Mixing
Determining if H is a subgroup of G
$begingroup$
I would like to know if I have gone off track on this problem. Given
$$G = langle mathbb{R}^*, * rangle$$ (set $mathbb{R}$ excluding zero, and operation of multiplication), and
$$H = { 2^n3^m : m, n in mathbb{Z} }$$
determine if H is a subgroup of G.
I know that, for H to be a subgroup, I must prove that
(i) for $$a, b in H, ab in H$$
(ii) for $$a in H, a^{-1} in H$$
For (i)
If I understand correctly, any a looks like $$a = 2^n3^n$$ So we could also say any b looks like $$b = 2^u3^v, u, v in mathbb{Z}$$ Then
$$ab = 2^n3^m2^u3^v$$
$$ab= 2^{n+u}3^{m+v}$$
and since $$ n + u, m + v in mathbb{Z} $$
The element ab must be in H.
For (ii)
$$a^{-1} = (2^n3^m)^{-1}$$
$$a^{-1} = 2^{-n}3^{-m}$$
and since $$ -n, -m in mathbb{Z} forall n, m in mathbb{Z} $$
There is an inverse for every element a.
So (i) and (ii) are satisfied, and H is a subgroup of G.
Math is not a strong subject for me, so I am concerned I might be misunderstanding basic points of this proof.
abstract-algebra group-theory
asked Jan 26 at 3:54
W. HongoW. Hongo
254
$endgroup$
add a comment |
$begingroup$
I would like to know if I have gone off track on this problem. Given
$$G = langle mathbb{R}^*, * rangle$$ (set $mathbb{R}$ excluding zero, and operation of multiplication), and
$$H = { 2^n3^m : m, n in mathbb{Z} }$$
determine if H is a subgroup of G.
I know that, for H to be a subgroup, I must prove that
(i) for $$a, b in H, ab in H$$
(ii) for $$a in H, a^{-1} in H$$
For (i)
If I understand correctly, any a looks like $$a = 2^n3^n$$ So we could also say any b looks like $$b = 2^u3^v, u, v in mathbb{Z}$$ Then
$$ab = 2^n3^m2^u3^v$$
$$ab= 2^{n+u}3^{m+v}$$
and since $$ n + u, m + v in mathbb{Z} $$
The element ab must be in H.
For (ii)
$$a^{-1} = (2^n3^m)^{-1}$$
$$a^{-1} = 2^{-n}3^{-m}$$
and since $$ -n, -m in mathbb{Z} forall n, m in mathbb{Z} $$
There is an inverse for every element a.
So (i) and (ii) are satisfied, and H is a subgroup of G.
Math is not a strong subject for me, so I am concerned I might be misunderstanding basic points of this proof.
abstract-algebra group-theory
asked Jan 26 at 3:54
W. HongoW. Hongo
254
$endgroup$
add a comment |
$begingroup$
I would like to know if I have gone off track on this problem. Given
$$G = langle mathbb{R}^*, * rangle$$ (set $mathbb{R}$ excluding zero, and operation of multiplication), and
$$H = { 2^n3^m : m, n in mathbb{Z} }$$
determine if H is a subgroup of G.
I know that, for H to be a subgroup, I must prove that
(i) for $$a, b in H, ab in H$$
(ii) for $$a in H, a^{-1} in H$$
For (i)
If I understand correctly, any a looks like $$a = 2^n3^n$$ So we could also say any b looks like $$b = 2^u3^v, u, v in mathbb{Z}$$ Then
$$ab = 2^n3^m2^u3^v$$
$$ab= 2^{n+u}3^{m+v}$$
and since $$ n + u, m + v in mathbb{Z} $$
The element ab must be in H.
For (ii)
$$a^{-1} = (2^n3^m)^{-1}$$
$$a^{-1} = 2^{-n}3^{-m}$$
and since $$ -n, -m in mathbb{Z} forall n, m in mathbb{Z} $$
There is an inverse for every element a.
So (i) and (ii) are satisfied, and H is a subgroup of G.
