Mixing
Hom and tensor are isomorphic for finite cyclic groups
$begingroup$
Something I noticed today: $mathsf{Ab}(mathbb{Z}/m, mathbb{Z}/n) cong mathbb{Z}/m mathop{otimes}limits_mathbb{Z} mathbb{Z}/n cong mathbb{Z}/operatorname{gcd}(m,n)$.
I'm not sure there's much deeper going on there, like for instance a nice consequence of the hom/tensor adjunction. I'd say the upshot is that that cyclic groups are really rigid. (Two important things here are that $mathbb{Z}/m to mathbb{Z}/m mathop{otimes}limits_mathbb{Z} mathbb{Z}/n$ is surjective, and everything in sight has to be cyclic, so same order implies isomorphic.
Does anyone have something to add to this? Does this reveal anything cool? Or have connections to anything?
abstract-algebra group-theory tensor-products
asked Dec 27 '16 at 22:13
Eric AuldEric Auld
13.2k432112
$endgroup$
add a comment |
$begingroup$
Something I noticed today: $mathsf{Ab}(mathbb{Z}/m, mathbb{Z}/n) cong mathbb{Z}/m mathop{otimes}limits_mathbb{Z} mathbb{Z}/n cong mathbb{Z}/operatorname{gcd}(m,n)$.
I'm not sure there's much deeper going on there, like for instance a nice consequence of the hom/tensor adjunction. I'd say the upshot is that that cyclic groups are really rigid. (Two important things here are that $mathbb{Z}/m to mathbb{Z}/m mathop{otimes}limits_mathbb{Z} mathbb{Z}/n$ is surjective, and everything in sight has to be cyclic, so same order implies isomorphic.
Does anyone have something to add to this? Does this reveal anything cool? Or have connections to anything?
abstract-algebra group-theory tensor-products
asked Dec 27 '16 at 22:13
Eric AuldEric Auld
13.2k432112
$endgroup$
$begingroup$
Ab being the abelianization?
$endgroup$
– Zelos Malum
Dec 28 '16 at 6:56
$begingroup$
@ZelosMalum Ab(A,B) meaning the mapping space.
$endgroup$
– Eric Auld
Dec 28 '16 at 7:00
$begingroup$
Do you mean Hom functor then?
$endgroup$
– Zelos Malum
Dec 28 '16 at 7:21
$begingroup$
@ZelosMalum Yes. In a category $mathscr{C}$, another notation for $operatorname{Hom}_mathscr{C}(a,b)$ is $mathscr{C}(a,b)$.
$endgroup$
– Eric Auld
Dec 28 '16 at 17:21
$begingroup$
I was not familiar with it, only seen "Ab" for Abelianization
$endgroup$
– Zelos Malum
Dec 28 '16 at 17:22
add a comment |
$begingroup$
Something I noticed today: $mathsf{Ab}(mathbb{Z}/m, mathbb{Z}/n) cong mathbb{Z}/m mathop{otimes}limits_mathbb{Z} mathbb{Z}/n cong mathbb{Z}/operatorname{gcd}(m,n)$.
I'm not sure there's much deeper going on there, like for instance a nice consequence of the hom/tensor adjunction. I'd say the upshot is that that cyclic groups are really rigid. (Two important things here are that $mathbb{Z}/m to mathbb{Z}/m mathop{otimes}limits_mathbb{Z} mathbb{Z}/n$ is surjective, and everything in sight has to be cyclic, so same order implies isomorphic.
Does anyone have something to add to this? Does this reveal anything cool? Or have connections to anything?
abstract-algebra group-theory tensor-products
asked Dec 27 '16 at 22:13
Eric AuldEric Auld
13.2k432112
$endgroup$
Something I noticed today: $mathsf{Ab}(mathbb{Z}/m, mathbb{Z}/n) cong mathbb{Z}/m mathop{otimes}limits_mathbb{Z} mathbb{Z}/n cong mathbb{Z}/operatorname{gcd}(m,n)$.
