show that it is in $Z_{i + 1}(G/H)$, which is defined recursively as
$$Z_{i + 1}(G/H)/Z_i(G/H) = Z((G/H)/Z_i(G/H)).$$
group-theory nilpotent-groups
edited Dec 7 '18 at 2:27
darij grinberg
10.5k33062
asked Dec 7 '18 at 2:25
RaekyeRaekye
24539
$endgroup$
Let $Z_0(G) = { 1 }$ and:
$$Z_{i + 1}(G)/Z_i(G) = Z(G/Z_i(G))$$
(as defined in dummit and foote).
I want to show that $Z_i(G)H/H subseteq Z_i(G/H)$.
I can see it's true for $i = 0$ and $i = 1$, but I'm stuck in the inductive case because I have to work with quotients of quotients (and not in the form $(G/H)/(K/H)$ so no 3rd isomorphism theorem).
In other words, I have an element $zH in Z_{i + 1}(G)H/H$.
I want to show that it is in $Z_{i + 1}(G/H)$, which is defined recursively as
$$Z_{i + 1}(G/H)/Z_i(G/H) = Z((G/H)/Z_i(G/H)).$$
group-theory nilpotent-groups
group-theory nilpotent-groups
edited Dec 7 '18 at 2:27
darij grinberg
10.5k33062
asked Dec 7 '18 at 2:25
RaekyeRaekye
24539
edited Dec 7 '18 at 2:27
darij grinberg
10.5k33062
asked Dec 7 '18 at 2:25
RaekyeRaekye
24539
edited Dec 7 '18 at 2:27
darij grinberg
10.5k33062
edited Dec 7 '18 at 2:27
darij grinberg
10.5k33062
edited Dec 7 '18 at 2:27
darij grinberg
10.5k33062
10.5k33062
asked Dec 7 '18 at 2:25
RaekyeRaekye
24539
asked Dec 7 '18 at 2:25
RaekyeRaekye
24539
asked Dec 7 '18 at 2:25
RaekyeRaekye
24539
24539
$begingroup$
Rewrite the recursive definition as $Z_{i+1}left(Gright) = left{g in G mid ghg^{-1}h^{-1} in Z_ileft(Gright) text{ for all } h in G right}$. Do you now see why each $Z_i$ is functorial wrt surjective homomorphisms (i.e., each surjective group homomorphism $f : G to H$ sends $Z_ileft(Gright)$ into $Z_ileft(Hright)$ for each $i$)?
$endgroup$
– darij grinberg
Dec 7 '18 at 2:44
add a comment |
$begingroup$
Rewrite the recursive definition as $Z_{i+1}left(Gright) = left{g in G mid ghg^{-1}h^{-1} in Z_ileft(Gright) text{ for all } h in G right}$. Do you now see why each $Z_i$ is functorial wrt surjective homomorphisms (i.e., each surjective group homomorphism $f : G to H$ sends $Z_ileft(Gright)$ into $Z_ileft(Hright)$ for each $i$)?
$endgroup$
– darij grinberg
Dec 7 '18 at 2:44
$begingroup$
Rewrite the recursive definition as $Z_{i+1}left(Gright) = left{g in G mid ghg^{-1}h^{-1} in Z_ileft(Gright) text{ for all } h in G right}$. Do you now see why each $Z_i$ is functorial wrt surjective homomorphisms (i.e., each surjective group homomorphism $f : G to H$ sends $Z_ileft(Gright)$ into $Z_ileft(Hright)$ for each $i$)?
$endgroup$
– darij grinberg
Dec 7 '18 at 2:44
$begingroup$
Rewrite the recursive definition as $Z_{i+1}left(Gright) = left{g in G mid ghg^{-1}h^{-1} in Z_ileft(Gright) text{ for all } h in G right}$. Do you now see why each $Z_i$ is functorial wrt surjective homomorphisms (i.e., each surjective group homomorphism $f : G to H$ sends $Z_ileft(Gright)$ into $Z_ileft(Hright)$ for each $i$)?
$endgroup$
– darij grinberg
Dec 7 '18 at 2:44
add a comment |
0
active
oldest
votes
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%2f3029387%2flet-z-ig-be-the-terms-of-the-upper-central-series-of-g-let-h-trianglele%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
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%2f3029387%2flet-z-ig-be-the-terms-of-the-upper-central-series-of-g-let-h-trianglele%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
$begingroup$
Rewrite the recursive definition as $Z_{i+1}left(Gright) = left{g in G mid ghg^{-1}h^{-1} in Z_ileft(Gright) text{ for all } h in G right}$. Do you now see why each $Z_i$ is functorial wrt surjective homomorphisms (i.e., each surjective group homomorphism $f : G to H$ sends $Z_ileft(Gright)$ into $Z_ileft(Hright)$ for each $i$)?
$endgroup$
– darij grinberg
Dec 7 '18 at 2:44