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What are the automorphism groups of direct products of dihedral group D4
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What is the automorphism group of direct sum of dihedral group of order $8$, $D_4$?
For example, $mathrmAut(D_4)$ is isomorphic to $D_4$. How about $mathrmAut(D_4times D_4)$, $mathrmAut(D_4times D_4times D_4)$, and $mathrmAut(D_4times D_4 times D_4 times D_4)$?
gr.group-theory finite-groups automorphism-groups
edited Aug 27 at 3:08
Martin Sleziak
2,74432028
asked Aug 26 at 22:43
Sirui Lu
312
add a comment |Â
up vote
6
down vote
favorite
What is the automorphism group of direct sum of dihedral group of order $8$, $D_4$?
For example, $mathrmAut(D_4)$ is isomorphic to $D_4$. How about $mathrmAut(D_4times D_4)$, $mathrmAut(D_4times D_4times D_4)$, and $mathrmAut(D_4times D_4 times D_4 times D_4)$?
gr.group-theory finite-groups automorphism-groups
edited Aug 27 at 3:08
Martin Sleziak
2,74432028
asked Aug 26 at 22:43
Sirui Lu
312
Is $D_4$ the dihedral group of order $8$?
– LSpice
Aug 26 at 23:47
@LSpice Yes it is.
– Sirui Lu
Aug 26 at 23:47
add a comment |Â
up vote
6
down vote
favorite
up vote
6
down vote
favorite
What is the automorphism group of direct sum of dihedral group of order $8$, $D_4$?
For example, $mathrmAut(D_4)$ is isomorphic to $D_4$. How about $mathrmAut(D_4times D_4)$, $mathrmAut(D_4times D_4times D_4)$, and $mathrmAut(D_4times D_4 times D_4 times D_4)$?
gr.group-theory finite-groups automorphism-groups
edited Aug 27 at 3:08
Martin Sleziak
2,74432028
asked Aug 26 at 22:43
Sirui Lu
312
What is the automorphism group of direct sum of dihedral group of order $8$, $D_4$?
For example, $mathrmAut(D_4)$ is isomorphic to $D_4$. How about $mathrmAut(D_4times D_4)$, $mathrmAut(D_4times D_4times D_4)$, and $mathrmAut(D_4times D_4 times D_4 times D_4)$?
gr.group-theory finite-groups automorphism-groups
edited Aug 27 at 3:08
Martin Sleziak
2,74432028
asked Aug 26 at 22:43
Sirui Lu
312
edited Aug 27 at 3:08
Martin Sleziak
2,74432028
edited Aug 27 at 3:08
Martin Sleziak
2,74432028
edited Aug 27 at 3:08
Martin Sleziak
2,74432028
2,74432028
asked Aug 26 at 22:43
Sirui Lu
312
asked Aug 26 at 22:43
Sirui Lu
312
asked Aug 26 at 22:43
Sirui Lu
312
312
Is $D_4$ the dihedral group of order $8$?
– LSpice
Aug 26 at 23:47
@LSpice Yes it is.
– Sirui Lu
Aug 26 at 23:47
add a comment |Â
Is $D_4$ the dihedral group of order $8$?
– LSpice
Aug 26 at 23:47
@LSpice Yes it is.
– Sirui Lu
Aug 26 at 23:47
Is $D_4$ the dihedral group of order $8$?
– LSpice
Aug 26 at 23:47
Is $D_4$ the dihedral group of order $8$?
– LSpice
Aug 26 at 23:47
@LSpice Yes it is.
– Sirui Lu
Aug 26 at 23:47
@LSpice Yes it is.
– Sirui Lu
Aug 26 at 23:47
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
9
down vote
The following papers are relevant:
[1] J. N. S. Bidwell, M. J. Curran, and D. J. McCaughan, Automorphisms of
direct products of finite groups, Arch. Math. 86, 481 – 489 (2006).
[2] J. N. S. Bidwell, Automorphisms of direct products of finite groups II, Arch. Math. 91, 111–121 (2008).
