Generators of $1+Delta (G)$, where $Delta(G)$ is augmentation ideal of group ring $FG.$
$begingroup$
Let $FG$ be a finite group ring of a finite non abelian $p$-group $G$ over finite field $F.$ It is well known that augmentation ideal $Delta(G)=J(FG)$ has basis as the set ${g-1:gin G, gne 1}$, being kernel of the augmentation map $f:FGrightarrow F.$ Now my question is what is a generating set of $1+Delta(G)?$ Can i say that its generating set is the set ${gin G:gne 1}?$
My real question is that as i proved that $1+Delta(G)$ is a finite non-abelian group. Can i say that exponent of this group will not exceed exponent of the group $G?$ Is there any way to find cardinality of $1+Delta(G)?$ Please help me . Thanks .
abstract-algebra finite-groups group-rings
asked Dec 7 '18 at 5:35
neelkanthneelkanth
2,0912928
$endgroup$
|
show 10 more comments
$begingroup$
Let $FG$ be a finite group ring of a finite non abelian $p$-group $G$ over finite field $F.$ It is well known that augmentation ideal $Delta(G)=J(FG)$ has basis as the set ${g-1:gin G, gne 1}$, being kernel of the augmentation map $f:FGrightarrow F.$ Now my question is what is a generating set of $1+Delta(G)?$ Can i say that its generating set is the set ${gin G:gne 1}?$
My real question is that as i proved that $1+Delta(G)$ is a finite non-abelian group. Can i say that exponent of this group will not exceed exponent of the group $G?$ Is there any way to find cardinality of $1+Delta(G)?$ Please help me . Thanks .
abstract-algebra finite-groups group-rings
asked Dec 7 '18 at 5:35
neelkanthneelkanth
2,0912928
$endgroup$
$begingroup$
the generating set? I think you must just mean a generating set. The set you gave clearly does not generate it: every finite product of elements and their inverses in $Gsetminus{1}$ is again an element of $G$, so you will never be able to generate $1+g_1-g_2 =1+(g_1-1)-(g_2-1)in 1+Delta(G)$ . It can't generate any more than $G$, which is only a small portion of $1+Delta(G)$. Are you under the misapprehension $1+Delta(G)$ is closed under more operations than $cdot$?
$endgroup$
– rschwieb
Dec 7 '18 at 15:08
$begingroup$
I've never really heard a proof that $1+Delta(G)$ is a group. It's obvious when $Delta(G)$ is nilpotent... but otherwise what do you do? I know that the search for units in group rings is a deep subject, so if this is a standard result I'd appreciate a pointer.
$endgroup$
– rschwieb
Dec 7 '18 at 15:14
$begingroup$
@rschwieb $1+J[GF]$ is a normal subgroup of unit group $U(FG)$ of group ring $RG$
$endgroup$
– neelkanth
Dec 7 '18 at 16:23
$begingroup$
and in case of $p$-group Jacobson radical is same as Augmentation ideal i.e. $Delta(G).$
$endgroup$
– neelkanth
Dec 7 '18 at 16:26
1
$begingroup$
I will edit the question
$endgroup$
– neelkanth
Dec 7 '18 at 16:44
|
show 10 more comments
$begingroup$
Let $FG$ be a finite group ring of a finite non abelian $p$-group $G$ over finite field $F.$ It is well known that augmentation ideal $Delta(G)=J(FG)$ has basis as the set ${g-1:gin G, gne 1}$, being kernel of the augmentation map $f:FGrightarrow F.$ Now my question is what is a generating set of $1+Delta(G)?$ Can i say that its generating set is the set ${gin G:gne 1}?$
My real question is that as i proved that $1+Delta(G)$ is a finite non-abelian group. Can i say that exponent of this group will not exceed exponent of the group $G?$ Is there any way to find cardinality of $1+Delta(G)?$ Please help me . Thanks .
