Taking the automorphism group of a group is not functorial.
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Once upon a time I proved that there is no functorial 'association'
$$F: mathbfGrp longrightarrow mathbfGrp: G longmapsto operatornameAut(G).$$
A few days ago I casually mentioned this to someone, and was asked for a proof. Unfortunately I could not and still can not recall how I proved it. Here is how much of my proof I do recall:
Suppose such a functor does exist. Choose some group $G$ wisely, and let $finoperatornameHom(V_4,G)$ and $ginoperatornameHom(G,V_4)$ be such that $gcirc f=operatornameid_V_4$. Then, because $F$ is a co- or contravariant functor we have
$$F(g)circ F(f)=F(gcirc f)=F(operatornameid_V_4)=operatornameid_operatornameAut(V_4),$$
or
$$F(f)circ F(g)=F(gcirc f)=F(operatornameid_V_4)=operatornameid_operatornameAut(V_4),$$
where $operatornameAut(V_4)cong S_3$. In particular $operatornameAut(G)$ contains a subgroup isomorphic to $S_3$. Then something about the order of $operatornameAut(G)$ leads to a contradiction.
I cannot for the life of me find which goup $G$ would do the trick. Any ideas?
abstract-algebra group-theory finite-groups category-theory examples-counterexamples
asked Nov 17 '13 at 9:18
Servaes
1
 |Â
show 7 more comments
up vote
24
down vote
favorite
Once upon a time I proved that there is no functorial 'association'
$$F: mathbfGrp longrightarrow mathbfGrp: G longmapsto operatornameAut(G).$$
A few days ago I casually mentioned this to someone, and was asked for a proof. Unfortunately I could not and still can not recall how I proved it. Here is how much of my proof I do recall:
Suppose such a functor does exist. Choose some group $G$ wisely, and let $finoperatornameHom(V_4,G)$ and $ginoperatornameHom(G,V_4)$ be such that $gcirc f=operatornameid_V_4$. Then, because $F$ is a co- or contravariant functor we have
$$F(g)circ F(f)=F(gcirc f)=F(operatornameid_V_4)=operatornameid_operatornameAut(V_4),$$
or
$$F(f)circ F(g)=F(gcirc f)=F(operatornameid_V_4)=operatornameid_operatornameAut(V_4),$$
where $operatornameAut(V_4)cong S_3$. In particular $operatornameAut(G)$ contains a subgroup isomorphic to $S_3$. Then something about the order of $operatornameAut(G)$ leads to a contradiction.
I cannot for the life of me find which goup $G$ would do the trick. Any ideas?
abstract-algebra group-theory finite-groups category-theory examples-counterexamples
asked Nov 17 '13 at 9:18
Servaes
1
3
There is no obvious candidate for what the induced maps should be. The point is to prove that for any choice of induced maps, the association is not functorial.
– Servaes
Nov 17 '13 at 9:55
2
It's a nice idea, but seems to require a group $G$ with a rather unusual property. It appears you are looking for a group $G$ with a subgroup $H$ such that the inclusion $H hookrightarrow G$ has a retraction, while $mathrmAut(H)$ is a group of greater cardinality than $mathrmAut(G)$.
– Zhen Lin
Nov 17 '13 at 9:57
5
If $NH$ is a semi-direct product, then $H to NH$ is a split monomorphism, and this property is preserved by any functor. Therefore I would try to find an example of a semi-direct product such that there is no split monomorphism $mathrmAut(H) to mathrmAut(NH)$.
– Martin Brandenburg
Nov 17 '13 at 10:44
6
You could take $NH$ to be a Frobenius group of order 56, with $|N|=8$, $|H|=7$. Then $rm Aut(H)$ is cyclic of order 6, whereas $rm Aut(NH) = NHT$ with $T$ cyclic of order 3. So a generator of $rm Aut(H)$ does not extend to an automorphism of $NH$. $rm Aut(NH)$ does have elements of order 6, but they are not complemented.
– Derek Holt
Nov 17 '13 at 11:54
2
@Derek Holt: Your example would do the trick if the identity on $operatornameAut(H)$ cannot factor over $operatornameAut(NH)$. In this case it can, though as far as I know there is no 'nice' way for it to factor. Your general idea is a good one though. I have understood $NHcongoperatornameGA(1,8)$, the group of affine transformations of $BbbF_8$. The group $operatornameGA(1,32)$ does work; we have $operatornameGA(1,32)=Nrtimes H$ with $|N|=32$, $|H|=31$. Then $operatornameAut(H)$ is cyclic of order $30$, and $|operatornameAut(NH)|=NHT$ with $T$ cyclic of order $5$.