I'm not sure there's much deeper going on there, like for instance a nice consequence of the hom/tensor adjunction. I'd say the upshot is that that cyclic groups are really rigid. (Two important things here are that $mathbb{Z}/m to mathbb{Z}/m mathop{otimes}limits_mathbb{Z} mathbb{Z}/n$ is surjective, and everything in sight has to be cyclic, so same order implies isomorphic.
Does anyone have something to add to this? Does this reveal anything cool? Or have connections to anything?
abstract-algebra group-theory tensor-products
abstract-algebra group-theory tensor-products
asked Dec 27 '16 at 22:13
Eric AuldEric Auld
13.2k432112
asked Dec 27 '16 at 22:13
Eric AuldEric Auld
13.2k432112
asked Dec 27 '16 at 22:13
Eric AuldEric Auld
13.2k432112
asked Dec 27 '16 at 22:13
Eric AuldEric Auld
13.2k432112
asked Dec 27 '16 at 22:13
Eric AuldEric Auld
13.2k432112
13.2k432112
$begingroup$
Ab being the abelianization?
$endgroup$
– Zelos Malum
Dec 28 '16 at 6:56
$begingroup$
@ZelosMalum Ab(A,B) meaning the mapping space.
$endgroup$
– Eric Auld
Dec 28 '16 at 7:00
$begingroup$
Do you mean Hom functor then?
$endgroup$
– Zelos Malum
Dec 28 '16 at 7:21
$begingroup$
@ZelosMalum Yes. In a category $mathscr{C}$, another notation for $operatorname{Hom}_mathscr{C}(a,b)$ is $mathscr{C}(a,b)$.
$endgroup$
– Eric Auld
Dec 28 '16 at 17:21
$begingroup$
I was not familiar with it, only seen "Ab" for Abelianization
$endgroup$
– Zelos Malum
Dec 28 '16 at 17:22
add a comment |
$begingroup$
Ab being the abelianization?
$endgroup$
– Zelos Malum
Dec 28 '16 at 6:56
$begingroup$
@ZelosMalum Ab(A,B) meaning the mapping space.
$endgroup$
– Eric Auld
Dec 28 '16 at 7:00
$begingroup$
Do you mean Hom functor then?
$endgroup$
– Zelos Malum
Dec 28 '16 at 7:21
$begingroup$
@ZelosMalum Yes. In a category $mathscr{C}$, another notation for $operatorname{Hom}_mathscr{C}(a,b)$ is $mathscr{C}(a,b)$.
$endgroup$
– Eric Auld
Dec 28 '16 at 17:21
$begingroup$
I was not familiar with it, only seen "Ab" for Abelianization
$endgroup$
– Zelos Malum
Dec 28 '16 at 17:22
$begingroup$
Ab being the abelianization?
$endgroup$
– Zelos Malum
Dec 28 '16 at 6:56
$begingroup$
Ab being the abelianization?
$endgroup$
– Zelos Malum
Dec 28 '16 at 6:56
$begingroup$
@ZelosMalum Ab(A,B) meaning the mapping space.
$endgroup$
– Eric Auld
Dec 28 '16 at 7:00
$begingroup$
@ZelosMalum Ab(A,B) meaning the mapping space.
$endgroup$
– Eric Auld
Dec 28 '16 at 7:00
$begingroup$
Do you mean Hom functor then?
$endgroup$
– Zelos Malum
Dec 28 '16 at 7:21
$begingroup$
Do you mean Hom functor then?
$endgroup$
– Zelos Malum
Dec 28 '16 at 7:21
$begingroup$
@ZelosMalum Yes. In a category $mathscr{C}$, another notation for $operatorname{Hom}_mathscr{C}(a,b)$ is $mathscr{C}(a,b)$.
$endgroup$
– Eric Auld
Dec 28 '16 at 17:21
$begingroup$
@ZelosMalum Yes. In a category $mathscr{C}$, another notation for $operatorname{Hom}_mathscr{C}(a,b)$ is $mathscr{C}(a,b)$.