abstract-algebra finite-groups group-rings
asked Dec 7 '18 at 5:35
neelkanthneelkanth
2,0912928
$endgroup$
Let $FG$ be a finite group ring of a finite non abelian $p$-group $G$ over finite field $F.$ It is well known that augmentation ideal $Delta(G)=J(FG)$ has basis as the set ${g-1:gin G, gne 1}$, being kernel of the augmentation map $f:FGrightarrow F.$ Now my question is what is a generating set of $1+Delta(G)?$ Can i say that its generating set is the set ${gin G:gne 1}?$
My real question is that as i proved that $1+Delta(G)$ is a finite non-abelian group. Can i say that exponent of this group will not exceed exponent of the group $G?$ Is there any way to find cardinality of $1+Delta(G)?$ Please help me . Thanks .
abstract-algebra finite-groups group-rings
abstract-algebra finite-groups group-rings
asked Dec 7 '18 at 5:35
neelkanthneelkanth
2,0912928
asked Dec 7 '18 at 5:35
neelkanthneelkanth
2,0912928
edited Dec 7 '18 at 17:30
neelkanth
asked Dec 7 '18 at 5:35
neelkanthneelkanth
2,0912928
asked Dec 7 '18 at 5:35
neelkanthneelkanth
2,0912928
asked Dec 7 '18 at 5:35
neelkanthneelkanth
2,0912928
2,0912928
$begingroup$
the generating set? I think you must just mean a generating set. The set you gave clearly does not generate it: every finite product of elements and their inverses in $Gsetminus{1}$ is again an element of $G$, so you will never be able to generate $1+g_1-g_2 =1+(g_1-1)-(g_2-1)in 1+Delta(G)$ . It can't generate any more than $G$, which is only a small portion of $1+Delta(G)$. Are you under the misapprehension $1+Delta(G)$ is closed under more operations than $cdot$?
$endgroup$
– rschwieb
Dec 7 '18 at 15:08
$begingroup$
I've never really heard a proof that $1+Delta(G)$ is a group. It's obvious when $Delta(G)$ is nilpotent... but otherwise what do you do? I know that the search for units in group rings is a deep subject, so if this is a standard result I'd appreciate a pointer.
$endgroup$
– rschwieb
Dec 7 '18 at 15:14
$begingroup$
@rschwieb $1+J[GF]$ is a normal subgroup of unit group $U(FG)$ of group ring $RG$
$endgroup$
– neelkanth
Dec 7 '18 at 16:23
$begingroup$
and in case of $p$-group Jacobson radical is same as Augmentation ideal i.e. $Delta(G).$
$endgroup$
– neelkanth
Dec 7 '18 at 16:26
1
$begingroup$
I will edit the question
$endgroup$
– neelkanth
Dec 7 '18 at 16:44
|
show 10 more comments
$begingroup$
the generating set? I think you must just mean a generating set. The set you gave clearly does not generate it: every finite product of elements and their inverses in $Gsetminus{1}$ is again an element of $G$, so you will never be able to generate $1+g_1-g_2 =1+(g_1-1)-(g_2-1)in 1+Delta(G)$ . It can't generate any more than $G$, which is only a small portion of $1+Delta(G)$. Are you under the misapprehension $1+Delta(G)$ is closed under more operations than $cdot$?
$endgroup$
– rschwieb
Dec 7 '18 at 15:08
$begingroup$
I've never really heard a proof that $1+Delta(G)$ is a group. It's obvious when $Delta(G)$ is nilpotent... but otherwise what do you do? I know that the search for units in group rings is a deep subject, so if this is a standard result I'd appreciate a pointer.
$endgroup$
– rschwieb
Dec 7 '18 at 15:14
$begingroup$
@rschwieb $1+J[GF]$ is a normal subgroup of unit group $U(FG)$ of group ring $RG$
$endgroup$
– neelkanth
Dec 7 '18 at 16:23
$begingroup$
and in case of $p$-group Jacobson radical is same as Augmentation ideal i.e. $Delta(G).$
$endgroup$
– neelkanth
Dec 7 '18 at 16:26
1
$begingroup$
I will edit the question
$endgroup$
– neelkanth
Dec 7 '18 at 16:44
$begingroup$
the generating set? I think you must just mean a generating set. The set you gave clearly does not generate it: every finite product of elements and their inverses in $Gsetminus{1}$ is again an element of $G$, so you will never be able to generate $1+g_1-g_2 =1+(g_1-1)-(g_2-1)in 1+Delta(G)$ . It can't generate any more than $G$, which is only a small portion of $1+Delta(G)$. Are you under the misapprehension $1+Delta(G)$ is closed under more operations than $cdot$?
