For every group of odd order, is $F(omega(G))$ abelian?
$begingroup$
Here, $F(G)$ stands for the Fitting subgroup of a group, and $omega(G) = cap N_G(H)$, where $H$ are all the subnormal subgroups of $G$, is the Wielandt subgroup.
Since we're talking about groups of odd order, by the Feit-Thompson theorem $G$ is solvable, this must mean $omega(G)$ is solvable and has an abelian series $$1=G_0 lhd G_1 lhd cdots lhd G_n=omega(G),$$
where each $G_{i+1}/G_i$ is abelian.
We also have $F(omega(G))= cap C_{omega(G)}(G_{i+1}/G_{i})$. I get stuck here and can't go further. How can I finish the proof?
And is there an alternative method to show $F(omega(G))$ is abelian because I don't think it ought to be necessary to use such a strong result (Fiet-Thompson)?
group-theory finite-groups solvable-groups
$endgroup$
add a comment |
$begingroup$
Here, $F(G)$ stands for the Fitting subgroup of a group, and $omega(G) = cap N_G(H)$, where $H$ are all the subnormal subgroups of $G$, is the Wielandt subgroup.
Since we're talking about groups of odd order, by the Feit-Thompson theorem $G$ is solvable, this must mean $omega(G)$ is solvable and has an abelian series $$1=G_0 lhd G_1 lhd cdots lhd G_n=omega(G),$$
where each $G_{i+1}/G_i$ is abelian.
We also have $F(omega(G))= cap C_{omega(G)}(G_{i+1}/G_{i})$. I get stuck here and can't go further. How can I finish the proof?
And is there an alternative method to show $F(omega(G))$ is abelian because I don't think it ought to be necessary to use such a strong result (Fiet-Thompson)?
group-theory finite-groups solvable-groups
$endgroup$
$begingroup$
I am really not sure what you are asking. I have no idea what you should do next. Is there an alternate (alternative?) approach to what?
$endgroup$
– Derek Holt
Jan 8 '17 at 7:13
$begingroup$
@DerekHolt Sorry, I will edit the question.
$endgroup$
– Maria-0
Jan 8 '17 at 7:27
$begingroup$
But why do you think that $F(omega(G)$ is abelian when $|G|$ is odd. Have you been asked to prove it as an exercise? If not, what is the evidence for it being true?
$endgroup$
– Derek Holt
Jan 8 '17 at 11:19
$begingroup$
@DerekHolt It was mentioned in class. This isn't homework that will add to the grade, but our prof. did wanted us to check if it was true or not.
$endgroup$
– Maria-0
Jan 8 '17 at 11:28
add a comment |
$begingroup$
Here, $F(G)$ stands for the Fitting subgroup of a group, and $omega(G) = cap N_G(H)$, where $H$ are all the subnormal subgroups of $G$, is the Wielandt subgroup.
Since we're talking about groups of odd order, by the Feit-Thompson theorem $G$ is solvable, this must mean $omega(G)$ is solvable and has an abelian series $$1=G_0 lhd G_1 lhd cdots lhd G_n=omega(G),$$
where each $G_{i+1}/G_i$ is abelian.
We also have $F(omega(G))= cap C_{omega(G)}(G_{i+1}/G_{i})$. I get stuck here and can't go further. How can I finish the proof?
And is there an alternative method to show $F(omega(G))$ is abelian because I don't think it ought to be necessary to use such a strong result (Fiet-Thompson)?
group-theory finite-groups solvable-groups
$endgroup$
Here, $F(G)$ stands for the Fitting subgroup of a group, and $omega(G) = cap N_G(H)$, where $H$ are all the subnormal subgroups of $G$, is the Wielandt subgroup.
Since we're talking about groups of odd order, by the Feit-Thompson theorem $G$ is solvable, this must mean $omega(G)$ is solvable and has an abelian series $$1=G_0 lhd G_1 lhd cdots lhd G_n=omega(G),$$
where each $G_{i+1}/G_i$ is abelian.
We also have $F(omega(G))= cap C_{omega(G)}(G_{i+1}/G_{i})$. I get stuck here and can't go further. How can I finish the proof?
And is there an alternative method to show $F(omega(G))$ is abelian because I don't think it ought to be necessary to use such a strong result (Fiet-Thompson)?
group-theory finite-groups solvable-groups
group-theory finite-groups solvable-groups
edited Dec 28 '18 at 3:24
the_fox
2,90031538
2,90031538
asked Jan 8 '17 at 6:38
Maria-0Maria-0
224
224
$begingroup$
I am really not sure what you are asking. I have no idea what you should do next. Is there an alternate (alternative?) approach to what?
$endgroup$
– Derek Holt
Jan 8 '17 at 7:13
$begingroup$
@DerekHolt Sorry, I will edit the question.
$endgroup$
– Maria-0
Jan 8 '17 at 7:27
$begingroup$
But why do you think that $F(omega(G)$ is abelian when $|G|$ is odd. Have you been asked to prove it as an exercise? If not, what is the evidence for it being true?
$endgroup$
– Derek Holt
Jan 8 '17 at 11:19
$begingroup$
@DerekHolt It was mentioned in class. This isn't homework that will add to the grade, but our prof. did wanted us to check if it was true or not.
