$phi in operatornameAut(Bbb Z_50)$ via $phi(11) = 3$ Then $phi(x) = $? For any $x inBbb Z_50$ The Next CEO of Stack OverflowInterpretations of the first cohomology groupIs there a nontrivial semidirect product of two groups isomorphic to their direct product?An exact sequence of unit groupsConstructing a group automorphism making a diagram commuteExistence of an isomorphism $BbbZ_n^times rightarrow BbbZ_phi(n)^+$Question about split short exact sequences of groupsTo complete the proof that $operatornamePSL(2,Bbb F_5)cong A_5$By analyzing the map $G rightarrow$ Aut(G) given by $g rightarrow phi_g$, identify Inn(G) as a quotient of G.How to calculate the generators for the multiplicative group $Bbb Z_4^*$Proving Characteristic Subgroups are Transitive
Multi tool use
"In the right combination" vs "with the right combination"?
What does "Its cash flow is deeply negative" mean?
Is it my responsibility to learn a new technology in my own time my employer wants to implement?
Grabbing quick drinks
Why didn't Khan get resurrected in the Genesis Explosion?
Would a galaxy be visible from outside, but nearby?
Why don't programming languages automatically manage the synchronous/asynchronous problem?
What is "(CFMCC)" on an ILS approach chart?
What was the first Unix version to run on a microcomputer?
Can you replace a racial trait cantrip when leveling up?
Which tube will fit a -(700 x 25c) wheel?
How do I transpose the first and deepest levels of an arbitrarily nested array?
What happened in Rome, when the western empire "fell"?
Can we say or write : "No, it'sn't"?
Does it take more energy to get to Venus or to Mars?
How do I make a variable always equal to the result of some calculations?
How to start emacs in "nothing" mode (`fundamental-mode`)
What happens if you roll doubles 3 times then land on "Go to jail?"
How long to clear the 'suck zone' of a turbofan after start is initiated?
Written every which way
WOW air has ceased operation, can I get my tickets refunded?
How does the mv command work with external drives?
I believe this to be a fraud - hired, then asked to cash check and send cash as Bitcoin
Can I equip Skullclamp on a creature I am sacrificing?
$phi in operatornameAut(Bbb Z_50)$ via $phi(11) = 3$ Then $phi(x) = $? For any $x inBbb Z_50$
The Next CEO of Stack OverflowInterpretations of the first cohomology groupIs there a nontrivial semidirect product of two groups isomorphic to their direct product?An exact sequence of unit groupsConstructing a group automorphism making a diagram commuteExistence of an isomorphism $BbbZ_n^times rightarrow BbbZ_phi(n)^+$Question about split short exact sequences of groupsTo complete the proof that $operatornamePSL(2,Bbb F_5)cong A_5$By analyzing the map $G rightarrow$ Aut(G) given by $g rightarrow phi_g$, identify Inn(G) as a quotient of G.How to calculate the generators for the multiplicative group $Bbb Z_4^*$Proving Characteristic Subgroups are Transitive
$begingroup$
$phi in operatornameAut(Bbb Z_50)$ via $phi(11) = 3$ Then $phi(x) = $? For any $x in Bbb Z_50$
The answer is $23x$ but I'm not quite sure how to figure that out. Here's what I did:
$phi(11) = 3 Rightarrow$ The function maps $11x rightarrow 3x$
Adding $(11)^-1 = 39$ (mod 50) on both sides gives us
$x rightarrow 42x$
What am I not doing correct?
group-theory
edited Mar 18 at 22:38
Bernard
123k741117
asked Mar 18 at 22:12
user2965071user2965071
1887
$endgroup$
add a comment |
$begingroup$
$phi in operatornameAut(Bbb Z_50)$ via $phi(11) = 3$ Then $phi(x) = $? For any $x in Bbb Z_50$
The answer is $23x$ but I'm not quite sure how to figure that out. Here's what I did:
$phi(11) = 3 Rightarrow$ The function maps $11x rightarrow 3x$
Adding $(11)^-1 = 39$ (mod 50) on both sides gives us
$x rightarrow 42x$
What am I not doing correct?
group-theory
edited Mar 18 at 22:38
Bernard
123k741117
asked Mar 18 at 22:12
user2965071user2965071
1887
$endgroup$
add a comment |
$begingroup$
$phi in operatornameAut(Bbb Z_50)$ via $phi(11) = 3$ Then $phi(x) = $? For any $x in Bbb Z_50$
The answer is $23x$ but I'm not quite sure how to figure that out. Here's what I did:
$phi(11) = 3 Rightarrow$ The function maps $11x rightarrow 3x$
Adding $(11)^-1 = 39$ (mod 50) on both sides gives us
$x rightarrow 42x$
What am I not doing correct?
group-theory
edited Mar 18 at 22:38
Bernard
123k741117
asked Mar 18 at 22:12
user2965071user2965071
1887
$endgroup$
$phi in operatornameAut(Bbb Z_50)$ via $phi(11) = 3$ Then $phi(x) = $? For any $x in Bbb Z_50$
The answer is $23x$ but I'm not quite sure how to figure that out. Here's what I did:
$phi(11) = 3 Rightarrow$ The function maps $11x rightarrow 3x$
Adding $(11)^-1 = 39$ (mod 50) on both sides gives us
$x rightarrow 42x$
What am I not doing correct?
