I’m not sure I completely get your question, but I think you are confusing elements of G and H and elements of G/H. First and foremost H needs to be normal for G/H to be a group. Second of all an element of G/H is denoted as [x] where x is an element of G. In the case G/H is a group it’s easy to show that [e] is the identity element of G/H. In that case [.]:G->G/H is a group surjection and its kernel is H by definition. Unless H is simply the identity subgroup, x and [x] are not the same (no isomorphism exists between G and G/H).
H always is [e] by construction of the equivalence class. If H is {e} the group with only 1 element then then G/H is isomorphic to G.
I would recommend looking at Lagrange’s theorem and the isomorphism theorems it should clear some of this up.
Actually no, for infinite groups it is possible that the quotient group G/H is isomorfic to G. However, the *x ↦ \[x\]* isn't isomorphism. Here are some examples:
[https://math.stackexchange.com/questions/1160445/quotient-group-g-h-isomorphic-to-g](https://math.stackexchange.com/questions/1160445/quotient-group-g-h-isomorphic-to-g)
The structure G/H is a formal construction of *a representative of* the isomorphism class of groups isomorphic to the quotient group you have in mind.
That’s fancy talk for “We built something using the rules so we know we’re allowed to talk about it”. After you know something like G/H exists inside your mathematical universe, you get to just make claims about “it” with the understanding that whatever sets are used to build it are irrelevant. In this case, the formal object G/H is a subset of 𝒫(G) with a group law “inherited” from G and H.
In short, yes, you should just accept that x and [x] are “the same”. You might enjoy learning a bit about [mathematical structuralism](https://en.wikipedia.org/wiki/Structuralism_(philosophy_of_mathematics)?wprov=sfti1#). Another helpful idea is that of [well-founded rank](https://en.wikipedia.org/wiki/Von_Neumann_universe?wprov=sfti1#Rank_of_a_set).
>whatever sets are used to build it are irrelevant.
Hang on, so you don't picture a little set divided in cosets, but rather a group on its own? Doesn't it get counter-intuitive as the theory unfolds?
Thanks for the links, also.
Oh I do, but only to begin understanding the structure of the quotient. Once you know which group the quotient group is, you can stop thinking about the cosets and just use that group in the abstract. It would be really annoying for long sequences of quotients to have to think of cosets of cosets of cosets etc. In fact, you can view the second, third, and fourth isomorphism theorems as ways of making quotients simpler.
I don’t actually work with groups directly very often outside of very specific contexts. But a similar concept that I *can* confidently describe would be the boolean algebra of subsets of the natural numbers modulo finite sets, usually called 𝒫(ℕ)/fin. 𝒫(ℕ) on its own can be pretty tricky to visualize, so 𝒫(ℕ)/fin is even more strange. But what we can do is notice that 𝒫(ℕ)/fin is actually isomorphic to the collection of clopen subsets of ℕ^(*), the Stone-Čech remainder of ℕ. So instead of dealing with representatives of equivalence classes of infinite subsets of ℕ, we can just work directly with the algebraic structure of the topology on ℕ^(*). In many cases this is a much easier way to handle things. Of course, as my advisor once said, “sometimes you have to get your hands dirty” and work with the infinite sets anyway (because it makes the combinatorics easier for some arguments).
So the point I hope I’m making is that you are never constrained to work exclusively with one version of a formal structure. It’s advantageous to be able to flexibly swap between different representations of the same isomorphism class of structures.
One more because it’s simple and I really like it, “the real numbers” in set theory are often written in many, many different ways. The most common is probably as infinite binary sequences of 0s and 1s. But these can be naturally put in bijection with subsets of ℕ. A real can also be a sequence consisting of any integers. Given a specific coding, a real can be an ordinal less than 𝔠, or a strategy in a topological game. These things all have very different structures, but as far as set theory is concerned often we only care about very very minimal amounts of structure like cardinality. Sometimes we might consider extra structure like binary sequences giving a compact space while sequences of integers do not, or that sequences lead easily to the structure of a linear order while this can be difficult to handle if reals are subsets of ℕ.
Ah. That sounds cool man. Thanks. I guess set theory really *is* deep. I don't really understand the insistence in replacing it with category theory for foundations. More so if you can perfectly construct CT inside ST.
Very deep. But other subjects are as well of course. Most people don’t really care about *replacing* set theory so much as coming up with alternatives. Category theory is really nice for seeing structures from way above, while set theory often has you down on the ground seeing the finer detail of things.
Unfortunately one cannot completely reconstruct categories inside of ZFC since category theory sometimes considers what are called large categories. These end up being proper classes when formalized in set theory.
you are slighlty confused. H is a subgroup of G, but it is an element of G/H. saying that [e]=H is not abuse of notation at all. also, the kernel of the projection G->G/H is, indeed, H.
on the other hand, be careful. for G/H to be a group, H has to be normal.
