How many group homomorphisms from $Bbb Z_3$ to $text{Aut}(Bbb Z_7)$
$begingroup$
How many group homomorphism from $Bbb Z_3$ to $text{Aut}(Bbb Z_7)$?
My attempt: Since $Bbb Z_3$ is cyclic, hence the homomorphisms are all determined by $phi(overline{1})$. On the other hand $text{Aut}(Bbb Z_7)cong Bbb Z_6$
.
Since $o(phi(overline{1}))mid |Bbb Z_3|$, so $o(phi(overline{1}))$ can only be $1,~3$. If the order is $1$, it means $phi(overline{1})=overline{1}$. It is indeed a homo. If the order is $3$, it means $phi(overline{1})=overline{2}$ or $overline{4}$. So the total number of group homomorphisms from $Bbb Z_3$ to $text{Aut}(Bbb Z_7)$ is three, as I show above. Am I correct?
abstract-algebra
asked Dec 31 '18 at 9:55
EricEric
1,846615
$endgroup$
add a comment |
$begingroup$
How many group homomorphism from $Bbb Z_3$ to $text{Aut}(Bbb Z_7)$?
My attempt: Since $Bbb Z_3$ is cyclic, hence the homomorphisms are all determined by $phi(overline{1})$. On the other hand $text{Aut}(Bbb Z_7)cong Bbb Z_6$
.
Since $o(phi(overline{1}))mid |Bbb Z_3|$, so $o(phi(overline{1}))$ can only be $1,~3$. If the order is $1$, it means $phi(overline{1})=overline{1}$. It is indeed a homo. If the order is $3$, it means $phi(overline{1})=overline{2}$ or $overline{4}$. So the total number of group homomorphisms from $Bbb Z_3$ to $text{Aut}(Bbb Z_7)$ is three, as I show above. Am I correct?
abstract-algebra
asked Dec 31 '18 at 9:55
EricEric
1,846615
$endgroup$
1
$begingroup$
yes, looks right
$endgroup$
– Martín Vacas Vignolo
Dec 31 '18 at 10:00
1
$begingroup$
Yes, it is correct.
$endgroup$
– toric_actions
Dec 31 '18 at 10:05
add a comment |
$begingroup$
How many group homomorphism from $Bbb Z_3$ to $text{Aut}(Bbb Z_7)$?
My attempt: Since $Bbb Z_3$ is cyclic, hence the homomorphisms are all determined by $phi(overline{1})$. On the other hand $text{Aut}(Bbb Z_7)cong Bbb Z_6$
.
Since $o(phi(overline{1}))mid |Bbb Z_3|$, so $o(phi(overline{1}))$ can only be $1,~3$. If the order is $1$, it means $phi(overline{1})=overline{1}$. It is indeed a homo. If the order is $3$, it means $phi(overline{1})=overline{2}$ or $overline{4}$. So the total number of group homomorphisms from $Bbb Z_3$ to $text{Aut}(Bbb Z_7)$ is three, as I show above. Am I correct?
abstract-algebra
asked Dec 31 '18 at 9:55
EricEric
1,846615
$endgroup$
How many group homomorphism from $Bbb Z_3$ to $text{Aut}(Bbb Z_7)$?
My attempt: Since $Bbb Z_3$ is cyclic, hence the homomorphisms are all determined by $phi(overline{1})$. On the other hand $text{Aut}(Bbb Z_7)cong Bbb Z_6$
.
Since $o(phi(overline{1}))mid |Bbb Z_3|$, so $o(phi(overline{1}))$ can only be $1,~3$. If the order is $1$, it means $phi(overline{1})=overline{1}$. It is indeed a homo. If the order is $3$, it means $phi(overline{1})=overline{2}$ or $overline{4}$. So the total number of group homomorphisms from $Bbb Z_3$ to $text{Aut}(Bbb Z_7)$ is three, as I show above. Am I correct?
abstract-algebra
abstract-algebra
asked Dec 31 '18 at 9:55
EricEric
1,846615
asked Dec 31 '18 at 9:55
EricEric
1,846615
asked Dec 31 '18 at 9:55
EricEric
1,846615
asked Dec 31 '18 at 9:55
EricEric
1,846615
asked Dec 31 '18 at 9:55
EricEric
1,846615
1,846615
1
$begingroup$
yes, looks right
$endgroup$
– Martín Vacas Vignolo
Dec 31 '18 at 10:00
1
$begingroup$
Yes, it is correct.
