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Why do so many things have the three isomorphism theorems?
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At the moment, I'm in my second year of undergrad and have encountered the isomorphism theorems in groups, rings, and now lie algebras, I've also heard that they pop-up in something called "universal algebra" aswell. Is there any fundamental reason for all of this?
abstract-algebra
asked 2 days ago
user3201708
334
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up vote
2
down vote
favorite
At the moment, I'm in my second year of undergrad and have encountered the isomorphism theorems in groups, rings, and now lie algebras, I've also heard that they pop-up in something called "universal algebra" aswell. Is there any fundamental reason for all of this?
abstract-algebra
asked 2 days ago
user3201708
334
5
As they pop up in universal algebra, it's no wonder they arise in particular algebraic structures too.
– Lord Shark the Unknown
2 days ago
From a bit of googling I've seen that universal algebra is meant to be able to prove things about algebras without having to talk about the specifics of the structures they can have. But what constraints does this have such that it's strong enough to prove the isomorphism theorems but also weak enough to apply to so many things? I've tried finding out on my own, but I've gotten lost fairly quickly when trying to read about them.
– user3201708
2 days ago
2
I think it's essentially that these isomophism theorems (especially the first one, I never remember which is the second and third, and they're consequences of the first one anyway) are tautologies. Let me explain what I mean for the first one : you're saying that if you declare "$x=y$ whenever $f(x)=f(y)$" then your function is bijective onto its image. Well, it was already surjective (onto its image), and your declaration made it injective, so of course it's bijective !
– Max
2 days ago
Ah ok, that makes sense. Thanks @max
– user3201708
2 days ago
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
At the moment, I'm in my second year of undergrad and have encountered the isomorphism theorems in groups, rings, and now lie algebras, I've also heard that they pop-up in something called "universal algebra" aswell. Is there any fundamental reason for all of this?
abstract-algebra
asked 2 days ago
user3201708
334
At the moment, I'm in my second year of undergrad and have encountered the isomorphism theorems in groups, rings, and now lie algebras, I've also heard that they pop-up in something called "universal algebra" aswell. Is there any fundamental reason for all of this?
abstract-algebra
asked 2 days ago
user3201708
334
asked 2 days ago
user3201708
334
asked 2 days ago
user3201708
334
asked 2 days ago
user3201708
334
334
5
As they pop up in universal algebra, it's no wonder they arise in particular algebraic structures too.
– Lord Shark the Unknown
2 days ago
From a bit of googling I've seen that universal algebra is meant to be able to prove things about algebras without having to talk about the specifics of the structures they can have. But what constraints does this have such that it's strong enough to prove the isomorphism theorems but also weak enough to apply to so many things? I've tried finding out on my own, but I've gotten lost fairly quickly when trying to read about them.
– user3201708
2 days ago
2
I think it's essentially that these isomophism theorems (especially the first one, I never remember which is the second and third, and they're consequences of the first one anyway) are tautologies. Let me explain what I mean for the first one : you're saying that if you declare "$x=y$ whenever $f(x)=f(y)$" then your function is bijective onto its image. Well, it was already surjective (onto its image), and your declaration made it injective, so of course it's bijective !
– Max
2 days ago
Ah ok, that makes sense. Thanks @max
– user3201708
2 days ago
add a comment |Â
5
As they pop up in universal algebra, it's no wonder they arise in particular algebraic structures too.
– Lord Shark the Unknown
2 days ago
From a bit of googling I've seen that universal algebra is meant to be able to prove things about algebras without having to talk about the specifics of the structures they can have. But what constraints does this have such that it's strong enough to prove the isomorphism theorems but also weak enough to apply to so many things? I've tried finding out on my own, but I've gotten lost fairly quickly when trying to read about them.
– user3201708
2 days ago
2
I think it's essentially that these isomophism theorems (especially the first one, I never remember which is the second and third, and they're consequences of the first one anyway) are tautologies. Let me explain what I mean for the first one : you're saying that if you declare "$x=y$ whenever $f(x)=f(y)$" then your function is bijective onto its image. Well, it was already surjective (onto its image), and your declaration made it injective, so of course it's bijective !
– Max
2 days ago
Ah ok, that makes sense. Thanks @max
– user3201708
2 days ago
5
5
As they pop up in universal algebra, it's no wonder they arise in particular algebraic structures too.
– Lord Shark the Unknown
2 days ago
As they pop up in universal algebra, it's no wonder they arise in particular algebraic structures too.
– Lord Shark the Unknown
2 days ago
From a bit of googling I've seen that universal algebra is meant to be able to prove things about algebras without having to talk about the specifics of the structures they can have. But what constraints does this have such that it's strong enough to prove the isomorphism theorems but also weak enough to apply to so many things? I've tried finding out on my own, but I've gotten lost fairly quickly when trying to read about them.
– user3201708
2 days ago
From a bit of googling I've seen that universal algebra is meant to be able to prove things about algebras without having to talk about the specifics of the structures they can have. But what constraints does this have such that it's strong enough to prove the isomorphism theorems but also weak enough to apply to so many things? I've tried finding out on my own, but I've gotten lost fairly quickly when trying to read about them.
– user3201708
2 days ago
2
2
I think it's essentially that these isomophism theorems (especially the first one, I never remember which is the second and third, and they're consequences of the first one anyway) are tautologies. Let me explain what I mean for the first one : you're saying that if you declare "$x=y$ whenever $f(x)=f(y)$" then your function is bijective onto its image. Well, it was already surjective (onto its image), and your declaration made it injective, so of course it's bijective !
– Max
2 days ago
I think it's essentially that these isomophism theorems (especially the first one, I never remember which is the second and third, and they're consequences of the first one anyway) are tautologies. Let me explain what I mean for the first one : you're saying that if you declare "$x=y$ whenever $f(x)=f(y)$" then your function is bijective onto its image. Well, it was already surjective (onto its image), and your declaration made it injective, so of course it's bijective !
– Max
2 days ago
Ah ok, that makes sense. Thanks @max
– user3201708
2 days ago
Ah ok, that makes sense. Thanks @max
– user3201708
2 days ago
add a comment |Â
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5
As they pop up in universal algebra, it's no wonder they arise in particular algebraic structures too.
– Lord Shark the Unknown
2 days ago
From a bit of googling I've seen that universal algebra is meant to be able to prove things about algebras without having to talk about the specifics of the structures they can have. But what constraints does this have such that it's strong enough to prove the isomorphism theorems but also weak enough to apply to so many things? I've tried finding out on my own, but I've gotten lost fairly quickly when trying to read about them.
– user3201708
2 days ago
2
I think it's essentially that these isomophism theorems (especially the first one, I never remember which is the second and third, and they're consequences of the first one anyway) are tautologies. Let me explain what I mean for the first one : you're saying that if you declare "$x=y$ whenever $f(x)=f(y)$" then your function is bijective onto its image. Well, it was already surjective (onto its image), and your declaration made it injective, so of course it's bijective !
– Max
2 days ago
Ah ok, that makes sense. Thanks @max
– user3201708
2 days ago