Mixing
For what $n in mathbb{N}$ is $ operatorname{Hol}(C_2^n)$ complete?
$begingroup$
For what $n in mathbb{N}$ is $ operatorname{Hol}(C_2^n)$ complete? Here $ operatorname{Hol}$ stands for holomorph, and $C_2^n$ stands for direct product of $n$ isomorphic copies of $C^2$.
It for $n = 1$ it is not true, however if $n = 2$, then $ operatorname{Hol}(C_2^n) cong S_4$ is complete (proof that $ operatorname{Hol}(C_2^n) cong S_4$ can be found here: The holomorph of $Z_2 times Z_2$). However, that is based on a specific property of $n = 2$ and does not help us in general case.
One can also easily see, that $ operatorname{Hol}(C_2^n) = C_2^n rtimes operatorname{Aut}(C_2^n)$ is centerless for all $n > 1$, as it $ operatorname{Aut}(C_2^n)$ acts both transitively and effectively on $C_2^n$.
However, determining for what $n$ all automorphisms of $ operatorname{Hol}(C_2^n)$ are inner is a much harder task and I do not know how to solve it.
Any help will be appreciated.
abstract-algebra group-theory finite-groups holomorph complete-groups
asked Feb 3 at 10:54
Yanior WegYanior Weg
2,92121549
$endgroup$
add a comment |
$begingroup$
For what $n in mathbb{N}$ is $ operatorname{Hol}(C_2^n)$ complete? Here $ operatorname{Hol}$ stands for holomorph, and $C_2^n$ stands for direct product of $n$ isomorphic copies of $C^2$.
It for $n = 1$ it is not true, however if $n = 2$, then $ operatorname{Hol}(C_2^n) cong S_4$ is complete (proof that $ operatorname{Hol}(C_2^n) cong S_4$ can be found here: The holomorph of $Z_2 times Z_2$). However, that is based on a specific property of $n = 2$ and does not help us in general case.
One can also easily see, that $ operatorname{Hol}(C_2^n) = C_2^n rtimes operatorname{Aut}(C_2^n)$ is centerless for all $n > 1$, as it $ operatorname{Aut}(C_2^n)$ acts both transitively and effectively on $C_2^n$.
However, determining for what $n$ all automorphisms of $ operatorname{Hol}(C_2^n)$ are inner is a much harder task and I do not know how to solve it.
Any help will be appreciated.
abstract-algebra group-theory finite-groups holomorph complete-groups
asked Feb 3 at 10:54
Yanior WegYanior Weg
2,92121549
$endgroup$
3
$begingroup$
It is complete for $n ne 3$. In the case $n=3$ there is an outer automorphism of order $2$ arising from the fact that $H^1({rm GL}(n,2),C_2^n)$ (with the natural action on the module) has order $2$ when $n=3$, but is trivial for $n ne 3$. If nobody else does, then I will write out an answer at some stage, but I would feel more inclined to do so if you put some effort into solving the problem yourself, and said where you were stuck. You could also explain the context in which you are trying to solve this problem.
$endgroup$
– Derek Holt
Feb 3 at 13:21
add a comment |
$begingroup$
For what $n in mathbb{N}$ is $ operatorname{Hol}(C_2^n)$ complete? Here $ operatorname{Hol}$ stands for holomorph, and $C_2^n$ stands for direct product of $n$ isomorphic copies of $C^2$.
It for $n = 1$ it is not true, however if $n = 2$, then $ operatorname{Hol}(C_2^n) cong S_4$ is complete (proof that $ operatorname{Hol}(C_2^n) cong S_4$ can be found here: The holomorph of $Z_2 times Z_2$). However, that is based on a specific property of $n = 2$ and does not help us in general case.
One can also easily see, that $ operatorname{Hol}(C_2^n) = C_2^n rtimes operatorname{Aut}(C_2^n)$ is centerless for all $n > 1$, as it $ operatorname{Aut}(C_2^n)$ acts both transitively and effectively on $C_2^n$.
However, determining for what $n$ all automorphisms of $ operatorname{Hol}(C_2^n)$ are inner is a much harder task and I do not know how to solve it.
Any help will be appreciated.
abstract-algebra group-theory finite-groups holomorph complete-groups
asked Feb 3 at 10:54
Yanior WegYanior Weg
2,92121549
$endgroup$
For what $n in mathbb{N}$ is $ operatorname{Hol}(C_2^n)$ complete? Here $ operatorname{Hol}$ stands for holomorph, and $C_2^n$ stands for direct product of $n$ isomorphic copies of $C^2$.