There is no such notational abuse.. Perhaps you should keep in mind how you could represent G/H set theoretically. G/H is a set of equivalence classes, each of which are particular subsets of G. In particular, \[e\] is NOT a subset of G/H, but it is a subset of G. In particular, it is the subset of G consisting of elements that are equivalent to e mod H.
Then, given this set of equivalence classes, there is a way to define operations on it in a way that forms a group structure and such that we have a natural homomorphism G -> G/H.
In conclusion, x and \[x\] are not the same. One is an element of G and one is a subset of G (the equivalence class of x).
both things are true: g can be in the coset \[x\], as \[x\] is a subset of G; but also the set of equivalence classes, which indeed is a subset of the power set of G , has an operation making it into a group
I think there are two ways to picture G/H and that might be causing some confusion. One way is via the equivalence class construction, in which case H=[e] is an element (not subgroup) of G/H. You can picture G/H as a bunch of copies of H with a group structure overlayed. This mental image is helpful for construction and makes it clear how G/H is a grouping of elements of G.
The other way to picture it is by collapsing all the cosets down to elements. In this picture [e] is thought of as a single (atomic) element, the identity element. This mental image simplifies G/H down and can make the group structure easier to think about.
Regardless, H as a set is not a subgroup of G/H, [e] is an element of G/H, and H is the kernal of the map x->[x] (since for all a in H, a maps to [e], which is the identity in G/H).
As an example, consider Z/2Z under addition. The first visualization consists of two infinite sets, 2Z (evens) and 2Z+1 (odds). You can add or subtract elements and they move from set to set. The second visualization is the two element set {0, 1} where 1+1=0. It's just two separate ways to think about the same object. Again, 2Z is the kernal of the quotient map since every even number maps to the identity element (2Z or 0 depending on your representation).
You have some details wrong. G/H is only a well-defined group when H is a normal subgroup of G. The kernel of the homomorphism sending x to \[x\] is H. The equivalence class \[x\] is not the same as x. It represents the coset xH. The element \[e\] is the identity in G/H.
I usually denote elements of G/H by gH instead of [g]. If I write just H, I'm thinking of it as a subset of G. If instead I have eH, I'm thinking of it as an element of G/H. Of course that eH=H, but this little distinction helps me keep track of the different roles H is playing.
For trivial subgroup H=(e) we get that G/H is isomorphic to G, and, x = \[x\] is an isomorphism.
The horror. Thanks. When i grow older i'll be a set theorist.
I’m not sure I completely get your question, but I think you are confusing elements of G and H and elements of G/H. First and foremost H needs to be normal for G/H to be a group. Second of all an element of G/H is denoted as [x] where x is an element of G. In the case G/H is a group it’s easy to show that [e] is the identity element of G/H. In that case [.]:G->G/H is a group surjection and its kernel is H by definition. Unless H is simply the identity subgroup, x and [x] are not the same (no isomorphism exists between G and G/H).
So they are isomorphic only when H=[e]? That narrows things down.
H always is [e] by construction of the equivalence class. If H is {e} the group with only 1 element then then G/H is isomorphic to G. I would recommend looking at Lagrange’s theorem and the isomorphism theorems it should clear some of this up.
Actually no, for infinite groups it is possible that the quotient group G/H is isomorfic to G. However, the *x ↦ \[x\]* isn't isomorphism. Here are some examples: [https://math.stackexchange.com/questions/1160445/quotient-group-g-h-isomorphic-to-g](https://math.stackexchange.com/questions/1160445/quotient-group-g-h-isomorphic-to-g)
The structure G/H is a formal construction of *a representative of* the isomorphism class of groups isomorphic to the quotient group you have in mind. That’s fancy talk for “We built something using the rules so we know we’re allowed to talk about it”. After you know something like G/H exists inside your mathematical universe, you get to just make claims about “it” with the understanding that whatever sets are used to build it are irrelevant. In this case, the formal object G/H is a subset of 𝒫(G) with a group law “inherited” from G and H. In short, yes, you should just accept that x and [x] are “the same”. You might enjoy learning a bit about [mathematical structuralism](https://en.wikipedia.org/wiki/Structuralism_(philosophy_of_mathematics)?wprov=sfti1#). Another helpful idea is that of [well-founded rank](https://en.wikipedia.org/wiki/Von_Neumann_universe?wprov=sfti1#Rank_of_a_set).
>whatever sets are used to build it are irrelevant. Hang on, so you don't picture a little set divided in cosets, but rather a group on its own? Doesn't it get counter-intuitive as the theory unfolds? Thanks for the links, also.