– Servaes
Nov 18 '13 at 2:16
 |Â
show 7 more comments
up vote
24
down vote
favorite
up vote
24
down vote
favorite
Once upon a time I proved that there is no functorial 'association'
$$F: mathbfGrp longrightarrow mathbfGrp: G longmapsto operatornameAut(G).$$
A few days ago I casually mentioned this to someone, and was asked for a proof. Unfortunately I could not and still can not recall how I proved it. Here is how much of my proof I do recall:
Suppose such a functor does exist. Choose some group $G$ wisely, and let $finoperatornameHom(V_4,G)$ and $ginoperatornameHom(G,V_4)$ be such that $gcirc f=operatornameid_V_4$. Then, because $F$ is a co- or contravariant functor we have
$$F(g)circ F(f)=F(gcirc f)=F(operatornameid_V_4)=operatornameid_operatornameAut(V_4),$$
or
$$F(f)circ F(g)=F(gcirc f)=F(operatornameid_V_4)=operatornameid_operatornameAut(V_4),$$
where $operatornameAut(V_4)cong S_3$. In particular $operatornameAut(G)$ contains a subgroup isomorphic to $S_3$. Then something about the order of $operatornameAut(G)$ leads to a contradiction.
I cannot for the life of me find which goup $G$ would do the trick. Any ideas?
abstract-algebra group-theory finite-groups category-theory examples-counterexamples
asked Nov 17 '13 at 9:18
Servaes
1
Once upon a time I proved that there is no functorial 'association'
$$F: mathbfGrp longrightarrow mathbfGrp: G longmapsto operatornameAut(G).$$
A few days ago I casually mentioned this to someone, and was asked for a proof. Unfortunately I could not and still can not recall how I proved it. Here is how much of my proof I do recall:
Suppose such a functor does exist. Choose some group $G$ wisely, and let $finoperatornameHom(V_4,G)$ and $ginoperatornameHom(G,V_4)$ be such that $gcirc f=operatornameid_V_4$. Then, because $F$ is a co- or contravariant functor we have
$$F(g)circ F(f)=F(gcirc f)=F(operatornameid_V_4)=operatornameid_operatornameAut(V_4),$$
or
$$F(f)circ F(g)=F(gcirc f)=F(operatornameid_V_4)=operatornameid_operatornameAut(V_4),$$
where $operatornameAut(V_4)cong S_3$. In particular $operatornameAut(G)$ contains a subgroup isomorphic to $S_3$. Then something about the order of $operatornameAut(G)$ leads to a contradiction.
I cannot for the life of me find which goup $G$ would do the trick. Any ideas?
abstract-algebra group-theory finite-groups category-theory examples-counterexamples
asked Nov 17 '13 at 9:18
Servaes
1
edited Sep 22 '15 at 20:13
asked Nov 17 '13 at 9:18
Servaes
1
asked Nov 17 '13 at 9:18
Servaes
1
asked Nov 17 '13 at 9:18
Servaes
1
1
3
There is no obvious candidate for what the induced maps should be. The point is to prove that for any choice of induced maps, the association is not functorial.
– Servaes
Nov 17 '13 at 9:55
2
It's a nice idea, but seems to require a group $G$ with a rather unusual property. It appears you are looking for a group $G$ with a subgroup $H$ such that the inclusion $H hookrightarrow G$ has a retraction, while $mathrmAut(H)$ is a group of greater cardinality than $mathrmAut(G)$.
– Zhen Lin
Nov 17 '13 at 9:57
5
If $NH$ is a semi-direct product, then $H to NH$ is a split monomorphism, and this property is preserved by any functor. Therefore I would try to find an example of a semi-direct product such that there is no split monomorphism $mathrmAut(H) to mathrmAut(NH)$.
– Martin Brandenburg
Nov 17 '13 at 10:44
6
You could take $NH$ to be a Frobenius group of order 56, with $|N|=8$, $|H|=7$. Then $rm Aut(H)$ is cyclic of order 6, whereas $rm Aut(NH) = NHT$ with $T$ cyclic of order 3. So a generator of $rm Aut(H)$ does not extend to an automorphism of $NH$. $rm Aut(NH)$ does have elements of order 6, but they are not complemented.