It for $n = 1$ it is not true, however if $n = 2$, then $ operatorname{Hol}(C_2^n) cong S_4$ is complete (proof that $ operatorname{Hol}(C_2^n) cong S_4$ can be found here: The holomorph of $Z_2 times Z_2$). However, that is based on a specific property of $n = 2$ and does not help us in general case.
One can also easily see, that $ operatorname{Hol}(C_2^n) = C_2^n rtimes operatorname{Aut}(C_2^n)$ is centerless for all $n > 1$, as it $ operatorname{Aut}(C_2^n)$ acts both transitively and effectively on $C_2^n$.
However, determining for what $n$ all automorphisms of $ operatorname{Hol}(C_2^n)$ are inner is a much harder task and I do not know how to solve it.
Any help will be appreciated.
abstract-algebra group-theory finite-groups holomorph complete-groups
abstract-algebra group-theory finite-groups holomorph complete-groups
asked Feb 3 at 10:54
Yanior WegYanior Weg
2,92121549
asked Feb 3 at 10:54
Yanior WegYanior Weg
2,92121549
edited Feb 22 at 9:27
Yanior Weg
asked Feb 3 at 10:54
Yanior WegYanior Weg
2,92121549
asked Feb 3 at 10:54
Yanior WegYanior Weg
2,92121549
asked Feb 3 at 10:54
Yanior WegYanior Weg
2,92121549
2,92121549
3
$begingroup$
It is complete for $n ne 3$. In the case $n=3$ there is an outer automorphism of order $2$ arising from the fact that $H^1({rm GL}(n,2),C_2^n)$ (with the natural action on the module) has order $2$ when $n=3$, but is trivial for $n ne 3$. If nobody else does, then I will write out an answer at some stage, but I would feel more inclined to do so if you put some effort into solving the problem yourself, and said where you were stuck. You could also explain the context in which you are trying to solve this problem.
$endgroup$
– Derek Holt
Feb 3 at 13:21
add a comment |
3
$begingroup$
It is complete for $n ne 3$. In the case $n=3$ there is an outer automorphism of order $2$ arising from the fact that $H^1({rm GL}(n,2),C_2^n)$ (with the natural action on the module) has order $2$ when $n=3$, but is trivial for $n ne 3$. If nobody else does, then I will write out an answer at some stage, but I would feel more inclined to do so if you put some effort into solving the problem yourself, and said where you were stuck. You could also explain the context in which you are trying to solve this problem.
$endgroup$
– Derek Holt
Feb 3 at 13:21
3
3
$begingroup$
It is complete for $n ne 3$. In the case $n=3$ there is an outer automorphism of order $2$ arising from the fact that $H^1({rm GL}(n,2),C_2^n)$ (with the natural action on the module) has order $2$ when $n=3$, but is trivial for $n ne 3$. If nobody else does, then I will write out an answer at some stage, but I would feel more inclined to do so if you put some effort into solving the problem yourself, and said where you were stuck. You could also explain the context in which you are trying to solve this problem.
$endgroup$
– Derek Holt
Feb 3 at 13:21
$begingroup$
It is complete for $n ne 3$. In the case $n=3$ there is an outer automorphism of order $2$ arising from the fact that $H^1({rm GL}(n,2),C_2^n)$ (with the natural action on the module) has order $2$ when $n=3$, but is trivial for $n ne 3$. If nobody else does, then I will write out an answer at some stage, but I would feel more inclined to do so if you put some effort into solving the problem yourself, and said where you were stuck. You could also explain the context in which you are trying to solve this problem.
$endgroup$
– Derek Holt
Feb 3 at 13:21
add a comment |
1 Answer
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$begingroup$
Let $G={rm Hol}(C_2^n) = C_2^n rtimes {rm Aut}(C_2^n) = C_2^n rtimes {rm GL}(n,2)$. You have discussed $n=1,2$, so let's assume that $n ge 3$.
Let $alpha$ be an outer automorphism of $G$. The subgroup $C_2^n$ is clearly characteristic in $G$, and hence fixed by $alpha$.
The outer automorphism group of the simple group ${rm GL}(n,2)$ has order $2$, and $A mapsto (A^T)^{-1}$ defines an outer automorphism, for matrices $A$, where $A^T$ is the transpose of $A$. In the action on the natural module $C_2^n$, outer automorphisms interchange stabilizers of subspaces of dimension $r$ with those of dimension $n-r$. So they do not induce actions on $C_2^n$, and hence they do not extend to automorphisms of $G$.
