Mixing
Is the statement that $ operatorname{Aut}( operatorname{Hol}(Z_n)) cong operatorname{Hol}(Z_n)$ true for...
$begingroup$
Is the statement that $ operatorname{Aut}(operatorname{Hol}(Z_n)) cong operatorname{Hol}(Z_n)$ true for every odd $n$? $Hol$ stands here for group holomorph.
This problem appeared, when I stumbled upon the following MO question: https://mathoverflow.net/questions/258886/conditions-for-a-finite-group-to-be-isomorphic-to-its-automorphism-group
OP of that question provided us with the complete list of groups $G$, such that $|G| leq 506$ and $ operatorname{Aut}(G) cong G$. Among those groups there are some looking like holomorphs of all cyclic groups of odd orders up to $23$. Does anybody know, if that pattern continues or is it just a coincidence?
group-theory finite-groups cyclic-groups automorphism-group holomorph
asked Jul 26 '18 at 12:24
Yanior WegYanior Weg
2,92121549
$endgroup$
add a comment |
$begingroup$
Is the statement that $ operatorname{Aut}(operatorname{Hol}(Z_n)) cong operatorname{Hol}(Z_n)$ true for every odd $n$? $Hol$ stands here for group holomorph.
This problem appeared, when I stumbled upon the following MO question: https://mathoverflow.net/questions/258886/conditions-for-a-finite-group-to-be-isomorphic-to-its-automorphism-group
OP of that question provided us with the complete list of groups $G$, such that $|G| leq 506$ and $ operatorname{Aut}(G) cong G$. Among those groups there are some looking like holomorphs of all cyclic groups of odd orders up to $23$. Does anybody know, if that pattern continues or is it just a coincidence?
group-theory finite-groups cyclic-groups automorphism-group holomorph
asked Jul 26 '18 at 12:24
Yanior WegYanior Weg
2,92121549
$endgroup$
$begingroup$
Very interesting question! Running this code on MAGMA shows no "small" counterexamplesfor i in [1..100] do G:=CyclicGroup(2*i+1); H:=Holomorph(G); K:=AutomorphismGroup(H); if IsIsomorphic(PermutationGroup(K),H) then; else print i; end if; end for;
$endgroup$
– AnalysisStudent0414
Jul 26 '18 at 13:55
3
$begingroup$
This is true. The key is that $n$ is odd,so the centeris trivial. If $phi$ is an automorphism of your holomorph, find an element $k$ in $Aut(Z_n)$ that does the same thing to the generator. Now look at what $k^{-1}phi(k)$ does to $Z_n$.
$endgroup$
– Steve D
Jul 26 '18 at 14:02
add a comment |
$begingroup$
Is the statement that $ operatorname{Aut}(operatorname{Hol}(Z_n)) cong operatorname{Hol}(Z_n)$ true for every odd $n$? $Hol$ stands here for group holomorph.
This problem appeared, when I stumbled upon the following MO question: https://mathoverflow.net/questions/258886/conditions-for-a-finite-group-to-be-isomorphic-to-its-automorphism-group
OP of that question provided us with the complete list of groups $G$, such that $|G| leq 506$ and $ operatorname{Aut}(G) cong G$. Among those groups there are some looking like holomorphs of all cyclic groups of odd orders up to $23$. Does anybody know, if that pattern continues or is it just a coincidence?
group-theory finite-groups cyclic-groups automorphism-group holomorph
asked Jul 26 '18 at 12:24
Yanior WegYanior Weg
2,92121549
$endgroup$
Is the statement that $ operatorname{Aut}(operatorname{Hol}(Z_n)) cong operatorname{Hol}(Z_n)$ true for every odd $n$? $Hol$ stands here for group holomorph.
This problem appeared, when I stumbled upon the following MO question: https://mathoverflow.net/questions/258886/conditions-for-a-finite-group-to-be-isomorphic-to-its-automorphism-group
OP of that question provided us with the complete list of groups $G$, such that $|G| leq 506$ and $ operatorname{Aut}(G) cong G$. Among those groups there are some looking like holomorphs of all cyclic groups of odd orders up to $23$. Does anybody know, if that pattern continues or is it just a coincidence?