– Derek Holt
Nov 17 '13 at 11:54
2
@Derek Holt: Your example would do the trick if the identity on $operatornameAut(H)$ cannot factor over $operatornameAut(NH)$. In this case it can, though as far as I know there is no 'nice' way for it to factor. Your general idea is a good one though. I have understood $NHcongoperatornameGA(1,8)$, the group of affine transformations of $BbbF_8$. The group $operatornameGA(1,32)$ does work; we have $operatornameGA(1,32)=Nrtimes H$ with $|N|=32$, $|H|=31$. Then $operatornameAut(H)$ is cyclic of order $30$, and $|operatornameAut(NH)|=NHT$ with $T$ cyclic of order $5$.
– Servaes
Nov 18 '13 at 2:16
 |Â
show 7 more comments
3
There is no obvious candidate for what the induced maps should be. The point is to prove that for any choice of induced maps, the association is not functorial.
– Servaes
Nov 17 '13 at 9:55
2
It's a nice idea, but seems to require a group $G$ with a rather unusual property. It appears you are looking for a group $G$ with a subgroup $H$ such that the inclusion $H hookrightarrow G$ has a retraction, while $mathrmAut(H)$ is a group of greater cardinality than $mathrmAut(G)$.
– Zhen Lin
Nov 17 '13 at 9:57
5
If $NH$ is a semi-direct product, then $H to NH$ is a split monomorphism, and this property is preserved by any functor. Therefore I would try to find an example of a semi-direct product such that there is no split monomorphism $mathrmAut(H) to mathrmAut(NH)$.
– Martin Brandenburg
Nov 17 '13 at 10:44
6
You could take $NH$ to be a Frobenius group of order 56, with $|N|=8$, $|H|=7$. Then $rm Aut(H)$ is cyclic of order 6, whereas $rm Aut(NH) = NHT$ with $T$ cyclic of order 3. So a generator of $rm Aut(H)$ does not extend to an automorphism of $NH$. $rm Aut(NH)$ does have elements of order 6, but they are not complemented.
– Derek Holt
Nov 17 '13 at 11:54
2
@Derek Holt: Your example would do the trick if the identity on $operatornameAut(H)$ cannot factor over $operatornameAut(NH)$. In this case it can, though as far as I know there is no 'nice' way for it to factor. Your general idea is a good one though. I have understood $NHcongoperatornameGA(1,8)$, the group of affine transformations of $BbbF_8$. The group $operatornameGA(1,32)$ does work; we have $operatornameGA(1,32)=Nrtimes H$ with $|N|=32$, $|H|=31$. Then $operatornameAut(H)$ is cyclic of order $30$, and $|operatornameAut(NH)|=NHT$ with $T$ cyclic of order $5$.
– Servaes
Nov 18 '13 at 2:16
3
3
There is no obvious candidate for what the induced maps should be. The point is to prove that for any choice of induced maps, the association is not functorial.
– Servaes
Nov 17 '13 at 9:55
There is no obvious candidate for what the induced maps should be. The point is to prove that for any choice of induced maps, the association is not functorial.
– Servaes
Nov 17 '13 at 9:55
2
2
It's a nice idea, but seems to require a group $G$ with a rather unusual property. It appears you are looking for a group $G$ with a subgroup $H$ such that the inclusion $H hookrightarrow G$ has a retraction, while $mathrmAut(H)$ is a group of greater cardinality than $mathrmAut(G)$.
– Zhen Lin
Nov 17 '13 at 9:57
It's a nice idea, but seems to require a group $G$ with a rather unusual property. It appears you are looking for a group $G$ with a subgroup $H$ such that the inclusion $H hookrightarrow G$ has a retraction, while $mathrmAut(H)$ is a group of greater cardinality than $mathrmAut(G)$.
– Zhen Lin
Nov 17 '13 at 9:57
5
5
If $NH$ is a semi-direct product, then $H to NH$ is a split monomorphism, and this property is preserved by any functor. Therefore I would try to find an example of a semi-direct product such that there is no split monomorphism $mathrmAut(H) to mathrmAut(NH)$.
– Martin Brandenburg
Nov 17 '13 at 10:44
If $NH$ is a semi-direct product, then $H to NH$ is a split monomorphism, and this property is preserved by any functor. Therefore I would try to find an example of a semi-direct product such that there is no split monomorphism $mathrmAut(H) to mathrmAut(NH)$.
