Mixing
Confusion in the definition of direct product of finite groups
$begingroup$
Let $G$ be a finite group. We will say that $G=A times B times C$ if
- A,B,C are normal in $G$
- $Acap B cap C ={e}$
- $|G|=|A||B||C|$
Is the first condition ok? or should I say $A times B$ is normal in $G$ and $C$ is normal in $G$.
direct-sum
asked Jan 24 at 6:11
I_wil_break_wallI_wil_break_wall
805
$endgroup$
add a comment |
$begingroup$
Let $G$ be a finite group. We will say that $G=A times B times C$ if
- A,B,C are normal in $G$
- $Acap B cap C ={e}$
- $|G|=|A||B||C|$
Is the first condition ok? or should I say $A times B$ is normal in $G$ and $C$ is normal in $G$.
direct-sum
asked Jan 24 at 6:11
I_wil_break_wallI_wil_break_wall
805
$endgroup$
add a comment |
$begingroup$
Let $G$ be a finite group. We will say that $G=A times B times C$ if
- A,B,C are normal in $G$
- $Acap B cap C ={e}$
- $|G|=|A||B||C|$
Is the first condition ok? or should I say $A times B$ is normal in $G$ and $C$ is normal in $G$.
direct-sum
asked Jan 24 at 6:11
I_wil_break_wallI_wil_break_wall
805
$endgroup$
Let $G$ be a finite group. We will say that $G=A times B times C$ if
- A,B,C are normal in $G$
- $Acap B cap C ={e}$
- $|G|=|A||B||C|$
Is the first condition ok? or should I say $A times B$ is normal in $G$ and $C$ is normal in $G$.
direct-sum
direct-sum
asked Jan 24 at 6:11
I_wil_break_wallI_wil_break_wall
805
asked Jan 24 at 6:11
I_wil_break_wallI_wil_break_wall
805
asked Jan 24 at 6:11
I_wil_break_wallI_wil_break_wall
805
asked Jan 24 at 6:11
I_wil_break_wallI_wil_break_wall
805
asked Jan 24 at 6:11
I_wil_break_wallI_wil_break_wall
805
805
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
That is satisfied if $G$ is an elementary Abelian group of order $8$,
$A={e,a}$, $B={e,b}$ and $C={e,ab}$ where $a$ and $b$ are distinct elements
of order two. But in this case the natural homomorphism from $Atimes Btimes C$
to $G$ is not an isomorphism.
In this case the subgroups "$Atimes B$", "$Atimes C$" and "$Btimes C$"
are all normal in $G$ too.
answered Jan 24 at 6:19
Lord Shark the UnknownLord Shark the Unknown
106k1161133
$endgroup$
$begingroup$
I did not get your answer. In general, Is first condition fine?
$endgroup$
– I_wil_break_wall
Jan 24 at 6:25
$begingroup$
In an Abelian group all subgroups are normal. @I_wil_break_wall
$endgroup$
– Lord Shark the Unknown
Jan 24 at 6:26
$begingroup$
My question is what are the conditions for $G$ can be written as a direct product of three subgroups.
$endgroup$
– I_wil_break_wall
Jan 24 at 6:27
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%2f3085537%2fconfusion-in-the-definition-of-direct-product-of-finite-groups%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
$begingroup$
That is satisfied if $G$ is an elementary Abelian group of order $8$,
$A={e,a}$, $B={e,b}$ and $C={e,ab}$ where $a$ and $b$ are distinct elements
of order two. But in this case the natural homomorphism from $Atimes Btimes C$
to $G$ is not an isomorphism.
In this case the subgroups "$Atimes B$", "$Atimes C$" and "$Btimes C$"
are all normal in $G$ too.
answered Jan 24 at 6:19
Lord Shark the UnknownLord Shark the Unknown
106k1161133
$endgroup$
$begingroup$
I did not get your answer. In general, Is first condition fine?
$endgroup$
– I_wil_break_wall
Jan 24 at 6:25
$begingroup$
In an Abelian group all subgroups are normal. @I_wil_break_wall
$endgroup$
– Lord Shark the Unknown
Jan 24 at 6:26
$begingroup$
My question is what are the conditions for $G$ can be written as a direct product of three subgroups.