$endgroup$
– Eric Auld
Dec 28 '16 at 17:21
$begingroup$
I was not familiar with it, only seen "Ab" for Abelianization
$endgroup$
– Zelos Malum
Dec 28 '16 at 17:22
$begingroup$
I was not familiar with it, only seen "Ab" for Abelianization
$endgroup$
– Zelos Malum
Dec 28 '16 at 17:22
add a comment |
1 Answer
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$begingroup$
I know the question is old, but was wondering the same thing. A partial answer is that the Hom/tensor adjunction reduces the problem of showing, say, $mathrm{Hom}_{mathbb Z} (mathbb Z/m ,mathbb Z/n)cong mathbb Z /(m,n)$ if we know the corresponding statement about tensor. For short, let $d = (m,n)$. Then
begin{align*}
mathrm{Hom}_{mathbb Z}(mathbb Z /m,mathbb Z/n)&cong mathrm{Hom}_{mathbb Z}(mathbb Z/m,mathrm{Hom}_{mathbb Z}(mathbb Z/n,mathbb Z/n)) \
&cong mathrm{Hom}_{mathbb Z}(mathbb Z /motimes_{mathbb Z}mathbb Z/n,mathbb Z/n)\
&cong mathrm{Hom}_{mathbb Z}(mathbb Z/d,mathbb Z/n)\
&cong mathbb Z/d.
end{align*}
The first isomorphism is trivial and the second is the Hom/tensor adjunction. The final isomorphism is certainly not as bad in the general case (since here $d$ divides $n$).
Similar tricks probably work in the opposite direction, i.e. going from the statement about Hom to the one about tensor.
answered Jan 18 at 11:17
o.h.o.h.
4616
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1 Answer
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1 Answer
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$begingroup$
I know the question is old, but was wondering the same thing. A partial answer is that the Hom/tensor adjunction reduces the problem of showing, say, $mathrm{Hom}_{mathbb Z} (mathbb Z/m ,mathbb Z/n)cong mathbb Z /(m,n)$ if we know the corresponding statement about tensor. For short, let $d = (m,n)$. Then
begin{align*}
mathrm{Hom}_{mathbb Z}(mathbb Z /m,mathbb Z/n)&cong mathrm{Hom}_{mathbb Z}(mathbb Z/m,mathrm{Hom}_{mathbb Z}(mathbb Z/n,mathbb Z/n)) \
&cong mathrm{Hom}_{mathbb Z}(mathbb Z /motimes_{mathbb Z}mathbb Z/n,mathbb Z/n)\
&cong mathrm{Hom}_{mathbb Z}(mathbb Z/d,mathbb Z/n)\
&cong mathbb Z/d.
end{align*}
The first isomorphism is trivial and the second is the Hom/tensor adjunction. The final isomorphism is certainly not as bad in the general case (since here $d$ divides $n$).
Similar tricks probably work in the opposite direction, i.e. going from the statement about Hom to the one about tensor.
answered Jan 18 at 11:17
o.h.o.h.
4616
$endgroup$
add a comment |
$begingroup$
I know the question is old, but was wondering the same thing. A partial answer is that the Hom/tensor adjunction reduces the problem of showing, say, $mathrm{Hom}_{mathbb Z} (mathbb Z/m ,mathbb Z/n)cong mathbb Z /(m,n)$ if we know the corresponding statement about tensor. For short, let $d = (m,n)$. Then
begin{align*}
mathrm{Hom}_{mathbb Z}(mathbb Z /m,mathbb Z/n)&cong mathrm{Hom}_{mathbb Z}(mathbb Z/m,mathrm{Hom}_{mathbb Z}(mathbb Z/n,mathbb Z/n)) \
&cong mathrm{Hom}_{mathbb Z}(mathbb Z /motimes_{mathbb Z}mathbb Z/n,mathbb Z/n)\
&cong mathrm{Hom}_{mathbb Z}(mathbb Z/d,mathbb Z/n)\
&cong mathbb Z/d.
end{align*}
The first isomorphism is trivial and the second is the Hom/tensor adjunction. The final isomorphism is certainly not as bad in the general case (since here $d$ divides $n$).
