Suppose $G$ is a commutative group, and $H_1,H_2$ are subgroups, then $H_1H_2$ is a subgroup.
$begingroup$
Suppose $G$ is a commutative group, and $H_1,H_2$ are subgroups, then I believe that $H_1H_2$ is a subgroup, but I'm not sure so I try to prove it:
$H_1H_2 := {h_1h_2 : h_i in H_i}$. So if $h in H_1H_2$ then $h = h_1h_2$ for $h_i in H_i$.
I checked that it contains the identity, is closed under multiplication and each element has an inverse. I believe $H_1H_2$ being a group strongly depends upon the commutativity of $G$ is this correct?
group-theory
asked Dec 6 '18 at 6:56
trynalearntrynalearn
662314
$endgroup$
add a comment |
$begingroup$
Suppose $G$ is a commutative group, and $H_1,H_2$ are subgroups, then I believe that $H_1H_2$ is a subgroup, but I'm not sure so I try to prove it:
$H_1H_2 := {h_1h_2 : h_i in H_i}$. So if $h in H_1H_2$ then $h = h_1h_2$ for $h_i in H_i$.
I checked that it contains the identity, is closed under multiplication and each element has an inverse. I believe $H_1H_2$ being a group strongly depends upon the commutativity of $G$ is this correct?
group-theory
asked Dec 6 '18 at 6:56
trynalearntrynalearn
662314
$endgroup$
$begingroup$
That's correct. Write down the proof for closed under multiplication. There commutativity is used! (It's also used for the inverse elements)
$endgroup$
– Paul K
Dec 6 '18 at 7:02
1
$begingroup$
No, it doesn't depend strongly on commutativity of $G$. Any group $G$, subgroups $H_1,H_2$ with $H_2leqslant N_G(H_1)$ would work.
$endgroup$
– user10354138
Dec 6 '18 at 7:02
add a comment |
$begingroup$
Suppose $G$ is a commutative group, and $H_1,H_2$ are subgroups, then I believe that $H_1H_2$ is a subgroup, but I'm not sure so I try to prove it:
$H_1H_2 := {h_1h_2 : h_i in H_i}$. So if $h in H_1H_2$ then $h = h_1h_2$ for $h_i in H_i$.
I checked that it contains the identity, is closed under multiplication and each element has an inverse. I believe $H_1H_2$ being a group strongly depends upon the commutativity of $G$ is this correct?
group-theory
asked Dec 6 '18 at 6:56
trynalearntrynalearn
662314
$endgroup$
Suppose $G$ is a commutative group, and $H_1,H_2$ are subgroups, then I believe that $H_1H_2$ is a subgroup, but I'm not sure so I try to prove it:
$H_1H_2 := {h_1h_2 : h_i in H_i}$. So if $h in H_1H_2$ then $h = h_1h_2$ for $h_i in H_i$.
I checked that it contains the identity, is closed under multiplication and each element has an inverse. I believe $H_1H_2$ being a group strongly depends upon the commutativity of $G$ is this correct?
group-theory
group-theory
asked Dec 6 '18 at 6:56
trynalearntrynalearn
662314
asked Dec 6 '18 at 6:56
trynalearntrynalearn
662314
asked Dec 6 '18 at 6:56
trynalearntrynalearn
662314
asked Dec 6 '18 at 6:56
trynalearntrynalearn
662314
asked Dec 6 '18 at 6:56
trynalearntrynalearn
662314
662314
$begingroup$
That's correct. Write down the proof for closed under multiplication. There commutativity is used! (It's also used for the inverse elements)
$endgroup$
– Paul K
Dec 6 '18 at 7:02
1
$begingroup$
No, it doesn't depend strongly on commutativity of $G$. Any group $G$, subgroups $H_1,H_2$ with $H_2leqslant N_G(H_1)$ would work.
$endgroup$
– user10354138
Dec 6 '18 at 7:02
add a comment |
$begingroup$
That's correct. Write down the proof for closed under multiplication. There commutativity is used! (It's also used for the inverse elements)
$endgroup$
– Paul K
Dec 6 '18 at 7:02
1
$begingroup$
No, it doesn't depend strongly on commutativity of $G$. Any group $G$, subgroups $H_1,H_2$ with $H_2leqslant N_G(H_1)$ would work.
$endgroup$
– user10354138
Dec 6 '18 at 7:02
$begingroup$
That's correct. Write down the proof for closed under multiplication. There commutativity is used! (It's also used for the inverse elements)
$endgroup$
– Paul K
Dec 6 '18 at 7:02
$begingroup$
That's correct. Write down the proof for closed under multiplication. There commutativity is used! (It's also used for the inverse elements)
$endgroup$
– Paul K
Dec 6 '18 at 7:02
1
1
$begingroup$
No, it doesn't depend strongly on commutativity of $G$. Any group $G$, subgroups $H_1,H_2$ with $H_2leqslant N_G(H_1)$ would work.
$endgroup$
– user10354138
Dec 6 '18 at 7:02
$begingroup$
No, it doesn't depend strongly on commutativity of $G$. Any group $G$, subgroups $H_1,H_2$ with $H_2leqslant N_G(H_1)$ would work.
$endgroup$
– user10354138
Dec 6 '18 at 7:02
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%2f3028151%2fsuppose-g-is-a-commutative-group-and-h-1-h-2-are-subgroups-then-h-1h-2-i%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%2f3028151%2fsuppose-g-is-a-commutative-group-and-h-1-h-2-are-subgroups-then-h-1h-2-i%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
$begingroup$
That's correct. Write down the proof for closed under multiplication. There commutativity is used! (It's also used for the inverse elements)
$endgroup$
– Paul K
Dec 6 '18 at 7:02
1
$begingroup$
No, it doesn't depend strongly on commutativity of $G$. Any group $G$, subgroups $H_1,H_2$ with $H_2leqslant N_G(H_1)$ would work.
$endgroup$
– user10354138
Dec 6 '18 at 7:02