Diamond Isomorphism Theorem Puns

Undergrad here, taking my first abstract algebra class. Wow, this theorem is abstract and confusing at first but I can just feel something deep and beautiful here. Can't wait until I really grasp it intuitively :)

I posted the following on the simple questions thread on r/math but I didn’t get any responses, so I was hoping somebody here could help me out.

I’m going through the isomorphism theorems now. I understand the proof of the 1st isomorphism theorem, and I can see how incredibly useful it is, allowing one to easily show results like G/Z(G) iso to Inn(G), and GL_n(R)/SL_n(R) iso to R*, as well as the second and third isomorphism theorems.

I can also somewhat see the utility in the third isomorphism theorem, as I’d imagine a case like (G/N)/(H/N) would come up sometimes. Could anyone give me some specific examples of this?

I can’t however see the utility in the second isomorphism theorem. Wikipedia said something about projective linear groups, but that means nothing to me. Are there any other special cases of the second isomorphism theorem that are seen?

So I was driving home from work, and for no particular reason, the first isomorphism theorem from group theory suddenly clicked in my head!

It made that sublime transition from a strange collection of symbols and axioms, to a coherent and logical statement - obvious, even! How could I have misunderstood it?

I was so excited on the drive home, on my arrival I was compelled to draw this image that had emblazened itself in my mind. The homomorphism (might) "compress" our group into a smaller one, and the "measure" of how much the group is compressed is the kernel of the homomorphism.

(Precisely, the fiber of each element of H under φ is a coset of K in G.)

I guess I wanted to share my excitement in this moment, even if the result is rather basic. I have been self studying and sometimes I doubt myself, but small moments like this keep me slowly pushing forward.

I understand if the post is inappropriate for the subreddit, and I apologize if so. I just couldn't help it! :)

(Edit: My self-assigned homework for tomorrow is to figure out all consistent ways to complete the multiplication tables of G and H as drawn.)

I'm not sure how to use the First Isomorphism Theorem to prove that Q[x]/(x^2 - 1) is isomorphic to Q[sqrt(2)]. Any ideas?

So I want to show that f: H -> HN/N is well-defined, where H is a subgroup of G and N is a normal subgroup of G. I know that well-defined means that x = y implies f(x) = f(y). But now if we take two elements h and h' from H, we get f(h) = hN and f(h') = h'N. And that's about it. I'm not sure how to progress any further.

I guess that one way to argument this is that f sends elements of H to the left cosets of N. Cosets in general are completely defined by their representatives. Thus, if h = h', the left-coset has to be the same, i.e.
hN = h'N

Hi, my question is in this video around 6min 14s. How does the fact that xh⁻¹=hkh⁻¹ ∈ K imply that
Kx = Kh? I don't see any connection between those two things.

A couple months back I was first learning about the First Isomorphism Theorem for Groups, and I thought it was kinda neat but overall I just brushed it off. Then a couple days ago I learned there is essentially an analogy of the theorem to rings, and it got me thinking: what other analogies are there? Specifically I was wondering about topological spaces. If we have a continuous function f:X-->Y between two spaces X and Y, is there a way to quotient X to get a homeomorphism to Y, or maybe the image of the original function f? I have not been able to find any hard evidence online, but one of my professors suggested letting the original continuous function be surjective, and defining the equivalence relation on X as: x~y if f(x)=f(y).

Hullo all!

The theorem i'm trying to prove is the following:

 

Let [;\mathcal{LT}(U,V);] be the vector space of all linear transformations with domain U and codomain V. Suppose [;dim(U) = n;], and [;dim(V) = m;]. Then [;\mathcal{LT}(U,V) \cong M_{mn} ;], where [; M_{mn};] is the vector space of all mxn matrices.

 

I understand how that might be true, since linear transformations can have a matrix representation with respect to some defined bases of U and V. I need to show that there exists an invertible linear transformation S such that [; S:\mathcal{LT}(U,V) \rightarrow M_{mn} ;] , or somehow that [; dim(\mathcal{LT}(U,V)) = dim(M_{mn});] but I'm unsure on where to begin. Some small hints on how to solve this problem would be greatly appreciated.

