Mixing
group objects in the category of groups and Eckmann–Hilton argument
$begingroup$
Let $mathscr C$ be the category of groups and $Gin mathscr C$. Let
$$m:Gtimes G to G,e:1to G, i:Gto G$$
be associated morphisms satisfying the commutativity diagrams.
I wonder whether it is true that the binary operation $otimes$ induced by $m$ is necessarily the binary operation $circ$ given by the definition of group $G$.
To show this, I am considering using the Eckmann–Hilton argument, but I don't know how to show (and now I am not even sure) that
$$(a otimes b) circ (c otimes d) = (a circ c) otimes (b circ d)$$
by playing with the commutative diagrams (?).
Thanks in advance!
abstract-algebra category-theory
asked Jan 19 at 20:58
No OneNo One
2,0491519
$endgroup$
add a comment |
$begingroup$
Let $mathscr C$ be the category of groups and $Gin mathscr C$. Let
$$m:Gtimes G to G,e:1to G, i:Gto G$$
be associated morphisms satisfying the commutativity diagrams.
I wonder whether it is true that the binary operation $otimes$ induced by $m$ is necessarily the binary operation $circ$ given by the definition of group $G$.
To show this, I am considering using the Eckmann–Hilton argument, but I don't know how to show (and now I am not even sure) that
$$(a otimes b) circ (c otimes d) = (a circ c) otimes (b circ d)$$
by playing with the commutative diagrams (?).
Thanks in advance!
abstract-algebra category-theory
asked Jan 19 at 20:58
No OneNo One
2,0491519
$endgroup$
1
$begingroup$
Hom(H,G) naturally has the structure of a group (i.e., the functor Hom(-,G) factors through the category of groups, which means that G is a group object).
$endgroup$
– Karl Kronenfeld
Jan 19 at 21:05
$begingroup$
@KarlKronenfeld Yes, I can show that $Hom(H,G)$ is a group when $G$ is a group object, but does this help showing that two operations coincide?
$endgroup$
– No One
Jan 19 at 21:08
$begingroup$
I was thinking of a Yoneda's Lemma argument. But can't you just use the fact that m is a group homomorphism to get Eckmann-Hilton going.
$endgroup$
– Karl Kronenfeld
Jan 19 at 21:35
$begingroup$
@KarlKronenfeld Do you think we need any extra conditions on $m$? It's really hard for me to see any relationship between $m$ and $circ$
$endgroup$
– No One
Jan 19 at 21:37
3
$begingroup$
No. $m$ is a morphism from the product group $Gtimes G$ (which can be shown to have the operation $(g,h)(k,l)=(gcirc k,hcirc l)$) to $G$ in the category of groups, and thus is a group homomorphism. By definition $$m(gcirc k,hcirc l)=m((g,h)(k,l)) = m(g,h)circ m(k,l)$$ On the left it is $(gcirc k)otimes (hcirc l)$, on the right it is $(gotimes h)circ (kotimes l)$.
$endgroup$
– Karl Kronenfeld
Jan 19 at 21:46
add a comment |
$begingroup$
Let $mathscr C$ be the category of groups and $Gin mathscr C$. Let
$$m:Gtimes G to G,e:1to G, i:Gto G$$
be associated morphisms satisfying the commutativity diagrams.
I wonder whether it is true that the binary operation $otimes$ induced by $m$ is necessarily the binary operation $circ$ given by the definition of group $G$.
To show this, I am considering using the Eckmann–Hilton argument, but I don't know how to show (and now I am not even sure) that
$$(a otimes b) circ (c otimes d) = (a circ c) otimes (b circ d)$$
by playing with the commutative diagrams (?).
Thanks in advance!
abstract-algebra category-theory
asked Jan 19 at 20:58
No OneNo One
2,0491519
$endgroup$
Let $mathscr C$ be the category of groups and $Gin mathscr C$. Let
$$m:Gtimes G to G,e:1to G, i:Gto G$$
be associated morphisms satisfying the commutativity diagrams.
I wonder whether it is true that the binary operation $otimes$ induced by $m$ is necessarily the binary operation $circ$ given by the definition of group $G$.