$endgroup$
– rschwieb
Dec 7 '18 at 15:08
$begingroup$
the generating set? I think you must just mean a generating set. The set you gave clearly does not generate it: every finite product of elements and their inverses in $Gsetminus{1}$ is again an element of $G$, so you will never be able to generate $1+g_1-g_2 =1+(g_1-1)-(g_2-1)in 1+Delta(G)$ . It can't generate any more than $G$, which is only a small portion of $1+Delta(G)$. Are you under the misapprehension $1+Delta(G)$ is closed under more operations than $cdot$?
$endgroup$
– rschwieb
Dec 7 '18 at 15:08
$begingroup$
I've never really heard a proof that $1+Delta(G)$ is a group. It's obvious when $Delta(G)$ is nilpotent... but otherwise what do you do? I know that the search for units in group rings is a deep subject, so if this is a standard result I'd appreciate a pointer.
$endgroup$
– rschwieb
Dec 7 '18 at 15:14
$begingroup$
I've never really heard a proof that $1+Delta(G)$ is a group. It's obvious when $Delta(G)$ is nilpotent... but otherwise what do you do? I know that the search for units in group rings is a deep subject, so if this is a standard result I'd appreciate a pointer.
$endgroup$
– rschwieb
Dec 7 '18 at 15:14
$begingroup$
@rschwieb $1+J[GF]$ is a normal subgroup of unit group $U(FG)$ of group ring $RG$
$endgroup$
– neelkanth
Dec 7 '18 at 16:23
$begingroup$
@rschwieb $1+J[GF]$ is a normal subgroup of unit group $U(FG)$ of group ring $RG$
$endgroup$
– neelkanth
Dec 7 '18 at 16:23
$begingroup$
and in case of $p$-group Jacobson radical is same as Augmentation ideal i.e. $Delta(G).$
$endgroup$
– neelkanth
Dec 7 '18 at 16:26
$begingroup$
and in case of $p$-group Jacobson radical is same as Augmentation ideal i.e. $Delta(G).$
$endgroup$
– neelkanth
Dec 7 '18 at 16:26
1
1
$begingroup$
I will edit the question
$endgroup$
– neelkanth
Dec 7 '18 at 16:44
$begingroup$
I will edit the question
$endgroup$
– neelkanth
Dec 7 '18 at 16:44
|
show 10 more comments
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%2f3029518%2fgenerators-of-1-delta-g-where-deltag-is-augmentation-ideal-of-group-r%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%2f3029518%2fgenerators-of-1-delta-g-where-deltag-is-augmentation-ideal-of-group-r%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
$begingroup$
the generating set? I think you must just mean a generating set. The set you gave clearly does not generate it: every finite product of elements and their inverses in $Gsetminus{1}$ is again an element of $G$, so you will never be able to generate $1+g_1-g_2 =1+(g_1-1)-(g_2-1)in 1+Delta(G)$ . It can't generate any more than $G$, which is only a small portion of $1+Delta(G)$. Are you under the misapprehension $1+Delta(G)$ is closed under more operations than $cdot$?
$endgroup$
– rschwieb
Dec 7 '18 at 15:08
$begingroup$
I've never really heard a proof that $1+Delta(G)$ is a group. It's obvious when $Delta(G)$ is nilpotent... but otherwise what do you do? I know that the search for units in group rings is a deep subject, so if this is a standard result I'd appreciate a pointer.
$endgroup$
– rschwieb
Dec 7 '18 at 15:14
$begingroup$
@rschwieb $1+J[GF]$ is a normal subgroup of unit group $U(FG)$ of group ring $RG$
$endgroup$
– neelkanth
Dec 7 '18 at 16:23
$begingroup$
and in case of $p$-group Jacobson radical is same as Augmentation ideal i.e. $Delta(G).$
$endgroup$
– neelkanth
Dec 7 '18 at 16:26
1
$begingroup$
I will edit the question
$endgroup$
– neelkanth
Dec 7 '18 at 16:44