For your question you want to look at [2]. This paper describes the automorphism group of $G = H^n = H times cdots times H$ where $H$ is an indecomposable non-abelian group. In this case $operatornameAut(G)$ has a normal subgroup $mathscrA$ isomorphic to the group formed by the matrices $$left beginpmatrix alpha_11 & cdots & alpha_1n \ vdots & ddots & vdots \ alpha_n1 & cdots & alpha_nnendpmatrix : beginalignalpha_ii &in operatornameAut(H) text for all 1 leq i leq n \ alpha_ij &in operatornameHom(H, Z(H)) text for all i $neq$ j endalignright.$$
(The group operation is matrix multiplication, with multiplication defined by composition and addition defined by $(alpha+beta)(x) = alpha(x)beta(x)$.)
Theorem 3.1 of [2] states that $operatornameAut(G) = mathscrA rtimes S_n$, where $S_n$ is the symmetric group acting on $G$ by permuting the direct factors. Thus $|operatornameAut(G)| = |operatornameAut(H)|^n |operatornameHom(H, Z(H))|^n^2-n n!$
In your case $operatornameAut(H) cong D_4$ and $operatornameHom(H, Z(H)) cong C_2 times C_2$, so $|operatornameAut(G)| = 2^2n^2+n n!$.
edited Aug 27 at 3:18
Martin Sleziak
2,74432028
answered Aug 27 at 2:36
Mikko Korhonen
903812
Thanks for your answer! Can we get things beyond order, like generators of G?
– Sirui Lu
Aug 27 at 2:39
1
@SiruiLu: Not sure what you mean. If you know $operatornameAut(H)$ and $operatornameHom(H, Z(H))$, you have an explicit description of all the automorphisms of $G$, not just the number of them.
– Mikko Korhonen
Aug 27 at 2:53
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
9
down vote
The following papers are relevant:
[1] J. N. S. Bidwell, M. J. Curran, and D. J. McCaughan, Automorphisms of
direct products of finite groups, Arch. Math. 86, 481 – 489 (2006).
[2] J. N. S. Bidwell, Automorphisms of direct products of finite groups II, Arch. Math. 91, 111–121 (2008).
For your question you want to look at [2]. This paper describes the automorphism group of $G = H^n = H times cdots times H$ where $H$ is an indecomposable non-abelian group. In this case $operatornameAut(G)$ has a normal subgroup $mathscrA$ isomorphic to the group formed by the matrices $$left beginpmatrix alpha_11 & cdots & alpha_1n \ vdots & ddots & vdots \ alpha_n1 & cdots & alpha_nnendpmatrix : beginalignalpha_ii &in operatornameAut(H) text for all 1 leq i leq n \ alpha_ij &in operatornameHom(H, Z(H)) text for all i $neq$ j endalignright.$$
(The group operation is matrix multiplication, with multiplication defined by composition and addition defined by $(alpha+beta)(x) = alpha(x)beta(x)$.)
Theorem 3.1 of [2] states that $operatornameAut(G) = mathscrA rtimes S_n$, where $S_n$ is the symmetric group acting on $G$ by permuting the direct factors. Thus $|operatornameAut(G)| = |operatornameAut(H)|^n |operatornameHom(H, Z(H))|^n^2-n n!$
In your case $operatornameAut(H) cong D_4$ and $operatornameHom(H, Z(H)) cong C_2 times C_2$, so $|operatornameAut(G)| = 2^2n^2+n n!$.
edited Aug 27 at 3:18
Martin Sleziak
2,74432028
answered Aug 27 at 2:36
Mikko Korhonen
903812
Thanks for your answer! Can we get things beyond order, like generators of G?
– Sirui Lu
Aug 27 at 2:39
1
@SiruiLu: Not sure what you mean. If you know $operatornameAut(H)$ and $operatornameHom(H, Z(H))$, you have an explicit description of all the automorphisms of $G$, not just the number of them.