To show this, I am considering using the Eckmann–Hilton argument, but I don't know how to show (and now I am not even sure) that
$$(a otimes b) circ (c otimes d) = (a circ c) otimes (b circ d)$$
by playing with the commutative diagrams (?).
Thanks in advance!
abstract-algebra category-theory
abstract-algebra category-theory
asked Jan 19 at 20:58
No OneNo One
2,0491519
asked Jan 19 at 20:58
No OneNo One
2,0491519
edited Jan 19 at 21:35
No One
asked Jan 19 at 20:58
No OneNo One
2,0491519
asked Jan 19 at 20:58
No OneNo One
2,0491519
asked Jan 19 at 20:58
No OneNo One
2,0491519
2,0491519
1
$begingroup$
Hom(H,G) naturally has the structure of a group (i.e., the functor Hom(-,G) factors through the category of groups, which means that G is a group object).
$endgroup$
– Karl Kronenfeld
Jan 19 at 21:05
$begingroup$
@KarlKronenfeld Yes, I can show that $Hom(H,G)$ is a group when $G$ is a group object, but does this help showing that two operations coincide?
$endgroup$
– No One
Jan 19 at 21:08
$begingroup$
I was thinking of a Yoneda's Lemma argument. But can't you just use the fact that m is a group homomorphism to get Eckmann-Hilton going.
$endgroup$
– Karl Kronenfeld
Jan 19 at 21:35
$begingroup$
@KarlKronenfeld Do you think we need any extra conditions on $m$? It's really hard for me to see any relationship between $m$ and $circ$
$endgroup$
– No One
Jan 19 at 21:37
3
$begingroup$
No. $m$ is a morphism from the product group $Gtimes G$ (which can be shown to have the operation $(g,h)(k,l)=(gcirc k,hcirc l)$) to $G$ in the category of groups, and thus is a group homomorphism. By definition $$m(gcirc k,hcirc l)=m((g,h)(k,l)) = m(g,h)circ m(k,l)$$ On the left it is $(gcirc k)otimes (hcirc l)$, on the right it is $(gotimes h)circ (kotimes l)$.
$endgroup$
– Karl Kronenfeld
Jan 19 at 21:46
add a comment |
1
$begingroup$
Hom(H,G) naturally has the structure of a group (i.e., the functor Hom(-,G) factors through the category of groups, which means that G is a group object).
$endgroup$
– Karl Kronenfeld
Jan 19 at 21:05
$begingroup$
@KarlKronenfeld Yes, I can show that $Hom(H,G)$ is a group when $G$ is a group object, but does this help showing that two operations coincide?
$endgroup$
– No One
Jan 19 at 21:08
$begingroup$
I was thinking of a Yoneda's Lemma argument. But can't you just use the fact that m is a group homomorphism to get Eckmann-Hilton going.
$endgroup$
– Karl Kronenfeld
Jan 19 at 21:35
$begingroup$
@KarlKronenfeld Do you think we need any extra conditions on $m$? It's really hard for me to see any relationship between $m$ and $circ$
$endgroup$
– No One
Jan 19 at 21:37
3
$begingroup$
No. $m$ is a morphism from the product group $Gtimes G$ (which can be shown to have the operation $(g,h)(k,l)=(gcirc k,hcirc l)$) to $G$ in the category of groups, and thus is a group homomorphism. By definition $$m(gcirc k,hcirc l)=m((g,h)(k,l)) = m(g,h)circ m(k,l)$$ On the left it is $(gcirc k)otimes (hcirc l)$, on the right it is $(gotimes h)circ (kotimes l)$.
$endgroup$
– Karl Kronenfeld
Jan 19 at 21:46
1
1
$begingroup$
Hom(H,G) naturally has the structure of a group (i.e., the functor Hom(-,G) factors through the category of groups, which means that G is a group object).
$endgroup$
– Karl Kronenfeld
Jan 19 at 21:05
$begingroup$
Hom(H,G) naturally has the structure of a group (i.e., the functor Hom(-,G) factors through the category of groups, which means that G is a group object).