In the proof for the above theorem (http://imgur.com/a/0QUYf) I'm a little confused when they show that phi is injective and surjective.

To me it doesn't really look like the proof of injectivity takes the form of phi(x)=phi(y) implies x=y for all x,y in G/K. Am I mistaken? If not, how does this show phi is injective?

In the first sentence of the proof of surjectivity it states "if y is in im(f) then there exists a in G/K such that f(a)=y." Don't we want to show there ezits an a such that f(a)=y? To me it looks like they're just assuming its surjective. Can somebody tell me what's going on here?

So I feel like I have a relatively decent grasp of the First Isomorphism Theorem for groups. A fairly literal statement is that the group mod the kernel of a homomorphism is isomorphic to the image. A looser expression is that, when you mod out by some normal subgroup, it's like you collapse that subgroup to a point (the element 0) and the quotient group is basically everything that's left over.

But I don't have anywhere near the same level of feeling for the Second Isomorphism Theorem. Part of it might be my discomfort with the appearance of the intersection with a normal subgroup and the product group. Why are these important and is there a more intuitive way to understand them?

With the third one, I think I get it, but I don't have a good way of saying it. My best attempt is: if you have subgroups N < H in G, what do you get if you first mod out by N and then mod out that by H? Same as just G/H.

Anyone have other perspectives or ways to internalize these theorems?

I've got the question; Let G be a group and let H (be normal in) G. Let ro belong to Aut(G). Show that G/H is isomorphic to G/ro(H).

So far I've taken theta to be the canonical map from G to G/H, ker(theta) = H, and then by the first isomorphism theorem im(theta) is isomorphic to G/ker(theta) = G/H. Now I just need to show that G/ro(H) = im(theta) but I'm unsure how exactly to do that, and what the Automorphism does here.

Many thanks for any help!

Let G, H be groups. Let K= {e in G} direct sum H. Show that (G direct sum H) / K is isomorphic to G. I'm supposed to use the first isomorphism theorem to prove this, which states, "If f: G->H is a homomorphism with ker (f) = K, then Im(f) is isomorphic to G/K."

So the issue I'm having is in a part of the proof where we suppose that G has no elements of order p^2 and conclude that all (non-identity) elements of G have order p. From there Gallian wants to show that for any 'a' in G, the subgroup <a> is normal by a proof by contradiction. Here's where I'm having trouble. So we suppose that <a> isn't normal, and Gallian claims there has to exist some 'b' in G such that bab^-1 is not an element of <a>, therefore <a> and <bab^-1 > are distinct subgroups.

I don't know why <a> not being normal implies that bab^-1 isn't an element of <a>, when the coset rules state that a subgroup, H, of G is normal if and only if xHx^-1 is contained in H for all x in G. So if Gallian is trying to show that <a> isn't normal, shouldn't he be trying to show that b<a>b^-1 isn't contained in <a>? I guess the question I'm asking is, how does he know specifically that 'a' itself doesn't commute with 'b'? Couldn't 'a' commute with 'b' and it's some other element of <a> that doesn't commute with 'b'? Like, wouldn't it be more rigorous for Gallian to argue something like, "Then there exists some 'b' in G such that b(a^i )b^-1 is not an element of <a>, where 0<i<p"? It just bothers me that he assumes it has to be 'a' without demonstrating it.

Edit: A secondary question: From the fact (which I'm still unsure about) that <a> being not normal implies bab^-1 not in <a>, and <a> and <bab^-1 > are both distinct subgroups of order p, Gallian states that the intersection of <a> and <bab^-1 > is {e}. I don't know exactly how he comes to that conclusion, but here is my reasoning, I'd like somebody to tell me if this is valid:

Since bab^-1 isn't an element of <a>, I can say that 'a' =/= bab^-1 . Therefore a^2 =/= (bab^-1 )^2 . Likewise a^3 =/= (bab^-1 )^3 .... and so on up to p-1: a^(p-1) =/= (bab^-1 )^(p-1). So the only time they are equal is when 'p' is the exponent and they both equal 'e'. I'm just not sure I'm allowed to treat the not-equals sign (=/=) like that.