– Martin Brandenburg
Nov 17 '13 at 10:44
6
6
You could take $NH$ to be a Frobenius group of order 56, with $|N|=8$, $|H|=7$. Then $rm Aut(H)$ is cyclic of order 6, whereas $rm Aut(NH) = NHT$ with $T$ cyclic of order 3. So a generator of $rm Aut(H)$ does not extend to an automorphism of $NH$. $rm Aut(NH)$ does have elements of order 6, but they are not complemented.
– Derek Holt
Nov 17 '13 at 11:54
You could take $NH$ to be a Frobenius group of order 56, with $|N|=8$, $|H|=7$. Then $rm Aut(H)$ is cyclic of order 6, whereas $rm Aut(NH) = NHT$ with $T$ cyclic of order 3. So a generator of $rm Aut(H)$ does not extend to an automorphism of $NH$. $rm Aut(NH)$ does have elements of order 6, but they are not complemented.
– Derek Holt
Nov 17 '13 at 11:54
2
2
@Derek Holt: Your example would do the trick if the identity on $operatornameAut(H)$ cannot factor over $operatornameAut(NH)$. In this case it can, though as far as I know there is no 'nice' way for it to factor. Your general idea is a good one though. I have understood $NHcongoperatornameGA(1,8)$, the group of affine transformations of $BbbF_8$. The group $operatornameGA(1,32)$ does work; we have $operatornameGA(1,32)=Nrtimes H$ with $|N|=32$, $|H|=31$. Then $operatornameAut(H)$ is cyclic of order $30$, and $|operatornameAut(NH)|=NHT$ with $T$ cyclic of order $5$.
– Servaes
Nov 18 '13 at 2:16
@Derek Holt: Your example would do the trick if the identity on $operatornameAut(H)$ cannot factor over $operatornameAut(NH)$. In this case it can, though as far as I know there is no 'nice' way for it to factor. Your general idea is a good one though. I have understood $NHcongoperatornameGA(1,8)$, the group of affine transformations of $BbbF_8$. The group $operatornameGA(1,32)$ does work; we have $operatornameGA(1,32)=Nrtimes H$ with $|N|=32$, $|H|=31$. Then $operatornameAut(H)$ is cyclic of order $30$, and $|operatornameAut(NH)|=NHT$ with $T$ cyclic of order $5$.
– Servaes
Nov 18 '13 at 2:16
 |Â
show 7 more comments
2 Answers
2
active
oldest
votes
up vote
4
down vote
accepted
Though the question has effectively been answered by the combined comments of Martin Brandenburg and Derek Holt, I thought I'd write up a complete answer for completeness' sake.
Let $N:=BbbF_11$ the finite field of $11$ elements and $H:=BbbF_11^times$ its unit group. Let $G:=Nrtimes H$ the semidirect product of $N$ and $H$ given by the multiplication action of $H$ on $N$. Note that $G$ is isomorphic to the group of affine transformations of $BbbF_11$. Then the maps
$$f: H longrightarrow G: h longmapsto (0,h)qquadtext and qquad g: G longrightarrow H: (n,h) longmapsto h,$$
are group homomorphisms and satisfy $gcirc f=operatornameid_H$. Now suppose there exists a covariant functor
$$F: mathbfGrp longrightarrow mathbfGrp: X longmapsto operatornameAut(X).$$
Then we have group homomorphisms
$$F(f): operatornameAut(H) longrightarrow operatornameAut(G)qquadtext and qquad F(g): operatornameAut(G) longrightarrow operatornameAut(H),$$
satisfying
$$F(g)circ F(f)=F(gcirc f)=F(operatornameid_H)=operatornameid_operatornameAut(H),$$
so the identity on $operatornameAut(H)$ factors over $operatornameAut(G)$, i.e. $operatornameAut(G)$ has a subgroup isomorphic to $operatornameAut(H)$.