Similar tricks probably work in the opposite direction, i.e. going from the statement about Hom to the one about tensor.
answered Jan 18 at 11:17
o.h.o.h.
4616
$endgroup$
add a comment |
$begingroup$
I know the question is old, but was wondering the same thing. A partial answer is that the Hom/tensor adjunction reduces the problem of showing, say, $mathrm{Hom}_{mathbb Z} (mathbb Z/m ,mathbb Z/n)cong mathbb Z /(m,n)$ if we know the corresponding statement about tensor. For short, let $d = (m,n)$. Then
begin{align*}
mathrm{Hom}_{mathbb Z}(mathbb Z /m,mathbb Z/n)&cong mathrm{Hom}_{mathbb Z}(mathbb Z/m,mathrm{Hom}_{mathbb Z}(mathbb Z/n,mathbb Z/n)) \
&cong mathrm{Hom}_{mathbb Z}(mathbb Z /motimes_{mathbb Z}mathbb Z/n,mathbb Z/n)\
&cong mathrm{Hom}_{mathbb Z}(mathbb Z/d,mathbb Z/n)\
&cong mathbb Z/d.
end{align*}
The first isomorphism is trivial and the second is the Hom/tensor adjunction. The final isomorphism is certainly not as bad in the general case (since here $d$ divides $n$).
Similar tricks probably work in the opposite direction, i.e. going from the statement about Hom to the one about tensor.
answered Jan 18 at 11:17
o.h.o.h.
4616
$endgroup$
I know the question is old, but was wondering the same thing. A partial answer is that the Hom/tensor adjunction reduces the problem of showing, say, $mathrm{Hom}_{mathbb Z} (mathbb Z/m ,mathbb Z/n)cong mathbb Z /(m,n)$ if we know the corresponding statement about tensor. For short, let $d = (m,n)$. Then
begin{align*}
mathrm{Hom}_{mathbb Z}(mathbb Z /m,mathbb Z/n)&cong mathrm{Hom}_{mathbb Z}(mathbb Z/m,mathrm{Hom}_{mathbb Z}(mathbb Z/n,mathbb Z/n)) \
&cong mathrm{Hom}_{mathbb Z}(mathbb Z /motimes_{mathbb Z}mathbb Z/n,mathbb Z/n)\
&cong mathrm{Hom}_{mathbb Z}(mathbb Z/d,mathbb Z/n)\
&cong mathbb Z/d.
end{align*}
The first isomorphism is trivial and the second is the Hom/tensor adjunction. The final isomorphism is certainly not as bad in the general case (since here $d$ divides $n$).
Similar tricks probably work in the opposite direction, i.e. going from the statement about Hom to the one about tensor.
answered Jan 18 at 11:17
o.h.o.h.
4616
edited Jan 20 at 20:31
answered Jan 18 at 11:17
o.h.o.h.
4616
answered Jan 18 at 11:17
o.h.o.h.
4616
answered Jan 18 at 11:17
o.h.o.h.
4616
4616
add a comment |
add a comment |
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$begingroup$
Ab being the abelianization?
$endgroup$
– Zelos Malum
Dec 28 '16 at 6:56
$begingroup$
@ZelosMalum Ab(A,B) meaning the mapping space.
$endgroup$
– Eric Auld
Dec 28 '16 at 7:00
$begingroup$
Do you mean Hom functor then?
$endgroup$
– Zelos Malum
Dec 28 '16 at 7:21
$begingroup$
@ZelosMalum Yes. In a category $mathscr{C}$, another notation for $operatorname{Hom}_mathscr{C}(a,b)$ is $mathscr{C}(a,b)$.
$endgroup$
– Eric Auld
Dec 28 '16 at 17:21
$begingroup$
I was not familiar with it, only seen "Ab" for Abelianization
$endgroup$
– Zelos Malum
Dec 28 '16 at 17:22