Edit 2: Third question. After showing that the intersection of <a> and <bab^-1 > is just {e}, Gallian says that the left cosets of <bab^-1 > in <a> are given by a<bab^-1 >, a^2 <bab^-1 >, a^3 <bab^-1 >,...,a^(p-1) <bab^-1 >. Then he says that b^-1 has to be contained in one of these cosets.

*Thm 1: Let G be a group of order 2p, where p is a prime greater than 2. Then G is isomorphic to Z_2p or D_p*

In the part of the proof where Gallian tries to prove that G~D_p (I am using ~ to represent "is isomorphic to"), he assumes G is non-cyclic so that its elements have either order 2 or order p, shows that it is impossible for all non-identity elements to be order 2, so he calls an element of order p "a". He then picks an element of G, "b", that is not an element of <a>. This is how he shows that |b|=2, and where I started having trouble following:

&gt;...we have that |b| = 2 or p. Because |<a>INTR<b>| divides |<a>|=p and <a>=/=<b> we have that |<a>INTR<b>|=1.

**I understand that |<a><b>|=|<a>||<b>|/|<a>INTR <b>| but I don't know how he got |<a>INTR<b>| divides |<a>| from that.**

Later in the proof, after showing |b|=2, he shows that |ab|=2, and therefore that ab=ba^-1 , he states:

&gt;...this relation completely determines the multiplication table for G... Since the multiplication table for all noncyclic groups of order 2p is uniquely determined by the relation ab=ba^-1 , all noncyclic groups of order 2p must be isomorphic to each other.

*I get that the relation allows any member of G (or any non-cyclic group of order 2p) to be written in the form a^i b^j or b^j a^-i but I don't get how that means all such groups are isomorphic.* With things like subgroups, I'm used to applying tests to determine that they are subgroups, and my instinct is to do the same when proving that something is an isomorphism. But Gallian seems to assume that it is trivial that an isomorphism exists, and I couldn't find one on my own.

As for the second theorem: *The group of rotations of a cube is isomorphic to S_4.*

Gallian starts out the proof by saying:

&gt;Since the group of rotations of a cube has the same order as S_4, we need only prove that the group of rotations is isomorphic to a subgroup of S_4

I don't understand why that is the case at all.

Gallian gives the designations 1,2,3,4 to four diagonals connecting the 8 corners of the cube (i.e. connecting the back right corner of the top face to the front left corner of the bottom face), and states that the rotation group induces a permutation group on the diagonals. He then goes on to try to show that there are 24 such permutations induced by the rotations. In d

I have been looking into k-uniform Euclidean tilings recently (https://en.wikipedia.org/wiki/List_of_k-uniform_tilings). As far as I know, their list is complete only to k=7.

I have made and implemented an algorithm (a variant of my previous tiling search approach) that can extend this list, and extend it significantly (I'm currently running it up to k=12, although this will take a few days to complete).

Here's the rub: I think that the algorithm is guaranteed to find every solution. (I haven't actually proven it, but the logic seems sound.) But the trouble is that the same solution can be (and usually is) found multiple times. Some solutions are actually found many times (particularly those that contain many similar vertex types such as the many, many solutions consisting of rows of squares and triangles alternating in some pattern).

I've been trying to go through the solutions by hand, but the potential for human error is too large. I managed to *almost* replicate the lists of 3-uniform and 4-uniform tilings from the Wikipedia, but I have always overlooked a few solutions (they were in the data set, I have just missed them).

I need help with devising some sort of pruning algorithm that could go over the result file and specifically point out unique solutions.

Some details: This is how a typical output looks:

Number of polygons: 10
(6,6,6)F, (3,3,6,6)F, (3,3,3,3,3,3)A2, (3,3,6,6)F, (3,3,3,3,6)F, (3,3,3,3,3,3)A2, (3,3,3,3,6)F, (3,3,3,3,6)F, (3,3,3,3,3,3)A2, (3,3,3,3,6)A
(6,6,6)F, (3,3,6,6)Fx2, (3,3,3,3,6)A, (3,3,3,3,6)Fx3, (3,3,3,3,3,3)A2x3
TES file: 10\10_36\3g 4e2 5a 5b3 6i3\eu raw 3g 4e2 5a 5b3 6i3 11.tes
(0 1')[1](2)(0' 2''')[2'](3' 2'')(0'' 2@4)(1'' 3''')(0''' 1@4)[1'''](0@4 1@6)[3@4](4@4 2@5)(0@5 4@7)(1@5 2@6)[0@6](3@6 3@7)(4@6 0@8)[0@7](1@7 0@9)(2@7 1@8)[2@8 2@9](3@9)
0: 0/1(6)-*1/*0(6)-*1'/*0'(6)-*2'''/*1'''(6)-1'''/2'''(6)-0'/1'(6)
1: 1/2(6)-2/0(6)-1'/2'(6)-*2'/*1'(6)-*0/*2(6)-*2/*1(6)
2: 2'/3'(3)-2''/*2''(3)-*3'/*2'(3)
3/4: 3'/0'(3)-2'''/3'''(3)-1''/2''(3)
*0'/*3'(3)-*2''/*1''(3)-*3'''/*2'''(3)
5/6: 0''/1''(3)-3'''/0'''(3)-1@4/2@4(3)
*1''/*0''(3)-*2@4/*1@4(3)-*0'''/*3'''(3)
7: *0''/0''(3)-2@4/3@4(3)-*3@4/*2@4(3)
8: 0'''/1'''(6)-*1'''/*0'''(6)-*1@4/*0@4(6)-*1@6/*0@6(6)-0@6/1@6(6)-0@4/1@4(6)
9: 3@4/4@4(3)-2@5/*2@5(3)-*4@4/*3@4(3)
10/11: 4@4/0@4(3)-1@6/2@6(3)-1@5/2@5(3)
*0@4/*4@4(3)-*2@5/*1@5(3)-*2@6/*1@6(3)
`12/1

So — a problem I’ve created for myself at work.

I have two graphs sets in different systems that were created based on the same dataset, but I’ve lost the master key. (As always, the real lesson here is to practice good data hygiene instead of what I did, which was a billion different versions all with various minor tweaks and no real version control.)

The original graph (G, E) is in GIS. Each vertex has between 0-10 neighbours, skewed to the low side (median degree is 2).

The copy (G’, E’) is in excel/SQL, and it wasn’t important to retain more than 5 neighbours for each vertex. So, it’s the same vertex set but the edge set is a strict subset (though a fairly large one).

Question: how can I best retrieve the mapping G <-> G’?

Obviously I don’t care about isolated vertices with degree 0. My data structures aren’t really meant to be graphs, and so it’s a pain to do graph traversal steps — I’d rather find a solution that’s solely based on inspecting lists of neighbours.

My boss asked what I was doing with my measurements and calculator - I proudly showed her. The new bulbs are a perfect “X” in the dining room. Thank you to the math teachers of Lake High School. 🤘

How come one can define isomorphisms in category theory without the concepts of injectiveness or surjectiveness ? How does one define the latter notions?

I'm not confident with my English, so pardon me if something is wrong. Also, I'm not sure if this is an off-topic post.

I only know few examples of morphisms: the homomorphisms of structures in model theory (knowing only the definition and some examples such as algebraic homomorphisms), isomorphisms of graphs and homeomorphisms, but I think I don't really understand the idea that's behind morphisms. Were isomorphisms meant to translate some kind of equality? If so, what was the idea behind morphisms? Intuitively, I think that isomorphisims are indeed trying to capture some idea of equality, but not full equality, if that means anything. But treat isomorphic objects as equal objects really bothers me because I don't understand what "to be equal" means. I'd like to know

1. The history of the idea of morphisms and isomorphisms. How do we arrive at the definition of morphism that we have now?
2. Were mathematicians trying to define a method of preserving certain properties? If so, how did they decide those properties?
3. Is there any philosophical explanation of the idea of morphism?
4. Do category theory try to abstract this idea of morphism?

I'll appreciate any references for these subjects.

Basically a theorem that says “all but some number of cases” satisfies the theorem

Is it not the case that HN/N = H/N? Each hnN ∈ HN/N is equal to hN ∈ H/N, for h ∈ H and n ∈ N. So why is the second isomorphism theorem not written as above? It seems like a more elegant result.