So we may assume that $alpha$ induces the identity on ${rm GL}(n,2)$. Since its action on the natural module is absolutely irreducible, using Schur's Lemma we see that $alpha$ must also induce the identity on $C_2^n$.
So $alpha$ interchanges the complements of $C_2^n$ in $G$. In fact there is a bijection between these complements and the group of automorphisms of $G$ that induce the identity on ${rm GL}(n,2)$, where inner automorphisms correspond to those complements that are conjugate to the principal complement in $G$. So the group of outer automorphisms of $G$ is isomorphic to the cohomology group $H^1({rm GL}(n,2),C_2^n)$.
It is known that this cohomology group has order $2$ for $n=3$ and is trivial for $n>3$. So $G$ is complete if and only if $n ge 4$ (and if $n=2$).
answered Feb 10 at 11:12
Derek HoltDerek Holt
54.6k53574
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add a comment |
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$begingroup$
Let $G={rm Hol}(C_2^n) = C_2^n rtimes {rm Aut}(C_2^n) = C_2^n rtimes {rm GL}(n,2)$. You have discussed $n=1,2$, so let's assume that $n ge 3$.
Let $alpha$ be an outer automorphism of $G$. The subgroup $C_2^n$ is clearly characteristic in $G$, and hence fixed by $alpha$.
The outer automorphism group of the simple group ${rm GL}(n,2)$ has order $2$, and $A mapsto (A^T)^{-1}$ defines an outer automorphism, for matrices $A$, where $A^T$ is the transpose of $A$. In the action on the natural module $C_2^n$, outer automorphisms interchange stabilizers of subspaces of dimension $r$ with those of dimension $n-r$. So they do not induce actions on $C_2^n$, and hence they do not extend to automorphisms of $G$.
So we may assume that $alpha$ induces the identity on ${rm GL}(n,2)$. Since its action on the natural module is absolutely irreducible, using Schur's Lemma we see that $alpha$ must also induce the identity on $C_2^n$.
So $alpha$ interchanges the complements of $C_2^n$ in $G$. In fact there is a bijection between these complements and the group of automorphisms of $G$ that induce the identity on ${rm GL}(n,2)$, where inner automorphisms correspond to those complements that are conjugate to the principal complement in $G$. So the group of outer automorphisms of $G$ is isomorphic to the cohomology group $H^1({rm GL}(n,2),C_2^n)$.
It is known that this cohomology group has order $2$ for $n=3$ and is trivial for $n>3$. So $G$ is complete if and only if $n ge 4$ (and if $n=2$).
answered Feb 10 at 11:12
Derek HoltDerek Holt
54.6k53574
$endgroup$
add a comment |
$begingroup$
Let $G={rm Hol}(C_2^n) = C_2^n rtimes {rm Aut}(C_2^n) = C_2^n rtimes {rm GL}(n,2)$. You have discussed $n=1,2$, so let's assume that $n ge 3$.
Let $alpha$ be an outer automorphism of $G$. The subgroup $C_2^n$ is clearly characteristic in $G$, and hence fixed by $alpha$.
The outer automorphism group of the simple group ${rm GL}(n,2)$ has order $2$, and $A mapsto (A^T)^{-1}$ defines an outer automorphism, for matrices $A$, where $A^T$ is the transpose of $A$. In the action on the natural module $C_2^n$, outer automorphisms interchange stabilizers of subspaces of dimension $r$ with those of dimension $n-r$. So they do not induce actions on $C_2^n$, and hence they do not extend to automorphisms of $G$.
So we may assume that $alpha$ induces the identity on ${rm GL}(n,2)$. Since its action on the natural module is absolutely irreducible, using Schur's Lemma we see that $alpha$ must also induce the identity on $C_2^n$.
So $alpha$ interchanges the complements of $C_2^n$ in $G$. In fact there is a bijection between these complements and the group of automorphisms of $G$ that induce the identity on ${rm GL}(n,2)$, where inner automorphisms correspond to those complements that are conjugate to the principal complement in $G$. So the group of outer automorphisms of $G$ is isomorphic to the cohomology group $H^1({rm GL}(n,2),C_2^n)$.