We have $operatornameAut(H)congBbbZ/4BbbZ$ because $H$ is abelian of order $10$. By this question we have $operatornameAut(G)cong G$. But $|operatornameAut(G)|=|G|=11times10=110$ is not divisible by $|operatornameAut(H)|=|BbbZ/4BbbZ|=4$, a contradiction. This shows that no such covariant functor exists. The contravariant case is entirely analogous.
answered Aug 12 '15 at 11:31
Servaes
1
add a comment |Â
up vote
0
down vote
Maybe, you can consider $G_1=G$ and $G_2=Goplus H$
Clearly, there are two canonical morphisms, the injection $i:G_1rightarrow G_2$ and the projection $pi:G_2rightarrow G_1$
Now if $Aut$ is a functor, no matter it's contravariant or covariant, a contradiction easily follows if you can construct $G, H$ satisfying
$Aut(G)$ has a generator $g_1$ of order $n$
$Aut(Goplus H)$ has a generator $g_2$ of order $m$
$(n,m)=1$
Then the Lagrange Theorem provides for contradiction
As $picirc i=1_G$, we get
1) covariant case
$mathrmord(Aut(i)(g_1))|mathrmord(g_1)=n$, $mathrmord(Aut(i)(g_1))|m$
Thus, $Aut(i)(g_1)=g_2^0=e_Aut(G_2)$, which contradicts to
$Aut(pi)circ Aut(i)=1_Aut(G)$
2) contravariant case
Change $Aut(i)(g_1)$ to $Aut(pi)(g_1)$ and it's OK !
answered Jun 16 '15 at 16:16
user18537
23016
Yah, that's the problem
– user18537
Jun 17 '15 at 12:51
Generally, if we can make $|Aut(G_1)|=n, |Aut(G_2)|=m$, it also can deduce a contradiction.
– user18537
Jun 17 '15 at 12:54
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
Though the question has effectively been answered by the combined comments of Martin Brandenburg and Derek Holt, I thought I'd write up a complete answer for completeness' sake.
Let $N:=BbbF_11$ the finite field of $11$ elements and $H:=BbbF_11^times$ its unit group. Let $G:=Nrtimes H$ the semidirect product of $N$ and $H$ given by the multiplication action of $H$ on $N$. Note that $G$ is isomorphic to the group of affine transformations of $BbbF_11$. Then the maps
$$f: H longrightarrow G: h longmapsto (0,h)qquadtext and qquad g: G longrightarrow H: (n,h) longmapsto h,$$
are group homomorphisms and satisfy $gcirc f=operatornameid_H$. Now suppose there exists a covariant functor
$$F: mathbfGrp longrightarrow mathbfGrp: X longmapsto operatornameAut(X).$$
Then we have group homomorphisms
$$F(f): operatornameAut(H) longrightarrow operatornameAut(G)qquadtext and qquad F(g): operatornameAut(G) longrightarrow operatornameAut(H),$$
satisfying
$$F(g)circ F(f)=F(gcirc f)=F(operatornameid_H)=operatornameid_operatornameAut(H),$$
so the identity on $operatornameAut(H)$ factors over $operatornameAut(G)$, i.e. $operatornameAut(G)$ has a subgroup isomorphic to $operatornameAut(H)$.
We have $operatornameAut(H)congBbbZ/4BbbZ$ because $H$ is abelian of order $10$. By this question we have $operatornameAut(G)cong G$. But $|operatornameAut(G)|=|G|=11times10=110$ is not divisible by $|operatornameAut(H)|=|BbbZ/4BbbZ|=4$, a contradiction. This shows that no such covariant functor exists. The contravariant case is entirely analogous.
answered Aug 12 '15 at 11:31
Servaes
1
add a comment |Â
up vote
4
down vote
accepted
Though the question has effectively been answered by the combined comments of Martin Brandenburg and Derek Holt, I thought I'd write up a complete answer for completeness' sake.
Let $N:=BbbF_11$ the finite field of $11$ elements and $H:=BbbF_11^times$ its unit group. Let $G:=Nrtimes H$ the semidirect product of $N$ and $H$ given by the multiplication action of $H$ on $N$. Note that $G$ is isomorphic to the group of affine transformations of $BbbF_11$. Then the maps
$$f: H longrightarrow G: h longmapsto (0,h)qquadtext and qquad g: G longrightarrow H: (n,h) longmapsto h,$$
are group homomorphisms and satisfy $gcirc f=operatornameid_H$. Now suppose there exists a covariant functor
$$F: mathbfGrp longrightarrow mathbfGrp: X longmapsto operatornameAut(X).$$
Then we have group homomorphisms
$$F(f): operatornameAut(H) longrightarrow operatornameAut(G)qquadtext and qquad F(g): operatornameAut(G) longrightarrow operatornameAut(H),$$
satisfying
$$F(g)circ F(f)=F(gcirc f)=F(operatornameid_H)=operatornameid_operatornameAut(H),$$
so the identity on $operatornameAut(H)$ factors over $operatornameAut(G)$, i.e. $operatornameAut(G)$ has a subgroup isomorphic to $operatornameAut(H)$.