group-theory finite-groups cyclic-groups automorphism-group holomorph
group-theory finite-groups cyclic-groups automorphism-group holomorph
asked Jul 26 '18 at 12:24
Yanior WegYanior Weg
2,92121549
asked Jul 26 '18 at 12:24
Yanior WegYanior Weg
2,92121549
edited Feb 3 at 11:31
Yanior Weg
asked Jul 26 '18 at 12:24
Yanior WegYanior Weg
2,92121549
asked Jul 26 '18 at 12:24
Yanior WegYanior Weg
2,92121549
asked Jul 26 '18 at 12:24
Yanior WegYanior Weg
2,92121549
2,92121549
$begingroup$
Very interesting question! Running this code on MAGMA shows no "small" counterexamplesfor i in [1..100] do G:=CyclicGroup(2*i+1); H:=Holomorph(G); K:=AutomorphismGroup(H); if IsIsomorphic(PermutationGroup(K),H) then; else print i; end if; end for;
$endgroup$
– AnalysisStudent0414
Jul 26 '18 at 13:55
3
$begingroup$
This is true. The key is that $n$ is odd,so the centeris trivial. If $phi$ is an automorphism of your holomorph, find an element $k$ in $Aut(Z_n)$ that does the same thing to the generator. Now look at what $k^{-1}phi(k)$ does to $Z_n$.
$endgroup$
– Steve D
Jul 26 '18 at 14:02
add a comment |
$begingroup$
Very interesting question! Running this code on MAGMA shows no "small" counterexamplesfor i in [1..100] do G:=CyclicGroup(2*i+1); H:=Holomorph(G); K:=AutomorphismGroup(H); if IsIsomorphic(PermutationGroup(K),H) then; else print i; end if; end for;
$endgroup$
– AnalysisStudent0414
Jul 26 '18 at 13:55
3
$begingroup$
This is true. The key is that $n$ is odd,so the centeris trivial. If $phi$ is an automorphism of your holomorph, find an element $k$ in $Aut(Z_n)$ that does the same thing to the generator. Now look at what $k^{-1}phi(k)$ does to $Z_n$.
$endgroup$
– Steve D
Jul 26 '18 at 14:02
$begingroup$
Very interesting question! Running this code on MAGMA shows no "small" counterexamples
for i in [1..100] do G:=CyclicGroup(2*i+1); H:=Holomorph(G); K:=AutomorphismGroup(H); if IsIsomorphic(PermutationGroup(K),H) then; else print i; end if; end for;$endgroup$
– AnalysisStudent0414
Jul 26 '18 at 13:55
$begingroup$
Very interesting question! Running this code on MAGMA shows no "small" counterexamples
for i in [1..100] do G:=CyclicGroup(2*i+1); H:=Holomorph(G); K:=AutomorphismGroup(H); if IsIsomorphic(PermutationGroup(K),H) then; else print i; end if; end for;$endgroup$
– AnalysisStudent0414
Jul 26 '18 at 13:55
3
3
$begingroup$
This is true. The key is that $n$ is odd,so the centeris trivial. If $phi$ is an automorphism of your holomorph, find an element $k$ in $Aut(Z_n)$ that does the same thing to the generator. Now look at what $k^{-1}phi(k)$ does to $Z_n$.
$endgroup$
– Steve D
Jul 26 '18 at 14:02
$begingroup$
This is true. The key is that $n$ is odd,so the centeris trivial. If $phi$ is an automorphism of your holomorph, find an element $k$ in $Aut(Z_n)$ that does the same thing to the generator. Now look at what $k^{-1}phi(k)$ does to $Z_n$.
$endgroup$
– Steve D
Jul 26 '18 at 14:02
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Here's a proof, which is really an expansion of my comment above.
Let $A=langle arangle$ be a cyclic group of order $n$, with $n$ odd. Let $K=Aut(A)$, and consider the holomorph $G=Artimes K$. Because $n$ is odd, $Z(G)$ is trivial: because the inversion map is in $K$, we see that $C_G(a)=A$, and no element of $A$ is central. Also, since $K$ is abelian, we see $A=[G,G]$.
Now let $phi$ be an automorphism of $G$. Then $phi(A)=A$, and so there exists $kin K$ such that $phi(a)=a^k$. Now $H=phi(K)$ is a self-centralizing subgroup of $G$ that is a complement to $A$. Because it's a complement, for every $gin K$, there's a unique element of the form $a^?gin H$. In particular, consider $iotain K$, the inversion map. If $a^riotain H$, then setting $m=-r(n+1)/2$, it is easy to check that $iotain a^mHa^{-m}$. By looking at $[iota,ga^s]$, we see that $C_G(iota)=K$, and so $a^mHa^{-m}=K$. Thus $phi$ acts on $G$ exactly like conjugation by $ka^m$, so $phi$ is inner. Combined with the triviality of $Z(G)$, we see $G$ is complete.
