Question about Isomorphism of Aut(G)
$begingroup$
In material supplied by my instructor there is a question which asks to pick the incorrect statement among the following:
If $text{Aut}(G_1)cong text{Aut}(G_2)$ and $G_1$ is infinite group then $G_2$ is also infinite
If $text{Aut}(G_1) cong text{Aut}(G_2)$ and $G_1$ is finite group then $G_2$ is also finite
If $G_1$ not isomorphic to $G_2$ then Aut($G_1$) not isomorphic to Aut($G_2$).
$G_1$ and $G_2$ are two groups, Aut(G) is group of their automorphisms and "$cong$" means isomorphism.
I know all three above are incorrect as Aut($Bbb Z_3$)=$U_3$
(that is, $Bbb Z_2$) and I also found this statement on groupprops: "Of the three endomorphisms, two are automorphisms: the identity map and the square map. These form a cyclic group of order two: the square map, applied twice, gives the identity map" and Aut($Bbb Z$) is isomorphic to $Bbb Z_2$
so
$G_1=Bbb Z$ and $G_2=Bbb Z_3$
$G_2=Bbb Z$ and $G_1=Bbb Z_3$
- We can easily see $Bbb Z$ and $Bbb Z_3$ are not isomorphic but Aut($Bbb Z$) and Aut($Bbb Z_3$) are.
But the given answer is that (1.) is only incorrect statement.
Is there any problem in my reasoning?
abstract-algebra automorphism-group
asked Dec 22 '18 at 7:22
onlymathonlymath
749
$endgroup$
add a comment |
$begingroup$
In material supplied by my instructor there is a question which asks to pick the incorrect statement among the following:
If $text{Aut}(G_1)cong text{Aut}(G_2)$ and $G_1$ is infinite group then $G_2$ is also infinite
If $text{Aut}(G_1) cong text{Aut}(G_2)$ and $G_1$ is finite group then $G_2$ is also finite
If $G_1$ not isomorphic to $G_2$ then Aut($G_1$) not isomorphic to Aut($G_2$).
$G_1$ and $G_2$ are two groups, Aut(G) is group of their automorphisms and "$cong$" means isomorphism.
I know all three above are incorrect as Aut($Bbb Z_3$)=$U_3$
(that is, $Bbb Z_2$) and I also found this statement on groupprops: "Of the three endomorphisms, two are automorphisms: the identity map and the square map. These form a cyclic group of order two: the square map, applied twice, gives the identity map" and Aut($Bbb Z$) is isomorphic to $Bbb Z_2$
so
$G_1=Bbb Z$ and $G_2=Bbb Z_3$
$G_2=Bbb Z$ and $G_1=Bbb Z_3$
- We can easily see $Bbb Z$ and $Bbb Z_3$ are not isomorphic but Aut($Bbb Z$) and Aut($Bbb Z_3$) are.
But the given answer is that (1.) is only incorrect statement.
Is there any problem in my reasoning?
abstract-algebra automorphism-group
asked Dec 22 '18 at 7:22
onlymathonlymath
749
$endgroup$
3
$begingroup$
Your reasoning is solid.
$endgroup$
– Cheerful Parsnip
Dec 22 '18 at 7:26
add a comment |
$begingroup$
In material supplied by my instructor there is a question which asks to pick the incorrect statement among the following:
If $text{Aut}(G_1)cong text{Aut}(G_2)$ and $G_1$ is infinite group then $G_2$ is also infinite
If $text{Aut}(G_1) cong text{Aut}(G_2)$ and $G_1$ is finite group then $G_2$ is also finite
If $G_1$ not isomorphic to $G_2$ then Aut($G_1$) not isomorphic to Aut($G_2$).
$G_1$ and $G_2$ are two groups, Aut(G) is group of their automorphisms and "$cong$" means isomorphism.
I know all three above are incorrect as Aut($Bbb Z_3$)=$U_3$
(that is, $Bbb Z_2$) and I also found this statement on groupprops: "Of the three endomorphisms, two are automorphisms: the identity map and the square map. These form a cyclic group of order two: the square map, applied twice, gives the identity map" and Aut($Bbb Z$) is isomorphic to $Bbb Z_2$
so
$G_1=Bbb Z$ and $G_2=Bbb Z_3$
$G_2=Bbb Z$ and $G_1=Bbb Z_3$
- We can easily see $Bbb Z$ and $Bbb Z_3$ are not isomorphic but Aut($Bbb Z$) and Aut($Bbb Z_3$) are.
But the given answer is that (1.) is only incorrect statement.
Is there any problem in my reasoning?
abstract-algebra automorphism-group
asked Dec 22 '18 at 7:22
onlymathonlymath
749
$endgroup$
In material supplied by my instructor there is a question which asks to pick the incorrect statement among the following:
If $text{Aut}(G_1)cong text{Aut}(G_2)$ and $G_1$ is infinite group then $G_2$ is also infinite
If $text{Aut}(G_1) cong text{Aut}(G_2)$ and $G_1$ is finite group then $G_2$ is also finite
If $G_1$ not isomorphic to $G_2$ then Aut($G_1$) not isomorphic to Aut($G_2$).