We have $operatornameAut(H)congBbbZ/4BbbZ$ because $H$ is abelian of order $10$. By this question we have $operatornameAut(G)cong G$. But $|operatornameAut(G)|=|G|=11times10=110$ is not divisible by $|operatornameAut(H)|=|BbbZ/4BbbZ|=4$, a contradiction. This shows that no such covariant functor exists. The contravariant case is entirely analogous.
answered Aug 12 '15 at 11:31
Servaes
1
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
Though the question has effectively been answered by the combined comments of Martin Brandenburg and Derek Holt, I thought I'd write up a complete answer for completeness' sake.
Let $N:=BbbF_11$ the finite field of $11$ elements and $H:=BbbF_11^times$ its unit group. Let $G:=Nrtimes H$ the semidirect product of $N$ and $H$ given by the multiplication action of $H$ on $N$. Note that $G$ is isomorphic to the group of affine transformations of $BbbF_11$. Then the maps
$$f: H longrightarrow G: h longmapsto (0,h)qquadtext and qquad g: G longrightarrow H: (n,h) longmapsto h,$$
are group homomorphisms and satisfy $gcirc f=operatornameid_H$. Now suppose there exists a covariant functor
$$F: mathbfGrp longrightarrow mathbfGrp: X longmapsto operatornameAut(X).$$
Then we have group homomorphisms
$$F(f): operatornameAut(H) longrightarrow operatornameAut(G)qquadtext and qquad F(g): operatornameAut(G) longrightarrow operatornameAut(H),$$
satisfying
$$F(g)circ F(f)=F(gcirc f)=F(operatornameid_H)=operatornameid_operatornameAut(H),$$
so the identity on $operatornameAut(H)$ factors over $operatornameAut(G)$, i.e. $operatornameAut(G)$ has a subgroup isomorphic to $operatornameAut(H)$.
We have $operatornameAut(H)congBbbZ/4BbbZ$ because $H$ is abelian of order $10$. By this question we have $operatornameAut(G)cong G$. But $|operatornameAut(G)|=|G|=11times10=110$ is not divisible by $|operatornameAut(H)|=|BbbZ/4BbbZ|=4$, a contradiction. This shows that no such covariant functor exists. The contravariant case is entirely analogous.
answered Aug 12 '15 at 11:31
Servaes
1
Though the question has effectively been answered by the combined comments of Martin Brandenburg and Derek Holt, I thought I'd write up a complete answer for completeness' sake.
Let $N:=BbbF_11$ the finite field of $11$ elements and $H:=BbbF_11^times$ its unit group. Let $G:=Nrtimes H$ the semidirect product of $N$ and $H$ given by the multiplication action of $H$ on $N$. Note that $G$ is isomorphic to the group of affine transformations of $BbbF_11$. Then the maps
$$f: H longrightarrow G: h longmapsto (0,h)qquadtext and qquad g: G longrightarrow H: (n,h) longmapsto h,$$
are group homomorphisms and satisfy $gcirc f=operatornameid_H$. Now suppose there exists a covariant functor
$$F: mathbfGrp longrightarrow mathbfGrp: X longmapsto operatornameAut(X).$$
Then we have group homomorphisms
$$F(f): operatornameAut(H) longrightarrow operatornameAut(G)qquadtext and qquad F(g): operatornameAut(G) longrightarrow operatornameAut(H),$$
satisfying
$$F(g)circ F(f)=F(gcirc f)=F(operatornameid_H)=operatornameid_operatornameAut(H),$$
so the identity on $operatornameAut(H)$ factors over $operatornameAut(G)$, i.e. $operatornameAut(G)$ has a subgroup isomorphic to $operatornameAut(H)$.