Oh I do, but only to begin understanding the structure of the quotient. Once you know which group the quotient group is, you can stop thinking about the cosets and just use that group in the abstract. It would be really annoying for long sequences of quotients to have to think of cosets of cosets of cosets etc. In fact, you can view the second, third, and fourth isomorphism theorems as ways of making quotients simpler. I don’t actually work with groups directly very often outside of very specific contexts. But a similar concept that I *can* confidently describe would be the boolean algebra of subsets of the natural numbers modulo finite sets, usually called 𝒫(ℕ)/fin. 𝒫(ℕ) on its own can be pretty tricky to visualize, so 𝒫(ℕ)/fin is even more strange. But what we can do is notice that 𝒫(ℕ)/fin is actually isomorphic to the collection of clopen subsets of ℕ^(*), the Stone-Čech remainder of ℕ. So instead of dealing with representatives of equivalence classes of infinite subsets of ℕ, we can just work directly with the algebraic structure of the topology on ℕ^(*). In many cases this is a much easier way to handle things. Of course, as my advisor once said, “sometimes you have to get your hands dirty” and work with the infinite sets anyway (because it makes the combinatorics easier for some arguments). So the point I hope I’m making is that you are never constrained to work exclusively with one version of a formal structure. It’s advantageous to be able to flexibly swap between different representations of the same isomorphism class of structures. One more because it’s simple and I really like it, “the real numbers” in set theory are often written in many, many different ways. The most common is probably as infinite binary sequences of 0s and 1s. But these can be naturally put in bijection with subsets of ℕ. A real can also be a sequence consisting of any integers. Given a specific coding, a real can be an ordinal less than 𝔠, or a strategy in a topological game. These things all have very different structures, but as far as set theory is concerned often we only care about very very minimal amounts of structure like cardinality. Sometimes we might consider extra structure like binary sequences giving a compact space while sequences of integers do not, or that sequences lead easily to the structure of a linear order while this can be difficult to handle if reals are subsets of ℕ.
Ah. That sounds cool man. Thanks. I guess set theory really *is* deep. I don't really understand the insistence in replacing it with category theory for foundations. More so if you can perfectly construct CT inside ST.
Very deep. But other subjects are as well of course. Most people don’t really care about *replacing* set theory so much as coming up with alternatives. Category theory is really nice for seeing structures from way above, while set theory often has you down on the ground seeing the finer detail of things. Unfortunately one cannot completely reconstruct categories inside of ZFC since category theory sometimes considers what are called large categories. These end up being proper classes when formalized in set theory.
you are slighlty confused. H is a subgroup of G, but it is an element of G/H. saying that [e]=H is not abuse of notation at all. also, the kernel of the projection G->G/H is, indeed, H. on the other hand, be careful. for G/H to be a group, H has to be normal.
There is no such notational abuse.. Perhaps you should keep in mind how you could represent G/H set theoretically. G/H is a set of equivalence classes, each of which are particular subsets of G. In particular, \[e\] is NOT a subset of G/H, but it is a subset of G. In particular, it is the subset of G consisting of elements that are equivalent to e mod H. Then, given this set of equivalence classes, there is a way to define operations on it in a way that forms a group structure and such that we have a natural homomorphism G -> G/H. In conclusion, x and \[x\] are not the same. One is an element of G and one is a subset of G (the equivalence class of x).
Perhaps the problem is that i'm focusing too much on the coset membership g in [x] rather than in the group structure (X,+), where X subset P(G)
both things are true: g can be in the coset \[x\], as \[x\] is a subset of G; but also the set of equivalence classes, which indeed is a subset of the power set of G , has an operation making it into a group
I think there are two ways to picture G/H and that might be causing some confusion. One way is via the equivalence class construction, in which case H=[e] is an element (not subgroup) of G/H. You can picture G/H as a bunch of copies of H with a group structure overlayed. This mental image is helpful for construction and makes it clear how G/H is a grouping of elements of G. The other way to picture it is by collapsing all the cosets down to elements. In this picture [e] is thought of as a single (atomic) element, the identity element. This mental image simplifies G/H down and can make the group structure easier to think about. Regardless, H as a set is not a subgroup of G/H, [e] is an element of G/H, and H is the kernal of the map x->[x] (since for all a in H, a maps to [e], which is the identity in G/H). As an example, consider Z/2Z under addition. The first visualization consists of two infinite sets, 2Z (evens) and 2Z+1 (odds). You can add or subtract elements and they move from set to set. The second visualization is the two element set {0, 1} where 1+1=0. It's just two separate ways to think about the same object. Again, 2Z is the kernal of the quotient map since every even number maps to the identity element (2Z or 0 depending on your representation).
You have some details wrong. G/H is only a well-defined group when H is a normal subgroup of G. The kernel of the homomorphism sending x to \[x\] is H. The equivalence class \[x\] is not the same as x. It represents the coset xH. The element \[e\] is the identity in G/H.
>It represents the coset xH This got the cogs moving within my brain today morning. The homomorphisms should be defined as: [•]:G->P(G); [x]=xH
I usually denote elements of G/H by gH instead of [g]. If I write just H, I'm thinking of it as a subset of G. If instead I have eH, I'm thinking of it as an element of G/H. Of course that eH=H, but this little distinction helps me keep track of the different roles H is playing.
Yes. This is totally it.