$G_1$ and $G_2$ are two groups, Aut(G) is group of their automorphisms and "$cong$" means isomorphism.
I know all three above are incorrect as Aut($Bbb Z_3$)=$U_3$
(that is, $Bbb Z_2$) and I also found this statement on groupprops: "Of the three endomorphisms, two are automorphisms: the identity map and the square map. These form a cyclic group of order two: the square map, applied twice, gives the identity map" and Aut($Bbb Z$) is isomorphic to $Bbb Z_2$
so
$G_1=Bbb Z$ and $G_2=Bbb Z_3$
$G_2=Bbb Z$ and $G_1=Bbb Z_3$
- We can easily see $Bbb Z$ and $Bbb Z_3$ are not isomorphic but Aut($Bbb Z$) and Aut($Bbb Z_3$) are.
But the given answer is that (1.) is only incorrect statement.
Is there any problem in my reasoning?
abstract-algebra automorphism-group
abstract-algebra automorphism-group
asked Dec 22 '18 at 7:22
onlymathonlymath
749
asked Dec 22 '18 at 7:22
onlymathonlymath
749
edited Dec 22 '18 at 9:51
onlymath
asked Dec 22 '18 at 7:22
onlymathonlymath
749
asked Dec 22 '18 at 7:22
onlymathonlymath
749
asked Dec 22 '18 at 7:22
onlymathonlymath
749
749
3
$begingroup$
Your reasoning is solid.
$endgroup$
– Cheerful Parsnip
Dec 22 '18 at 7:26
add a comment |
3
$begingroup$
Your reasoning is solid.
$endgroup$
– Cheerful Parsnip
Dec 22 '18 at 7:26
3
3
$begingroup$
Your reasoning is solid.
$endgroup$
– Cheerful Parsnip
Dec 22 '18 at 7:26
$begingroup$
Your reasoning is solid.
$endgroup$
– Cheerful Parsnip
Dec 22 '18 at 7:26
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You are correct in your logic. It is possible that the instructor made an error and meant that one should pick all statements that are incorrect when writing the problem, which happens to be all of them. Perhaps hastily compiling a solution sheet at some later time, they saw the first was incorrect and seeing that the problem was to pick "the" incorrect one they gave that as the answer without checking the others.
answered Dec 22 '18 at 11:57
Matt SamuelMatt Samuel
38k63666
$endgroup$
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%2f3049188%2fquestion-about-isomorphism-of-autg%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
$begingroup$
You are correct in your logic. It is possible that the instructor made an error and meant that one should pick all statements that are incorrect when writing the problem, which happens to be all of them. Perhaps hastily compiling a solution sheet at some later time, they saw the first was incorrect and seeing that the problem was to pick "the" incorrect one they gave that as the answer without checking the others.
answered Dec 22 '18 at 11:57
Matt SamuelMatt Samuel
38k63666
$endgroup$
add a comment |
$begingroup$
You are correct in your logic. It is possible that the instructor made an error and meant that one should pick all statements that are incorrect when writing the problem, which happens to be all of them. Perhaps hastily compiling a solution sheet at some later time, they saw the first was incorrect and seeing that the problem was to pick "the" incorrect one they gave that as the answer without checking the others.
answered Dec 22 '18 at 11:57
Matt SamuelMatt Samuel
38k63666
$endgroup$
add a comment |
$begingroup$
You are correct in your logic. It is possible that the instructor made an error and meant that one should pick all statements that are incorrect when writing the problem, which happens to be all of them. Perhaps hastily compiling a solution sheet at some later time, they saw the first was incorrect and seeing that the problem was to pick "the" incorrect one they gave that as the answer without checking the others.
answered Dec 22 '18 at 11:57
Matt SamuelMatt Samuel
38k63666
$endgroup$
You are correct in your logic. It is possible that the instructor made an error and meant that one should pick all statements that are incorrect when writing the problem, which happens to be all of them. Perhaps hastily compiling a solution sheet at some later time, they saw the first was incorrect and seeing that the problem was to pick "the" incorrect one they gave that as the answer without checking the others.
answered Dec 22 '18 at 11:57
Matt SamuelMatt Samuel
38k63666
answered Dec 22 '18 at 11:57
Matt SamuelMatt Samuel
38k63666
answered Dec 22 '18 at 11:57
Matt SamuelMatt Samuel
38k63666
answered Dec 22 '18 at 11:57
Matt SamuelMatt Samuel
38k63666
38k63666
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%2f3049188%2fquestion-about-isomorphism-of-autg%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
3
$begingroup$
Your reasoning is solid.
$endgroup$
– Cheerful Parsnip
Dec 22 '18 at 7:26