– Mikko Korhonen
Aug 27 at 2:53
add a comment |Â
up vote
9
down vote
The following papers are relevant:
[1] J. N. S. Bidwell, M. J. Curran, and D. J. McCaughan, Automorphisms of
direct products of finite groups, Arch. Math. 86, 481 – 489 (2006).
[2] J. N. S. Bidwell, Automorphisms of direct products of finite groups II, Arch. Math. 91, 111–121 (2008).
For your question you want to look at [2]. This paper describes the automorphism group of $G = H^n = H times cdots times H$ where $H$ is an indecomposable non-abelian group. In this case $operatornameAut(G)$ has a normal subgroup $mathscrA$ isomorphic to the group formed by the matrices $$left beginpmatrix alpha_11 & cdots & alpha_1n \ vdots & ddots & vdots \ alpha_n1 & cdots & alpha_nnendpmatrix : beginalignalpha_ii &in operatornameAut(H) text for all 1 leq i leq n \ alpha_ij &in operatornameHom(H, Z(H)) text for all i $neq$ j endalignright.$$
(The group operation is matrix multiplication, with multiplication defined by composition and addition defined by $(alpha+beta)(x) = alpha(x)beta(x)$.)
Theorem 3.1 of [2] states that $operatornameAut(G) = mathscrA rtimes S_n$, where $S_n$ is the symmetric group acting on $G$ by permuting the direct factors. Thus $|operatornameAut(G)| = |operatornameAut(H)|^n |operatornameHom(H, Z(H))|^n^2-n n!$
In your case $operatornameAut(H) cong D_4$ and $operatornameHom(H, Z(H)) cong C_2 times C_2$, so $|operatornameAut(G)| = 2^2n^2+n n!$.
edited Aug 27 at 3:18
Martin Sleziak
2,74432028
answered Aug 27 at 2:36
Mikko Korhonen
903812
Thanks for your answer! Can we get things beyond order, like generators of G?
– Sirui Lu
Aug 27 at 2:39
1
@SiruiLu: Not sure what you mean. If you know $operatornameAut(H)$ and $operatornameHom(H, Z(H))$, you have an explicit description of all the automorphisms of $G$, not just the number of them.
– Mikko Korhonen
Aug 27 at 2:53
add a comment |Â
up vote
9
down vote
up vote
9
down vote
The following papers are relevant:
[1] J. N. S. Bidwell, M. J. Curran, and D. J. McCaughan, Automorphisms of
direct products of finite groups, Arch. Math. 86, 481 – 489 (2006).
[2] J. N. S. Bidwell, Automorphisms of direct products of finite groups II, Arch. Math. 91, 111–121 (2008).
For your question you want to look at [2]. This paper describes the automorphism group of $G = H^n = H times cdots times H$ where $H$ is an indecomposable non-abelian group. In this case $operatornameAut(G)$ has a normal subgroup $mathscrA$ isomorphic to the group formed by the matrices $$left beginpmatrix alpha_11 & cdots & alpha_1n \ vdots & ddots & vdots \ alpha_n1 & cdots & alpha_nnendpmatrix : beginalignalpha_ii &in operatornameAut(H) text for all 1 leq i leq n \ alpha_ij &in operatornameHom(H, Z(H)) text for all i $neq$ j endalignright.$$
(The group operation is matrix multiplication, with multiplication defined by composition and addition defined by $(alpha+beta)(x) = alpha(x)beta(x)$.)
Theorem 3.1 of [2] states that $operatornameAut(G) = mathscrA rtimes S_n$, where $S_n$ is the symmetric group acting on $G$ by permuting the direct factors. Thus $|operatornameAut(G)| = |operatornameAut(H)|^n |operatornameHom(H, Z(H))|^n^2-n n!$
In your case $operatornameAut(H) cong D_4$ and $operatornameHom(H, Z(H)) cong C_2 times C_2$, so $|operatornameAut(G)| = 2^2n^2+n n!$.
edited Aug 27 at 3:18
Martin Sleziak
2,74432028
answered Aug 27 at 2:36
Mikko Korhonen
903812
The following papers are relevant:
[1] J. N. S. Bidwell, M. J. Curran, and D. J. McCaughan, Automorphisms of
direct products of finite groups, Arch. Math. 86, 481 – 489 (2006).