Math is not a strong subject for me, so I am concerned I might be misunderstanding basic points of this proof.
abstract-algebra group-theory
asked Jan 26 at 3:54
W. HongoW. Hongo
254
$endgroup$
I would like to know if I have gone off track on this problem. Given
$$G = langle mathbb{R}^*, * rangle$$ (set $mathbb{R}$ excluding zero, and operation of multiplication), and
$$H = { 2^n3^m : m, n in mathbb{Z} }$$
determine if H is a subgroup of G.
I know that, for H to be a subgroup, I must prove that
(i) for $$a, b in H, ab in H$$
(ii) for $$a in H, a^{-1} in H$$
For (i)
If I understand correctly, any a looks like $$a = 2^n3^n$$ So we could also say any b looks like $$b = 2^u3^v, u, v in mathbb{Z}$$ Then
$$ab = 2^n3^m2^u3^v$$
$$ab= 2^{n+u}3^{m+v}$$
and since $$ n + u, m + v in mathbb{Z} $$
The element ab must be in H.
For (ii)
$$a^{-1} = (2^n3^m)^{-1}$$
$$a^{-1} = 2^{-n}3^{-m}$$
and since $$ -n, -m in mathbb{Z} forall n, m in mathbb{Z} $$
There is an inverse for every element a.
So (i) and (ii) are satisfied, and H is a subgroup of G.
Math is not a strong subject for me, so I am concerned I might be misunderstanding basic points of this proof.
abstract-algebra group-theory
abstract-algebra group-theory
asked Jan 26 at 3:54
W. HongoW. Hongo
254
asked Jan 26 at 3:54
W. HongoW. Hongo
254
asked Jan 26 at 3:54
W. HongoW. Hongo
254
asked Jan 26 at 3:54
W. HongoW. Hongo
254
asked Jan 26 at 3:54
W. HongoW. Hongo
254
254
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Your proof is correct and everything is fine, as long as you show that $H$ is non-empty (because, groups are not empty: they always contain the identity).It is easy to show that $1 in H$, for example.
The point is, both statements $i,ii$ above are statements that already assume that something is in $H$, and then want you to show that something else is in $H$. Ifthere was nothing in $H$ in the first place, then those statements would be vacuously true, although $H$ does not contain the identity. Therefore, as an extra condition, you should also check that $e in H$, the identity.
Also, sometimes the condition for being a subgroup is stated as :
i' : For every $a,b in H$, we have $ab^{-1} in H$. Note that this captures both i and ii in your question.
ii' : $e in H$.
You can check that $H$ is a subgroup using just these conditions.
answered Jan 26 at 4:07
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
39.8k33477
$endgroup$
$begingroup$
I understand, thank you for your explanation.
$endgroup$
– W. Hongo
Jan 26 at 4:15
$begingroup$
You are welcome!
$endgroup$
– астон вілла олоф мэллбэрг
Jan 26 at 4:17
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%2f3087881%2fdetermining-if-h-is-a-subgroup-of-g%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
$begingroup$
Your proof is correct and everything is fine, as long as you show that $H$ is non-empty (because, groups are not empty: they always contain the identity).It is easy to show that $1 in H$, for example.
The point is, both statements $i,ii$ above are statements that already assume that something is in $H$, and then want you to show that something else is in $H$. Ifthere was nothing in $H$ in the first place, then those statements would be vacuously true, although $H$ does not contain the identity. Therefore, as an extra condition, you should also check that $e in H$, the identity.
Also, sometimes the condition for being a subgroup is stated as :
i' : For every $a,b in H$, we have $ab^{-1} in H$. Note that this captures both i and ii in your question.
ii' : $e in H$.
You can check that $H$ is a subgroup using just these conditions.
answered Jan 26 at 4:07
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
39.8k33477
$endgroup$
$begingroup$
I understand, thank you for your explanation.
$endgroup$
– W. Hongo
Jan 26 at 4:15
$begingroup$
You are welcome!