group-theory
group-theory
edited Mar 18 at 22:38
Bernard
123k741117
asked Mar 18 at 22:12
user2965071user2965071
1887
edited Mar 18 at 22:38
Bernard
123k741117
asked Mar 18 at 22:12
user2965071user2965071
1887
edited Mar 18 at 22:38
Bernard
123k741117
edited Mar 18 at 22:38
Bernard
123k741117
edited Mar 18 at 22:38
Bernard
123k741117
123k741117
asked Mar 18 at 22:12
user2965071user2965071
1887
asked Mar 18 at 22:12
user2965071user2965071
1887
asked Mar 18 at 22:12
user2965071user2965071
1887
1887
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
We have
$$1equiv -9cdot11pmod50 .$$
Therefore,
$$phi(x)equivphi(-9xcdot11)equiv-9xcdotphi(11)equiv-27xequiv23xpmod50 .$$
In the second equivalence we used that $phi(nx) = phi(x)+cdots+phi(x) = nphi(x)$.
Adding $30equiv-11pmod50$ to both sides just tells you that
$$phi(11x)-11equiv3x+39 ,$$
which is no help at all.
answered Mar 18 at 22:17
Daniel Robert-NicoudDaniel Robert-Nicoud
20.5k33797
$endgroup$
$begingroup$
Shouldn't the second equivalence be $-9 phi(x) cdot phi(11)$?
$endgroup$
– Robert Shore
Mar 18 at 22:38
1
$begingroup$
@RobertShore No, the group structure is given by addition, not multiplication.
$endgroup$
– Daniel Robert-Nicoud
Mar 19 at 6:20
add a comment |
$begingroup$
Any automorphism $phi$ of $Bbb Z / n Bbb Z$ is determined by $phi(1)$ because $phi(n) = n phi(1)$, so it suffices to determine $phi(1)$. As noted by Daniel Robert-Nicoud, $1 equiv -9 cdot 11 pmod50$, so $phi(1) = -9 phi(11) = -9 cdot 3 = -27 ~(mod 50) = 23$ and $phi(n) = n phi(1) = 23n$.
answered Mar 18 at 22:45
Robert ShoreRobert Shore
3,603324
$endgroup$
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%2f3153385%2fphi-in-operatornameaut-bbb-z-50-via-phi11-3-then-phix%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
We have
$$1equiv -9cdot11pmod50 .$$
Therefore,
$$phi(x)equivphi(-9xcdot11)equiv-9xcdotphi(11)equiv-27xequiv23xpmod50 .$$
In the second equivalence we used that $phi(nx) = phi(x)+cdots+phi(x) = nphi(x)$.
Adding $30equiv-11pmod50$ to both sides just tells you that
$$phi(11x)-11equiv3x+39 ,$$
which is no help at all.
answered Mar 18 at 22:17
Daniel Robert-NicoudDaniel Robert-Nicoud
20.5k33797
$endgroup$
$begingroup$
Shouldn't the second equivalence be $-9 phi(x) cdot phi(11)$?
$endgroup$
– Robert Shore
Mar 18 at 22:38
1
$begingroup$
@RobertShore No, the group structure is given by addition, not multiplication.
$endgroup$
– Daniel Robert-Nicoud
Mar 19 at 6:20
add a comment |
$begingroup$
We have
$$1equiv -9cdot11pmod50 .$$
Therefore,
$$phi(x)equivphi(-9xcdot11)equiv-9xcdotphi(11)equiv-27xequiv23xpmod50 .$$
In the second equivalence we used that $phi(nx) = phi(x)+cdots+phi(x) = nphi(x)$.
Adding $30equiv-11pmod50$ to both sides just tells you that
$$phi(11x)-11equiv3x+39 ,$$
which is no help at all.
answered Mar 18 at 22:17
Daniel Robert-NicoudDaniel Robert-Nicoud
20.5k33797
$endgroup$
$begingroup$
Shouldn't the second equivalence be $-9 phi(x) cdot phi(11)$?
$endgroup$
– Robert Shore
Mar 18 at 22:38
1
$begingroup$
@RobertShore No, the group structure is given by addition, not multiplication.
$endgroup$
– Daniel Robert-Nicoud
Mar 19 at 6:20
add a comment |
$begingroup$
We have
$$1equiv -9cdot11pmod50 .$$
Therefore,
$$phi(x)equivphi(-9xcdot11)equiv-9xcdotphi(11)equiv-27xequiv23xpmod50 .$$
In the second equivalence we used that $phi(nx) = phi(x)+cdots+phi(x) = nphi(x)$.