Edit: I might have glossed over one too many details in the end above. Let $psiin Aut(G)$ be conjugation by $ka^m$, and let $alpha=phipsi^{-1}$. Then we've shown $alpha$ fixes $A$ pointwise, and $K$ setwise. But then for any $gin K$, we have
begin{align}
a^g &= alpha(a^g)\
&= a^{alpha(g)}
end{align}
and thus $g$ and $alpha(g)$ are two automorphisms of $A$ with the same action, meaning $alpha(g)=g$. Thus $alpha$ fixes $K$ pointwise, and since $G=AK$, $alpha$ is the identity map.
answered Jul 26 '18 at 22:04
Steve DSteve D
2,4502623
$endgroup$
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%2f2863363%2fis-the-statement-that-operatornameaut-operatornameholz-n-cong-oper%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
$begingroup$
Here's a proof, which is really an expansion of my comment above.
Let $A=langle arangle$ be a cyclic group of order $n$, with $n$ odd. Let $K=Aut(A)$, and consider the holomorph $G=Artimes K$. Because $n$ is odd, $Z(G)$ is trivial: because the inversion map is in $K$, we see that $C_G(a)=A$, and no element of $A$ is central. Also, since $K$ is abelian, we see $A=[G,G]$.
Now let $phi$ be an automorphism of $G$. Then $phi(A)=A$, and so there exists $kin K$ such that $phi(a)=a^k$. Now $H=phi(K)$ is a self-centralizing subgroup of $G$ that is a complement to $A$. Because it's a complement, for every $gin K$, there's a unique element of the form $a^?gin H$. In particular, consider $iotain K$, the inversion map. If $a^riotain H$, then setting $m=-r(n+1)/2$, it is easy to check that $iotain a^mHa^{-m}$. By looking at $[iota,ga^s]$, we see that $C_G(iota)=K$, and so $a^mHa^{-m}=K$. Thus $phi$ acts on $G$ exactly like conjugation by $ka^m$, so $phi$ is inner. Combined with the triviality of $Z(G)$, we see $G$ is complete.
Edit: I might have glossed over one too many details in the end above. Let $psiin Aut(G)$ be conjugation by $ka^m$, and let $alpha=phipsi^{-1}$. Then we've shown $alpha$ fixes $A$ pointwise, and $K$ setwise. But then for any $gin K$, we have
begin{align}
a^g &= alpha(a^g)\
&= a^{alpha(g)}
end{align}
and thus $g$ and $alpha(g)$ are two automorphisms of $A$ with the same action, meaning $alpha(g)=g$. Thus $alpha$ fixes $K$ pointwise, and since $G=AK$, $alpha$ is the identity map.
answered Jul 26 '18 at 22:04
Steve DSteve D
2,4502623
$endgroup$
add a comment |
$begingroup$
Here's a proof, which is really an expansion of my comment above.
Let $A=langle arangle$ be a cyclic group of order $n$, with $n$ odd. Let $K=Aut(A)$, and consider the holomorph $G=Artimes K$. Because $n$ is odd, $Z(G)$ is trivial: because the inversion map is in $K$, we see that $C_G(a)=A$, and no element of $A$ is central. Also, since $K$ is abelian, we see $A=[G,G]$.
Now let $phi$ be an automorphism of $G$. Then $phi(A)=A$, and so there exists $kin K$ such that $phi(a)=a^k$. Now $H=phi(K)$ is a self-centralizing subgroup of $G$ that is a complement to $A$. Because it's a complement, for every $gin K$, there's a unique element of the form $a^?gin H$. In particular, consider $iotain K$, the inversion map. If $a^riotain H$, then setting $m=-r(n+1)/2$, it is easy to check that $iotain a^mHa^{-m}$. By looking at $[iota,ga^s]$, we see that $C_G(iota)=K$, and so $a^mHa^{-m}=K$. Thus $phi$ acts on $G$ exactly like conjugation by $ka^m$, so $phi$ is inner. Combined with the triviality of $Z(G)$, we see $G$ is complete.
Edit: I might have glossed over one too many details in the end above. Let $psiin Aut(G)$ be conjugation by $ka^m$, and let $alpha=phipsi^{-1}$. Then we've shown $alpha$ fixes $A$ pointwise, and $K$ setwise. But then for any $gin K$, we have
begin{align}
a^g &= alpha(a^g)\
&= a^{alpha(g)}
end{align}
and thus $g$ and $alpha(g)$ are two automorphisms of $A$ with the same action, meaning $alpha(g)=g$. Thus $alpha$ fixes $K$ pointwise, and since $G=AK$, $alpha$ is the identity map.
answered Jul 26 '18 at 22:04
Steve DSteve D
2,4502623
$endgroup$
add a comment |
$begingroup$
Here's a proof, which is really an expansion of my comment above.
Let $A=langle arangle$ be a cyclic group of order $n$, with $n$ odd. Let $K=Aut(A)$, and consider the holomorph $G=Artimes K$. Because $n$ is odd, $Z(G)$ is trivial: because the inversion map is in $K$, we see that $C_G(a)=A$, and no element of $A$ is central. Also, since $K$ is abelian, we see $A=[G,G]$.