$endgroup$
– астон вілла олоф мэллбэрг
Jan 26 at 4:17
add a comment |
$begingroup$
Your proof is correct and everything is fine, as long as you show that $H$ is non-empty (because, groups are not empty: they always contain the identity).It is easy to show that $1 in H$, for example.
The point is, both statements $i,ii$ above are statements that already assume that something is in $H$, and then want you to show that something else is in $H$. Ifthere was nothing in $H$ in the first place, then those statements would be vacuously true, although $H$ does not contain the identity. Therefore, as an extra condition, you should also check that $e in H$, the identity.
Also, sometimes the condition for being a subgroup is stated as :
i' : For every $a,b in H$, we have $ab^{-1} in H$. Note that this captures both i and ii in your question.
ii' : $e in H$.
You can check that $H$ is a subgroup using just these conditions.
answered Jan 26 at 4:07
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
39.8k33477
$endgroup$
$begingroup$
I understand, thank you for your explanation.
$endgroup$
– W. Hongo
Jan 26 at 4:15
$begingroup$
You are welcome!
$endgroup$
– астон вілла олоф мэллбэрг
Jan 26 at 4:17
add a comment |
$begingroup$
Your proof is correct and everything is fine, as long as you show that $H$ is non-empty (because, groups are not empty: they always contain the identity).It is easy to show that $1 in H$, for example.
The point is, both statements $i,ii$ above are statements that already assume that something is in $H$, and then want you to show that something else is in $H$. Ifthere was nothing in $H$ in the first place, then those statements would be vacuously true, although $H$ does not contain the identity. Therefore, as an extra condition, you should also check that $e in H$, the identity.
Also, sometimes the condition for being a subgroup is stated as :
i' : For every $a,b in H$, we have $ab^{-1} in H$. Note that this captures both i and ii in your question.
ii' : $e in H$.
You can check that $H$ is a subgroup using just these conditions.
answered Jan 26 at 4:07
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
39.8k33477
$endgroup$
Your proof is correct and everything is fine, as long as you show that $H$ is non-empty (because, groups are not empty: they always contain the identity).It is easy to show that $1 in H$, for example.
The point is, both statements $i,ii$ above are statements that already assume that something is in $H$, and then want you to show that something else is in $H$. Ifthere was nothing in $H$ in the first place, then those statements would be vacuously true, although $H$ does not contain the identity. Therefore, as an extra condition, you should also check that $e in H$, the identity.
Also, sometimes the condition for being a subgroup is stated as :
i' : For every $a,b in H$, we have $ab^{-1} in H$. Note that this captures both i and ii in your question.
ii' : $e in H$.
You can check that $H$ is a subgroup using just these conditions.
answered Jan 26 at 4:07
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
39.8k33477
answered Jan 26 at 4:07
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
39.8k33477
answered Jan 26 at 4:07
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
39.8k33477
answered Jan 26 at 4:07
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
39.8k33477
39.8k33477
$begingroup$
I understand, thank you for your explanation.
$endgroup$
– W. Hongo
Jan 26 at 4:15
$begingroup$
You are welcome!
$endgroup$
– астон вілла олоф мэллбэрг
Jan 26 at 4:17
add a comment |
$begingroup$
I understand, thank you for your explanation.
$endgroup$
– W. Hongo
Jan 26 at 4:15
$begingroup$
You are welcome!
$endgroup$
– астон вілла олоф мэллбэрг
Jan 26 at 4:17
$begingroup$
I understand, thank you for your explanation.
$endgroup$
– W. Hongo
Jan 26 at 4:15
$begingroup$
I understand, thank you for your explanation.
$endgroup$
– W. Hongo
Jan 26 at 4:15
$begingroup$
You are welcome!
$endgroup$
– астон вілла олоф мэллбэрг
Jan 26 at 4:17
$begingroup$
You are welcome!
$endgroup$
– астон вілла олоф мэллбэрг
Jan 26 at 4:17
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%2f3087881%2fdetermining-if-h-is-a-subgroup-of-g%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