Adding $30equiv-11pmod50$ to both sides just tells you that
$$phi(11x)-11equiv3x+39 ,$$
which is no help at all.
answered Mar 18 at 22:17
Daniel Robert-NicoudDaniel Robert-Nicoud
20.5k33797
$endgroup$
We have
$$1equiv -9cdot11pmod50 .$$
Therefore,
$$phi(x)equivphi(-9xcdot11)equiv-9xcdotphi(11)equiv-27xequiv23xpmod50 .$$
In the second equivalence we used that $phi(nx) = phi(x)+cdots+phi(x) = nphi(x)$.
Adding $30equiv-11pmod50$ to both sides just tells you that
$$phi(11x)-11equiv3x+39 ,$$
which is no help at all.
answered Mar 18 at 22:17
Daniel Robert-NicoudDaniel Robert-Nicoud
20.5k33797
answered Mar 18 at 22:17
Daniel Robert-NicoudDaniel Robert-Nicoud
20.5k33797
answered Mar 18 at 22:17
Daniel Robert-NicoudDaniel Robert-Nicoud
20.5k33797
answered Mar 18 at 22:17
Daniel Robert-NicoudDaniel Robert-Nicoud
20.5k33797
20.5k33797
$begingroup$
Shouldn't the second equivalence be $-9 phi(x) cdot phi(11)$?
$endgroup$
– Robert Shore
Mar 18 at 22:38
1
$begingroup$
@RobertShore No, the group structure is given by addition, not multiplication.
$endgroup$
– Daniel Robert-Nicoud
Mar 19 at 6:20
add a comment |
$begingroup$
Shouldn't the second equivalence be $-9 phi(x) cdot phi(11)$?
$endgroup$
– Robert Shore
Mar 18 at 22:38
1
$begingroup$
@RobertShore No, the group structure is given by addition, not multiplication.
$endgroup$
– Daniel Robert-Nicoud
Mar 19 at 6:20
$begingroup$
Shouldn't the second equivalence be $-9 phi(x) cdot phi(11)$?
$endgroup$
– Robert Shore
Mar 18 at 22:38
$begingroup$
Shouldn't the second equivalence be $-9 phi(x) cdot phi(11)$?
$endgroup$
– Robert Shore
Mar 18 at 22:38
1
1
$begingroup$
@RobertShore No, the group structure is given by addition, not multiplication.
$endgroup$
– Daniel Robert-Nicoud
Mar 19 at 6:20
$begingroup$
@RobertShore No, the group structure is given by addition, not multiplication.
$endgroup$
– Daniel Robert-Nicoud
Mar 19 at 6:20
add a comment |
$begingroup$
Any automorphism $phi$ of $Bbb Z / n Bbb Z$ is determined by $phi(1)$ because $phi(n) = n phi(1)$, so it suffices to determine $phi(1)$. As noted by Daniel Robert-Nicoud, $1 equiv -9 cdot 11 pmod50$, so $phi(1) = -9 phi(11) = -9 cdot 3 = -27 ~(mod 50) = 23$ and $phi(n) = n phi(1) = 23n$.
answered Mar 18 at 22:45
Robert ShoreRobert Shore
3,603324
$endgroup$
add a comment |
$begingroup$
Any automorphism $phi$ of $Bbb Z / n Bbb Z$ is determined by $phi(1)$ because $phi(n) = n phi(1)$, so it suffices to determine $phi(1)$. As noted by Daniel Robert-Nicoud, $1 equiv -9 cdot 11 pmod50$, so $phi(1) = -9 phi(11) = -9 cdot 3 = -27 ~(mod 50) = 23$ and $phi(n) = n phi(1) = 23n$.
answered Mar 18 at 22:45
Robert ShoreRobert Shore
3,603324
$endgroup$
add a comment |
$begingroup$
Any automorphism $phi$ of $Bbb Z / n Bbb Z$ is determined by $phi(1)$ because $phi(n) = n phi(1)$, so it suffices to determine $phi(1)$. As noted by Daniel Robert-Nicoud, $1 equiv -9 cdot 11 pmod50$, so $phi(1) = -9 phi(11) = -9 cdot 3 = -27 ~(mod 50) = 23$ and $phi(n) = n phi(1) = 23n$.
answered Mar 18 at 22:45
Robert ShoreRobert Shore
3,603324
$endgroup$
Any automorphism $phi$ of $Bbb Z / n Bbb Z$ is determined by $phi(1)$ because $phi(n) = n phi(1)$, so it suffices to determine $phi(1)$. As noted by Daniel Robert-Nicoud, $1 equiv -9 cdot 11 pmod50$, so $phi(1) = -9 phi(11) = -9 cdot 3 = -27 ~(mod 50) = 23$ and $phi(n) = n phi(1) = 23n$.
answered Mar 18 at 22:45
Robert ShoreRobert Shore
3,603324
answered Mar 18 at 22:45
Robert ShoreRobert Shore
3,603324
answered Mar 18 at 22:45
Robert ShoreRobert Shore
3,603324
answered Mar 18 at 22:45
Robert ShoreRobert Shore
3,603324
3,603324
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%2f3153385%2fphi-in-operatornameaut-bbb-z-50-via-phi11-3-then-phix%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
9sNCzGayz FP hynwgDqG