$endgroup$
– Karl Kronenfeld
Jan 19 at 21:05
$begingroup$
@KarlKronenfeld Yes, I can show that $Hom(H,G)$ is a group when $G$ is a group object, but does this help showing that two operations coincide?
$endgroup$
– No One
Jan 19 at 21:08
$begingroup$
@KarlKronenfeld Yes, I can show that $Hom(H,G)$ is a group when $G$ is a group object, but does this help showing that two operations coincide?
$endgroup$
– No One
Jan 19 at 21:08
$begingroup$
I was thinking of a Yoneda's Lemma argument. But can't you just use the fact that m is a group homomorphism to get Eckmann-Hilton going.
$endgroup$
– Karl Kronenfeld
Jan 19 at 21:35
$begingroup$
I was thinking of a Yoneda's Lemma argument. But can't you just use the fact that m is a group homomorphism to get Eckmann-Hilton going.
$endgroup$
– Karl Kronenfeld
Jan 19 at 21:35
$begingroup$
@KarlKronenfeld Do you think we need any extra conditions on $m$? It's really hard for me to see any relationship between $m$ and $circ$
$endgroup$
– No One
Jan 19 at 21:37
$begingroup$
@KarlKronenfeld Do you think we need any extra conditions on $m$? It's really hard for me to see any relationship between $m$ and $circ$
$endgroup$
– No One
Jan 19 at 21:37
3
3
$begingroup$
No. $m$ is a morphism from the product group $Gtimes G$ (which can be shown to have the operation $(g,h)(k,l)=(gcirc k,hcirc l)$) to $G$ in the category of groups, and thus is a group homomorphism. By definition $$m(gcirc k,hcirc l)=m((g,h)(k,l)) = m(g,h)circ m(k,l)$$ On the left it is $(gcirc k)otimes (hcirc l)$, on the right it is $(gotimes h)circ (kotimes l)$.
$endgroup$
– Karl Kronenfeld
Jan 19 at 21:46
$begingroup$
No. $m$ is a morphism from the product group $Gtimes G$ (which can be shown to have the operation $(g,h)(k,l)=(gcirc k,hcirc l)$) to $G$ in the category of groups, and thus is a group homomorphism. By definition $$m(gcirc k,hcirc l)=m((g,h)(k,l)) = m(g,h)circ m(k,l)$$ On the left it is $(gcirc k)otimes (hcirc l)$, on the right it is $(gotimes h)circ (kotimes l)$.
$endgroup$
– Karl Kronenfeld
Jan 19 at 21:46
add a comment |
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1
$begingroup$
Hom(H,G) naturally has the structure of a group (i.e., the functor Hom(-,G) factors through the category of groups, which means that G is a group object).
$endgroup$
– Karl Kronenfeld
Jan 19 at 21:05
$begingroup$
@KarlKronenfeld Yes, I can show that $Hom(H,G)$ is a group when $G$ is a group object, but does this help showing that two operations coincide?
$endgroup$
– No One
Jan 19 at 21:08
$begingroup$
I was thinking of a Yoneda's Lemma argument. But can't you just use the fact that m is a group homomorphism to get Eckmann-Hilton going.
$endgroup$
– Karl Kronenfeld
Jan 19 at 21:35
$begingroup$
@KarlKronenfeld Do you think we need any extra conditions on $m$? It's really hard for me to see any relationship between $m$ and $circ$
$endgroup$
– No One
Jan 19 at 21:37
3
$begingroup$
No. $m$ is a morphism from the product group $Gtimes G$ (which can be shown to have the operation $(g,h)(k,l)=(gcirc k,hcirc l)$) to $G$ in the category of groups, and thus is a group homomorphism. By definition $$m(gcirc k,hcirc l)=m((g,h)(k,l)) = m(g,h)circ m(k,l)$$ On the left it is $(gcirc k)otimes (hcirc l)$, on the right it is $(gotimes h)circ (kotimes l)$.
$endgroup$
– Karl Kronenfeld
Jan 19 at 21:46