$endgroup$
– toric_actions
Dec 31 '18 at 10:05
add a comment |
1
$begingroup$
yes, looks right
$endgroup$
– Martín Vacas Vignolo
Dec 31 '18 at 10:00
1
$begingroup$
Yes, it is correct.
$endgroup$
– toric_actions
Dec 31 '18 at 10:05
1
1
$begingroup$
yes, looks right
$endgroup$
– Martín Vacas Vignolo
Dec 31 '18 at 10:00
$begingroup$
yes, looks right
$endgroup$
– Martín Vacas Vignolo
Dec 31 '18 at 10:00
1
1
$begingroup$
Yes, it is correct.
$endgroup$
– toric_actions
Dec 31 '18 at 10:05
$begingroup$
Yes, it is correct.
$endgroup$
– toric_actions
Dec 31 '18 at 10:05
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
$$operatorname{Hom}(Bbb Z/3Bbb Z, operatorname{Aut}(Bbb Z/7Bbb Z)) = operatorname{Hom}(Bbb Z/3Bbb Z, C_6) = C_6[3] = {e, g^2, g^4}$$
answered Dec 31 '18 at 10:04
Kenny LauKenny Lau
20k2260
$endgroup$
add a comment |
Your Answer
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%2f3057559%2fhow-many-group-homomorphisms-from-bbb-z-3-to-textaut-bbb-z-7%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
$begingroup$
$$operatorname{Hom}(Bbb Z/3Bbb Z, operatorname{Aut}(Bbb Z/7Bbb Z)) = operatorname{Hom}(Bbb Z/3Bbb Z, C_6) = C_6[3] = {e, g^2, g^4}$$
answered Dec 31 '18 at 10:04
Kenny LauKenny Lau
20k2260
$endgroup$
add a comment |
$begingroup$
$$operatorname{Hom}(Bbb Z/3Bbb Z, operatorname{Aut}(Bbb Z/7Bbb Z)) = operatorname{Hom}(Bbb Z/3Bbb Z, C_6) = C_6[3] = {e, g^2, g^4}$$
answered Dec 31 '18 at 10:04
Kenny LauKenny Lau
20k2260
$endgroup$
add a comment |
$begingroup$
$$operatorname{Hom}(Bbb Z/3Bbb Z, operatorname{Aut}(Bbb Z/7Bbb Z)) = operatorname{Hom}(Bbb Z/3Bbb Z, C_6) = C_6[3] = {e, g^2, g^4}$$
answered Dec 31 '18 at 10:04
Kenny LauKenny Lau
20k2260
$endgroup$
$$operatorname{Hom}(Bbb Z/3Bbb Z, operatorname{Aut}(Bbb Z/7Bbb Z)) = operatorname{Hom}(Bbb Z/3Bbb Z, C_6) = C_6[3] = {e, g^2, g^4}$$
answered Dec 31 '18 at 10:04
Kenny LauKenny Lau
20k2260
answered Dec 31 '18 at 10:04
Kenny LauKenny Lau
20k2260
answered Dec 31 '18 at 10:04
Kenny LauKenny Lau
20k2260
answered Dec 31 '18 at 10:04
Kenny LauKenny Lau
20k2260
20k2260
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%2f3057559%2fhow-many-group-homomorphisms-from-bbb-z-3-to-textaut-bbb-z-7%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
1
$begingroup$
yes, looks right
$endgroup$
– Martín Vacas Vignolo
Dec 31 '18 at 10:00
1
$begingroup$
Yes, it is correct.
$endgroup$
– toric_actions
Dec 31 '18 at 10:05