[2] J. N. S. Bidwell, Automorphisms of direct products of finite groups II, Arch. Math. 91, 111–121 (2008).
For your question you want to look at [2]. This paper describes the automorphism group of $G = H^n = H times cdots times H$ where $H$ is an indecomposable non-abelian group. In this case $operatornameAut(G)$ has a normal subgroup $mathscrA$ isomorphic to the group formed by the matrices $$left beginpmatrix alpha_11 & cdots & alpha_1n \ vdots & ddots & vdots \ alpha_n1 & cdots & alpha_nnendpmatrix : beginalignalpha_ii &in operatornameAut(H) text for all 1 leq i leq n \ alpha_ij &in operatornameHom(H, Z(H)) text for all i $neq$ j endalignright.$$
(The group operation is matrix multiplication, with multiplication defined by composition and addition defined by $(alpha+beta)(x) = alpha(x)beta(x)$.)
Theorem 3.1 of [2] states that $operatornameAut(G) = mathscrA rtimes S_n$, where $S_n$ is the symmetric group acting on $G$ by permuting the direct factors. Thus $|operatornameAut(G)| = |operatornameAut(H)|^n |operatornameHom(H, Z(H))|^n^2-n n!$
In your case $operatornameAut(H) cong D_4$ and $operatornameHom(H, Z(H)) cong C_2 times C_2$, so $|operatornameAut(G)| = 2^2n^2+n n!$.
edited Aug 27 at 3:18
Martin Sleziak
2,74432028
answered Aug 27 at 2:36
Mikko Korhonen
903812
edited Aug 27 at 3:18
Martin Sleziak
2,74432028
edited Aug 27 at 3:18
Martin Sleziak
2,74432028
edited Aug 27 at 3:18
Martin Sleziak
2,74432028
2,74432028
answered Aug 27 at 2:36
Mikko Korhonen
903812
answered Aug 27 at 2:36
Mikko Korhonen
903812
answered Aug 27 at 2:36
Mikko Korhonen
903812
903812
Thanks for your answer! Can we get things beyond order, like generators of G?
– Sirui Lu
Aug 27 at 2:39
1
@SiruiLu: Not sure what you mean. If you know $operatornameAut(H)$ and $operatornameHom(H, Z(H))$, you have an explicit description of all the automorphisms of $G$, not just the number of them.
– Mikko Korhonen
Aug 27 at 2:53
add a comment |Â
Thanks for your answer! Can we get things beyond order, like generators of G?
– Sirui Lu
Aug 27 at 2:39
1
@SiruiLu: Not sure what you mean. If you know $operatornameAut(H)$ and $operatornameHom(H, Z(H))$, you have an explicit description of all the automorphisms of $G$, not just the number of them.
– Mikko Korhonen
Aug 27 at 2:53
Thanks for your answer! Can we get things beyond order, like generators of G?
– Sirui Lu
Aug 27 at 2:39
Thanks for your answer! Can we get things beyond order, like generators of G?
– Sirui Lu
Aug 27 at 2:39
1
1
@SiruiLu: Not sure what you mean. If you know $operatornameAut(H)$ and $operatornameHom(H, Z(H))$, you have an explicit description of all the automorphisms of $G$, not just the number of them.
– Mikko Korhonen
Aug 27 at 2:53
@SiruiLu: Not sure what you mean. If you know $operatornameAut(H)$ and $operatornameHom(H, Z(H))$, you have an explicit description of all the automorphisms of $G$, not just the number of them.
– Mikko Korhonen
Aug 27 at 2:53
add a comment |Â
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Is $D_4$ the dihedral group of order $8$?
– LSpice
Aug 26 at 23:47
@LSpice Yes it is.
– Sirui Lu
Aug 26 at 23:47