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Is the statement that $ operatornameAut( operatornameHol(Z_n)) cong operatornameHol(Z_n)$ true for every odd $n$?
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Is the statement that $ operatornameAut(operatornameHol(Z_n)) cong operatornameHol(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 $ operatornameAut(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?
abstract-algebra group-theory finite-groups cyclic-groups automorphism-group
edited Jul 26 at 13:18
Bernard
110k635102
asked Jul 26 at 12:24
Yanior Weg
1,0791527
add a comment |Â
up vote
5
down vote
favorite
Is the statement that $ operatornameAut(operatornameHol(Z_n)) cong operatornameHol(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 $ operatornameAut(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?
abstract-algebra group-theory finite-groups cyclic-groups automorphism-group
edited Jul 26 at 13:18
Bernard
110k635102
asked Jul 26 at 12:24
Yanior Weg
1,0791527
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;
– AnalysisStudent0414
Jul 26 at 13:55
2
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 $k^-1phi(k)$ does to $Z_n$.
– Steve D
Jul 26 at 14:02
add a comment |Â
up vote
5
down vote
favorite
up vote
5
down vote
favorite
Is the statement that $ operatornameAut(operatornameHol(Z_n)) cong operatornameHol(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 $ operatornameAut(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?
abstract-algebra group-theory finite-groups cyclic-groups automorphism-group
edited Jul 26 at 13:18
Bernard
110k635102
asked Jul 26 at 12:24
Yanior Weg
1,0791527
Is the statement that $ operatornameAut(operatornameHol(Z_n)) cong operatornameHol(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 $ operatornameAut(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?
abstract-algebra group-theory finite-groups cyclic-groups automorphism-group
edited Jul 26 at 13:18
Bernard
110k635102
asked Jul 26 at 12:24
Yanior Weg
1,0791527
edited Jul 26 at 13:18
Bernard
110k635102
edited Jul 26 at 13:18
Bernard
110k635102
edited Jul 26 at 13:18
Bernard
110k635102
110k635102
asked Jul 26 at 12:24
Yanior Weg
1,0791527
asked Jul 26 at 12:24
Yanior Weg
1,0791527
asked Jul 26 at 12:24
Yanior Weg
1,0791527
1,0791527
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;
– AnalysisStudent0414
Jul 26 at 13:55
2
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 $k^-1phi(k)$ does to $Z_n$.
– Steve D
Jul 26 at 14:02
add a comment |Â
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;
– AnalysisStudent0414
Jul 26 at 13:55
2
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 $k^-1phi(k)$ does to $Z_n$.
– Steve D
Jul 26 at 14:02
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;– AnalysisStudent0414
Jul 26 at 13:55
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;– AnalysisStudent0414
Jul 26 at 13:55
2
2
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 $k^-1phi(k)$ does to $Z_n$.
– Steve D
Jul 26 at 14:02
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 $k^-1phi(k)$ does to $Z_n$.
– Steve D
Jul 26 at 14:02
add a comment |Â
1 Answer
1
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up vote
3
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accepted
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
beginalign
a^g &= alpha(a^g)\
&= a^alpha(g)
endalign
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 at 22:04
Steve D
2,022419
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
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
beginalign
a^g &= alpha(a^g)\
&= a^alpha(g)
endalign
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 at 22:04
Steve D
2,022419
add a comment |Â
up vote
3
down vote
accepted
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
beginalign
a^g &= alpha(a^g)\
&= a^alpha(g)
endalign
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 at 22:04
Steve D
2,022419
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
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
beginalign
a^g &= alpha(a^g)\
&= a^alpha(g)
endalign
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 at 22:04
Steve D
2,022419
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
beginalign
a^g &= alpha(a^g)\
&= a^alpha(g)
endalign
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 at 22:04
Steve D
2,022419
edited Jul 26 at 22:26
answered Jul 26 at 22:04
Steve D
2,022419
answered Jul 26 at 22:04
Steve D
2,022419
answered Jul 26 at 22:04
Steve D
2,022419
2,022419
add a comment |Â
add a comment |Â
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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;– AnalysisStudent0414
Jul 26 at 13:55
2
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 $k^-1phi(k)$ does to $Z_n$.
– Steve D
Jul 26 at 14:02