$endgroup$
– I_wil_break_wall
Jan 24 at 6:27
add a comment |
$begingroup$
That is satisfied if $G$ is an elementary Abelian group of order $8$,
$A={e,a}$, $B={e,b}$ and $C={e,ab}$ where $a$ and $b$ are distinct elements
of order two. But in this case the natural homomorphism from $Atimes Btimes C$
to $G$ is not an isomorphism.
In this case the subgroups "$Atimes B$", "$Atimes C$" and "$Btimes C$"
are all normal in $G$ too.
answered Jan 24 at 6:19
Lord Shark the UnknownLord Shark the Unknown
106k1161133
$endgroup$
$begingroup$
I did not get your answer. In general, Is first condition fine?
$endgroup$
– I_wil_break_wall
Jan 24 at 6:25
$begingroup$
In an Abelian group all subgroups are normal. @I_wil_break_wall
$endgroup$
– Lord Shark the Unknown
Jan 24 at 6:26
$begingroup$
My question is what are the conditions for $G$ can be written as a direct product of three subgroups.
$endgroup$
– I_wil_break_wall
Jan 24 at 6:27
add a comment |
$begingroup$
That is satisfied if $G$ is an elementary Abelian group of order $8$,
$A={e,a}$, $B={e,b}$ and $C={e,ab}$ where $a$ and $b$ are distinct elements
of order two. But in this case the natural homomorphism from $Atimes Btimes C$
to $G$ is not an isomorphism.
In this case the subgroups "$Atimes B$", "$Atimes C$" and "$Btimes C$"
are all normal in $G$ too.
answered Jan 24 at 6:19
Lord Shark the UnknownLord Shark the Unknown
106k1161133
$endgroup$
That is satisfied if $G$ is an elementary Abelian group of order $8$,
$A={e,a}$, $B={e,b}$ and $C={e,ab}$ where $a$ and $b$ are distinct elements
of order two. But in this case the natural homomorphism from $Atimes Btimes C$
to $G$ is not an isomorphism.
In this case the subgroups "$Atimes B$", "$Atimes C$" and "$Btimes C$"
are all normal in $G$ too.
answered Jan 24 at 6:19
Lord Shark the UnknownLord Shark the Unknown
106k1161133
answered Jan 24 at 6:19
Lord Shark the UnknownLord Shark the Unknown
106k1161133
answered Jan 24 at 6:19
Lord Shark the UnknownLord Shark the Unknown
106k1161133
answered Jan 24 at 6:19
Lord Shark the UnknownLord Shark the Unknown
106k1161133
106k1161133
$begingroup$
I did not get your answer. In general, Is first condition fine?
$endgroup$
– I_wil_break_wall
Jan 24 at 6:25
$begingroup$
In an Abelian group all subgroups are normal. @I_wil_break_wall
$endgroup$
– Lord Shark the Unknown
Jan 24 at 6:26
$begingroup$
My question is what are the conditions for $G$ can be written as a direct product of three subgroups.
$endgroup$
– I_wil_break_wall
Jan 24 at 6:27
add a comment |
$begingroup$
I did not get your answer. In general, Is first condition fine?
$endgroup$
– I_wil_break_wall
Jan 24 at 6:25
$begingroup$
In an Abelian group all subgroups are normal. @I_wil_break_wall
$endgroup$
– Lord Shark the Unknown
Jan 24 at 6:26
$begingroup$
My question is what are the conditions for $G$ can be written as a direct product of three subgroups.
$endgroup$
– I_wil_break_wall
Jan 24 at 6:27
$begingroup$
I did not get your answer. In general, Is first condition fine?
$endgroup$
– I_wil_break_wall
Jan 24 at 6:25
$begingroup$
I did not get your answer. In general, Is first condition fine?
$endgroup$
– I_wil_break_wall
Jan 24 at 6:25
$begingroup$
In an Abelian group all subgroups are normal. @I_wil_break_wall
$endgroup$
– Lord Shark the Unknown
Jan 24 at 6:26
$begingroup$
In an Abelian group all subgroups are normal. @I_wil_break_wall
$endgroup$
– Lord Shark the Unknown
Jan 24 at 6:26
$begingroup$
My question is what are the conditions for $G$ can be written as a direct product of three subgroups.
$endgroup$
– I_wil_break_wall
Jan 24 at 6:27
$begingroup$
My question is what are the conditions for $G$ can be written as a direct product of three subgroups.
$endgroup$
– I_wil_break_wall
Jan 24 at 6:27
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%2f3085537%2fconfusion-in-the-definition-of-direct-product-of-finite-groups%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