We have $operatornameAut(H)congBbbZ/4BbbZ$ because $H$ is abelian of order $10$. By this question we have $operatornameAut(G)cong G$. But $|operatornameAut(G)|=|G|=11times10=110$ is not divisible by $|operatornameAut(H)|=|BbbZ/4BbbZ|=4$, a contradiction. This shows that no such covariant functor exists. The contravariant case is entirely analogous.
answered Aug 12 '15 at 11:31
Servaes
1
edited Aug 22 at 7:54
answered Aug 12 '15 at 11:31
Servaes
1
answered Aug 12 '15 at 11:31
Servaes
1
answered Aug 12 '15 at 11:31
Servaes
1
1
add a comment |Â
add a comment |Â
up vote
0
down vote
Maybe, you can consider $G_1=G$ and $G_2=Goplus H$
Clearly, there are two canonical morphisms, the injection $i:G_1rightarrow G_2$ and the projection $pi:G_2rightarrow G_1$
Now if $Aut$ is a functor, no matter it's contravariant or covariant, a contradiction easily follows if you can construct $G, H$ satisfying
$Aut(G)$ has a generator $g_1$ of order $n$
$Aut(Goplus H)$ has a generator $g_2$ of order $m$
$(n,m)=1$
Then the Lagrange Theorem provides for contradiction
As $picirc i=1_G$, we get
1) covariant case
$mathrmord(Aut(i)(g_1))|mathrmord(g_1)=n$, $mathrmord(Aut(i)(g_1))|m$
Thus, $Aut(i)(g_1)=g_2^0=e_Aut(G_2)$, which contradicts to
$Aut(pi)circ Aut(i)=1_Aut(G)$
2) contravariant case
Change $Aut(i)(g_1)$ to $Aut(pi)(g_1)$ and it's OK !
answered Jun 16 '15 at 16:16
user18537
23016
Yah, that's the problem
– user18537
Jun 17 '15 at 12:51
Generally, if we can make $|Aut(G_1)|=n, |Aut(G_2)|=m$, it also can deduce a contradiction.
– user18537
Jun 17 '15 at 12:54
add a comment |Â
up vote
0
down vote
Maybe, you can consider $G_1=G$ and $G_2=Goplus H$
Clearly, there are two canonical morphisms, the injection $i:G_1rightarrow G_2$ and the projection $pi:G_2rightarrow G_1$
Now if $Aut$ is a functor, no matter it's contravariant or covariant, a contradiction easily follows if you can construct $G, H$ satisfying
$Aut(G)$ has a generator $g_1$ of order $n$
$Aut(Goplus H)$ has a generator $g_2$ of order $m$
$(n,m)=1$
Then the Lagrange Theorem provides for contradiction
As $picirc i=1_G$, we get
1) covariant case
$mathrmord(Aut(i)(g_1))|mathrmord(g_1)=n$, $mathrmord(Aut(i)(g_1))|m$
Thus, $Aut(i)(g_1)=g_2^0=e_Aut(G_2)$, which contradicts to
$Aut(pi)circ Aut(i)=1_Aut(G)$
2) contravariant case
Change $Aut(i)(g_1)$ to $Aut(pi)(g_1)$ and it's OK !
answered Jun 16 '15 at 16:16
user18537
23016
Yah, that's the problem
– user18537
Jun 17 '15 at 12:51
Generally, if we can make $|Aut(G_1)|=n, |Aut(G_2)|=m$, it also can deduce a contradiction.
– user18537
Jun 17 '15 at 12:54
add a comment |Â
up vote
0
down vote
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Maybe, you can consider $G_1=G$ and $G_2=Goplus H$
Clearly, there are two canonical morphisms, the injection $i:G_1rightarrow G_2$ and the projection $pi:G_2rightarrow G_1$
Now if $Aut$ is a functor, no matter it's contravariant or covariant, a contradiction easily follows if you can construct $G, H$ satisfying
$Aut(G)$ has a generator $g_1$ of order $n$
$Aut(Goplus H)$ has a generator $g_2$ of order $m$
$(n,m)=1$
Then the Lagrange Theorem provides for contradiction
As $picirc i=1_G$, we get
1) covariant case
$mathrmord(Aut(i)(g_1))|mathrmord(g_1)=n$, $mathrmord(Aut(i)(g_1))|m$
Thus, $Aut(i)(g_1)=g_2^0=e_Aut(G_2)$, which contradicts to
$Aut(pi)circ Aut(i)=1_Aut(G)$
2) contravariant case
Change $Aut(i)(g_1)$ to $Aut(pi)(g_1)$ and it's OK !