Now let $phi$ be an automorphism of $G$. Then $phi(A)=A$, and so there exists $kin K$ such that $phi(a)=a^k$. Now $H=phi(K)$ is a self-centralizing subgroup of $G$ that is a complement to $A$. Because it's a complement, for every $gin K$, there's a unique element of the form $a^?gin H$. In particular, consider $iotain K$, the inversion map. If $a^riotain H$, then setting $m=-r(n+1)/2$, it is easy to check that $iotain a^mHa^{-m}$. By looking at $[iota,ga^s]$, we see that $C_G(iota)=K$, and so $a^mHa^{-m}=K$. Thus $phi$ acts on $G$ exactly like conjugation by $ka^m$, so $phi$ is inner. Combined with the triviality of $Z(G)$, we see $G$ is complete.
Edit: I might have glossed over one too many details in the end above. Let $psiin Aut(G)$ be conjugation by $ka^m$, and let $alpha=phipsi^{-1}$. Then we've shown $alpha$ fixes $A$ pointwise, and $K$ setwise. But then for any $gin K$, we have
begin{align}
a^g &= alpha(a^g)\
&= a^{alpha(g)}
end{align}
and thus $g$ and $alpha(g)$ are two automorphisms of $A$ with the same action, meaning $alpha(g)=g$. Thus $alpha$ fixes $K$ pointwise, and since $G=AK$, $alpha$ is the identity map.
answered Jul 26 '18 at 22:04
Steve DSteve D
2,4502623
$endgroup$
Here's a proof, which is really an expansion of my comment above.
Let $A=langle arangle$ be a cyclic group of order $n$, with $n$ odd. Let $K=Aut(A)$, and consider the holomorph $G=Artimes K$. Because $n$ is odd, $Z(G)$ is trivial: because the inversion map is in $K$, we see that $C_G(a)=A$, and no element of $A$ is central. Also, since $K$ is abelian, we see $A=[G,G]$.
Now let $phi$ be an automorphism of $G$. Then $phi(A)=A$, and so there exists $kin K$ such that $phi(a)=a^k$. Now $H=phi(K)$ is a self-centralizing subgroup of $G$ that is a complement to $A$. Because it's a complement, for every $gin K$, there's a unique element of the form $a^?gin H$. In particular, consider $iotain K$, the inversion map. If $a^riotain H$, then setting $m=-r(n+1)/2$, it is easy to check that $iotain a^mHa^{-m}$. By looking at $[iota,ga^s]$, we see that $C_G(iota)=K$, and so $a^mHa^{-m}=K$. Thus $phi$ acts on $G$ exactly like conjugation by $ka^m$, so $phi$ is inner. Combined with the triviality of $Z(G)$, we see $G$ is complete.
Edit: I might have glossed over one too many details in the end above. Let $psiin Aut(G)$ be conjugation by $ka^m$, and let $alpha=phipsi^{-1}$. Then we've shown $alpha$ fixes $A$ pointwise, and $K$ setwise. But then for any $gin K$, we have
begin{align}
a^g &= alpha(a^g)\
&= a^{alpha(g)}
end{align}
and thus $g$ and $alpha(g)$ are two automorphisms of $A$ with the same action, meaning $alpha(g)=g$. Thus $alpha$ fixes $K$ pointwise, and since $G=AK$, $alpha$ is the identity map.
answered Jul 26 '18 at 22:04
Steve DSteve D
2,4502623
edited Jul 26 '18 at 22:26
answered Jul 26 '18 at 22:04
Steve DSteve D
2,4502623
answered Jul 26 '18 at 22:04
Steve DSteve D
2,4502623
answered Jul 26 '18 at 22:04
Steve DSteve D
2,4502623
2,4502623
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%2f2863363%2fis-the-statement-that-operatornameaut-operatornameholz-n-cong-oper%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
$begingroup$
Very interesting question! Running this code on MAGMA shows no "small" counterexamples
for i in [1..100] do G:=CyclicGroup(2*i+1); H:=Holomorph(G); K:=AutomorphismGroup(H); if IsIsomorphic(PermutationGroup(K),H) then; else print i; end if; end for;$endgroup$
– AnalysisStudent0414
Jul 26 '18 at 13:55
3
$begingroup$
This is true. The key is that $n$ is odd,so the centeris trivial. If $phi$ is an automorphism of your holomorph, find an element $k$ in $Aut(Z_n)$ that does the same thing to the generator. Now look at what $k^{-1}phi(k)$ does to $Z_n$.
$endgroup$
– Steve D
Jul 26 '18 at 14:02