$endgroup$
– Maria-0
Jan 8 '17 at 11:28
add a comment |
$begingroup$
I am really not sure what you are asking. I have no idea what you should do next. Is there an alternate (alternative?) approach to what?
$endgroup$
– Derek Holt
Jan 8 '17 at 7:13
$begingroup$
@DerekHolt Sorry, I will edit the question.
$endgroup$
– Maria-0
Jan 8 '17 at 7:27
$begingroup$
But why do you think that $F(omega(G)$ is abelian when $|G|$ is odd. Have you been asked to prove it as an exercise? If not, what is the evidence for it being true?
$endgroup$
– Derek Holt
Jan 8 '17 at 11:19
$begingroup$
@DerekHolt It was mentioned in class. This isn't homework that will add to the grade, but our prof. did wanted us to check if it was true or not.
$endgroup$
– Maria-0
Jan 8 '17 at 11:28
$begingroup$
I am really not sure what you are asking. I have no idea what you should do next. Is there an alternate (alternative?) approach to what?
$endgroup$
– Derek Holt
Jan 8 '17 at 7:13
$begingroup$
I am really not sure what you are asking. I have no idea what you should do next. Is there an alternate (alternative?) approach to what?
$endgroup$
– Derek Holt
Jan 8 '17 at 7:13
$begingroup$
@DerekHolt Sorry, I will edit the question.
$endgroup$
– Maria-0
Jan 8 '17 at 7:27
$begingroup$
@DerekHolt Sorry, I will edit the question.
$endgroup$
– Maria-0
Jan 8 '17 at 7:27
$begingroup$
But why do you think that $F(omega(G)$ is abelian when $|G|$ is odd. Have you been asked to prove it as an exercise? If not, what is the evidence for it being true?
$endgroup$
– Derek Holt
Jan 8 '17 at 11:19
$begingroup$
But why do you think that $F(omega(G)$ is abelian when $|G|$ is odd. Have you been asked to prove it as an exercise? If not, what is the evidence for it being true?
$endgroup$
– Derek Holt
Jan 8 '17 at 11:19
$begingroup$
@DerekHolt It was mentioned in class. This isn't homework that will add to the grade, but our prof. did wanted us to check if it was true or not.
$endgroup$
– Maria-0
Jan 8 '17 at 11:28
$begingroup$
@DerekHolt It was mentioned in class. This isn't homework that will add to the grade, but our prof. did wanted us to check if it was true or not.
$endgroup$
– Maria-0
Jan 8 '17 at 11:28
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
All subgroups of $F(omega(G))$ are subnormal in $G$, and hence all such subgroups are normal in $F(omega(G))$. So $F(omega(G))$ is a Dedekind group. The structure of these groups is fully understood, and in particular all Dedekind groups of odd order are abelian. Note that this argument does not make any use of the Feit-Thompson theorem.
$endgroup$
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%2f2088429%2ffor-every-group-of-odd-order-is-f-omegag-abelian%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
All subgroups of $F(omega(G))$ are subnormal in $G$, and hence all such subgroups are normal in $F(omega(G))$. So $F(omega(G))$ is a Dedekind group. The structure of these groups is fully understood, and in particular all Dedekind groups of odd order are abelian. Note that this argument does not make any use of the Feit-Thompson theorem.
$endgroup$
add a comment |
$begingroup$
All subgroups of $F(omega(G))$ are subnormal in $G$, and hence all such subgroups are normal in $F(omega(G))$. So $F(omega(G))$ is a Dedekind group. The structure of these groups is fully understood, and in particular all Dedekind groups of odd order are abelian. Note that this argument does not make any use of the Feit-Thompson theorem.
$endgroup$
add a comment |
$begingroup$
All subgroups of $F(omega(G))$ are subnormal in $G$, and hence all such subgroups are normal in $F(omega(G))$. So $F(omega(G))$ is a Dedekind group. The structure of these groups is fully understood, and in particular all Dedekind groups of odd order are abelian. Note that this argument does not make any use of the Feit-Thompson theorem.
$endgroup$
All subgroups of $F(omega(G))$ are subnormal in $G$, and hence all such subgroups are normal in $F(omega(G))$. So $F(omega(G))$ is a Dedekind group. The structure of these groups is fully understood, and in particular all Dedekind groups of odd order are abelian. Note that this argument does not make any use of the Feit-Thompson theorem.
answered Jan 8 '17 at 21:12
Derek HoltDerek Holt
54.3k53573
54.3k53573
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%2f2088429%2ffor-every-group-of-odd-order-is-f-omegag-abelian%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
$begingroup$
I am really not sure what you are asking. I have no idea what you should do next. Is there an alternate (alternative?) approach to what?
$endgroup$
– Derek Holt
Jan 8 '17 at 7:13
$begingroup$
@DerekHolt Sorry, I will edit the question.
$endgroup$
– Maria-0
Jan 8 '17 at 7:27
$begingroup$
But why do you think that $F(omega(G)$ is abelian when $|G|$ is odd. Have you been asked to prove it as an exercise? If not, what is the evidence for it being true?
$endgroup$
– Derek Holt
Jan 8 '17 at 11:19
$begingroup$
@DerekHolt It was mentioned in class. This isn't homework that will add to the grade, but our prof. did wanted us to check if it was true or not.
$endgroup$
– Maria-0
Jan 8 '17 at 11:28