answered Jun 16 '15 at 16:16
user18537
23016
Maybe, you can consider $G_1=G$ and $G_2=Goplus H$
Clearly, there are two canonical morphisms, the injection $i:G_1rightarrow G_2$ and the projection $pi:G_2rightarrow G_1$
Now if $Aut$ is a functor, no matter it's contravariant or covariant, a contradiction easily follows if you can construct $G, H$ satisfying
$Aut(G)$ has a generator $g_1$ of order $n$
$Aut(Goplus H)$ has a generator $g_2$ of order $m$
$(n,m)=1$
Then the Lagrange Theorem provides for contradiction
As $picirc i=1_G$, we get
1) covariant case
$mathrmord(Aut(i)(g_1))|mathrmord(g_1)=n$, $mathrmord(Aut(i)(g_1))|m$
Thus, $Aut(i)(g_1)=g_2^0=e_Aut(G_2)$, which contradicts to
$Aut(pi)circ Aut(i)=1_Aut(G)$
2) contravariant case
Change $Aut(i)(g_1)$ to $Aut(pi)(g_1)$ and it's OK !
answered Jun 16 '15 at 16:16
user18537
23016
edited Jun 16 '15 at 18:54
answered Jun 16 '15 at 16:16
user18537
23016
answered Jun 16 '15 at 16:16
user18537
23016
answered Jun 16 '15 at 16:16
user18537
23016
23016
Yah, that's the problem
– user18537
Jun 17 '15 at 12:51
Generally, if we can make $|Aut(G_1)|=n, |Aut(G_2)|=m$, it also can deduce a contradiction.
– user18537
Jun 17 '15 at 12:54
add a comment |Â
Yah, that's the problem
– user18537
Jun 17 '15 at 12:51
Generally, if we can make $|Aut(G_1)|=n, |Aut(G_2)|=m$, it also can deduce a contradiction.
– user18537
Jun 17 '15 at 12:54
Yah, that's the problem
– user18537
Jun 17 '15 at 12:51
Yah, that's the problem
– user18537
Jun 17 '15 at 12:51
Generally, if we can make $|Aut(G_1)|=n, |Aut(G_2)|=m$, it also can deduce a contradiction.
– user18537
Jun 17 '15 at 12:54
Generally, if we can make $|Aut(G_1)|=n, |Aut(G_2)|=m$, it also can deduce a contradiction.
– user18537
Jun 17 '15 at 12:54
add a comment |Â
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3
There is no obvious candidate for what the induced maps should be. The point is to prove that for any choice of induced maps, the association is not functorial.
– Servaes
Nov 17 '13 at 9:55
2
It's a nice idea, but seems to require a group $G$ with a rather unusual property. It appears you are looking for a group $G$ with a subgroup $H$ such that the inclusion $H hookrightarrow G$ has a retraction, while $mathrmAut(H)$ is a group of greater cardinality than $mathrmAut(G)$.
– Zhen Lin
Nov 17 '13 at 9:57
5
If $NH$ is a semi-direct product, then $H to NH$ is a split monomorphism, and this property is preserved by any functor. Therefore I would try to find an example of a semi-direct product such that there is no split monomorphism $mathrmAut(H) to mathrmAut(NH)$.
– Martin Brandenburg
Nov 17 '13 at 10:44
6
You could take $NH$ to be a Frobenius group of order 56, with $|N|=8$, $|H|=7$. Then $rm Aut(H)$ is cyclic of order 6, whereas $rm Aut(NH) = NHT$ with $T$ cyclic of order 3. So a generator of $rm Aut(H)$ does not extend to an automorphism of $NH$. $rm Aut(NH)$ does have elements of order 6, but they are not complemented.
– Derek Holt
Nov 17 '13 at 11:54
2
@Derek Holt: Your example would do the trick if the identity on $operatornameAut(H)$ cannot factor over $operatornameAut(NH)$. In this case it can, though as far as I know there is no 'nice' way for it to factor. Your general idea is a good one though. I have understood $NHcongoperatornameGA(1,8)$, the group of affine transformations of $BbbF_8$. The group $operatornameGA(1,32)$ does work; we have $operatornameGA(1,32)=Nrtimes H$ with $|N|=32$, $|H|=31$. Then $operatornameAut(H)$ is cyclic of order $30$, and $|operatornameAut(NH)|=NHT$ with $T$ cyclic of order $5$.
– Servaes
Nov 18 '13 at 2:16