It is known that this cohomology group has order $2$ for $n=3$ and is trivial for $n>3$. So $G$ is complete if and only if $n ge 4$ (and if $n=2$).
answered Feb 10 at 11:12
Derek HoltDerek Holt
54.6k53574
$endgroup$
add a comment |
$begingroup$
Let $G={rm Hol}(C_2^n) = C_2^n rtimes {rm Aut}(C_2^n) = C_2^n rtimes {rm GL}(n,2)$. You have discussed $n=1,2$, so let's assume that $n ge 3$.
Let $alpha$ be an outer automorphism of $G$. The subgroup $C_2^n$ is clearly characteristic in $G$, and hence fixed by $alpha$.
The outer automorphism group of the simple group ${rm GL}(n,2)$ has order $2$, and $A mapsto (A^T)^{-1}$ defines an outer automorphism, for matrices $A$, where $A^T$ is the transpose of $A$. In the action on the natural module $C_2^n$, outer automorphisms interchange stabilizers of subspaces of dimension $r$ with those of dimension $n-r$. So they do not induce actions on $C_2^n$, and hence they do not extend to automorphisms of $G$.
So we may assume that $alpha$ induces the identity on ${rm GL}(n,2)$. Since its action on the natural module is absolutely irreducible, using Schur's Lemma we see that $alpha$ must also induce the identity on $C_2^n$.
So $alpha$ interchanges the complements of $C_2^n$ in $G$. In fact there is a bijection between these complements and the group of automorphisms of $G$ that induce the identity on ${rm GL}(n,2)$, where inner automorphisms correspond to those complements that are conjugate to the principal complement in $G$. So the group of outer automorphisms of $G$ is isomorphic to the cohomology group $H^1({rm GL}(n,2),C_2^n)$.
It is known that this cohomology group has order $2$ for $n=3$ and is trivial for $n>3$. So $G$ is complete if and only if $n ge 4$ (and if $n=2$).
answered Feb 10 at 11:12
Derek HoltDerek Holt
54.6k53574
$endgroup$
Let $G={rm Hol}(C_2^n) = C_2^n rtimes {rm Aut}(C_2^n) = C_2^n rtimes {rm GL}(n,2)$. You have discussed $n=1,2$, so let's assume that $n ge 3$.
Let $alpha$ be an outer automorphism of $G$. The subgroup $C_2^n$ is clearly characteristic in $G$, and hence fixed by $alpha$.
The outer automorphism group of the simple group ${rm GL}(n,2)$ has order $2$, and $A mapsto (A^T)^{-1}$ defines an outer automorphism, for matrices $A$, where $A^T$ is the transpose of $A$. In the action on the natural module $C_2^n$, outer automorphisms interchange stabilizers of subspaces of dimension $r$ with those of dimension $n-r$. So they do not induce actions on $C_2^n$, and hence they do not extend to automorphisms of $G$.
So we may assume that $alpha$ induces the identity on ${rm GL}(n,2)$. Since its action on the natural module is absolutely irreducible, using Schur's Lemma we see that $alpha$ must also induce the identity on $C_2^n$.
So $alpha$ interchanges the complements of $C_2^n$ in $G$. In fact there is a bijection between these complements and the group of automorphisms of $G$ that induce the identity on ${rm GL}(n,2)$, where inner automorphisms correspond to those complements that are conjugate to the principal complement in $G$. So the group of outer automorphisms of $G$ is isomorphic to the cohomology group $H^1({rm GL}(n,2),C_2^n)$.
It is known that this cohomology group has order $2$ for $n=3$ and is trivial for $n>3$. So $G$ is complete if and only if $n ge 4$ (and if $n=2$).
answered Feb 10 at 11:12
Derek HoltDerek Holt
54.6k53574
answered Feb 10 at 11:12
Derek HoltDerek Holt
54.6k53574
answered Feb 10 at 11:12
Derek HoltDerek Holt
54.6k53574
answered Feb 10 at 11:12
Derek HoltDerek Holt
54.6k53574
54.6k53574
add a comment |
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3
$begingroup$
It is complete for $n ne 3$. In the case $n=3$ there is an outer automorphism of order $2$ arising from the fact that $H^1({rm GL}(n,2),C_2^n)$ (with the natural action on the module) has order $2$ when $n=3$, but is trivial for $n ne 3$. If nobody else does, then I will write out an answer at some stage, but I would feel more inclined to do so if you put some effort into solving the problem yourself, and said where you were stuck. You could also explain the context in which you are trying to solve this problem.
$endgroup$
– Derek Holt
Feb 3 at 13:21