1. ## Subgroup proof check

Here's the problem:
Given that H, K, and N are subgroups of G with H < N, $\displaystyle H \cap K = K \cap N$ and HK = HK. Prove H = N.
Okay, here we go ladies and gents!

$\displaystyle K \cap N = H \cap K$

Since
$\displaystyle H(K \cap N) = HK \cap N$ <--From the previous problem in the set

$\displaystyle H(H \cap K) = (HK) \cap N = (NK) \cap N$ <-- HK = NK

and
$\displaystyle (NK) \cap N = N$ <--N, K < G are groups

Thus we have that
$\displaystyle H(H \cap K) = N$

Since $\displaystyle N = H(H \cap K) \subset H$

$\displaystyle N \subset H$

But we know that H < N, so $\displaystyle H \subset N$. Thus N = H.

Are there any assumptions or (gasp) errors that I have made in the above? Thanks!

-Dan

Edit: Actually I think I can just go $\displaystyle N = H(H \cap K) = H$ since $\displaystyle e \in H \cap K$ and $\displaystyle H \cap K \subset H$ couldn't I?

2. Originally Posted by topsquark
Here's the problem:
Okay, here we go ladies and gents!

$\displaystyle K \cap N = H \cap K$

Since
$\displaystyle H(K \cap N) = HK \cap N$ <--From the previous problem in the set

$\displaystyle H(H \cap K) = (HK) \cap N = (NK) \cap N$ <-- HK = NK

and
$\displaystyle (NK) \cap N = N$ <--N, K < G are groups

Thus we have that
$\displaystyle H(H \cap K) = N$

Since $\displaystyle N = H(H \cap K) \subset H$

$\displaystyle N \subset H$

But we know that H < N, so $\displaystyle H \subset N$. Thus N = H.

Are there any assumptions or (gasp) errors that I have made in the above? Thanks!

-Dan
This looks good to me.

Originally Posted by topsquark
Edit: Actually I think I can just go $\displaystyle N = H(H \cap K) = H$ since $\displaystyle e \in H \cap K$ and $\displaystyle H \cap K \subset H$ couldn't I?
You could...but the way you did it originally is clearer (the original way is a very common argument, so people would just skip over it as they know what your doing, while your new way requires some thought to verify).

3. looks fine to me. H∩K = N∩K, tells you that the part of H in K is the same part of N that is in K.

HK = NK, tells you that when you "expand by K" you get the same results.

so the proviso that H∩K = N∩K is exactly what you need to "cancel the K's". in other words if N is bigger than H, N∩K will be bigger than H∩K.

it is the fact that G is a group that makes this happen, group multiplication introduces a certain regularity to the subsets of G.

4. Yay! I finally got one! (This section is whipping my, ummm...unmentionables.)

Thanks!

-Dan

5. sometimes the extreme abstraction of group theory overwhelms people. it involves a bit of "un-learning", because operations like + or * no longer act in the ways we have been drilled into accepting as "self-evident".

the integers modulo n are somewhat helpful, because they are a concrete representation of cyclic groups, and ordinary things we learned from 3rd grade (or so) are still helpful in guiding our intuition (ok, 3 divides 6, so i believe that {0,3} is a subgroup of {0,1,2,3,4,5}).

although cyclic groups are important (they are the "prime numbers" of the abelian groups), they are still rather specialized. abelian groups are also rather specialized, they are "almost" vector spaces (the technical term here is Z-module). non-abelian groups are, in some sense, the most "typical" (the worst case).

the smallest non-abelian group is S3, which has only a scant 6 elements. it is a worth-while exercise to do the following:

version 1: imagine you have a set of 3 elements {a,b,c}. describe all possible bijections on this set.

version 2: imagine you have an equilateral triangle in R^2, with vertices at (1,0), (-1/2, √3/2), and (-1/2, -√3/2). describe all 2x2 matrices that fix these 3 points.

version 3: imagine you have the free group on 2 letters (formally, these are all strings such as aabbbab*a*a*ba, etc, with the proviso that aa* = bb* = a*a = b*b = [ ], the empty word, the multiplication is "concatenation"), subject to the following rules for "reduction": aaa = bb = [ ], ba = aab. describe all possible words.

now, convince yourself these are all "the same thing". are they all groups? what are the possible subgroups?

if any subgroups exist, explicitly calculate left and right cosets, xH and Hx. for which subgroup(s) are these two cosets always the same? express this in terms of:

1) how many elements are "fixed" by the elements of the subgroup.

2) the "kind" of geometric operation it is.

3) a's and b's.

these examples were not chosen at random. (1) corresponds to an "action" view, (2) corresponds to a "matrix" view (i think you can see the physical significance of this), and (3) corresponds to an "elements" view. in order of abstraction, we have (2), (1), (3). with any group, all 3 views are possible, and sometimes we can gain unique information by getting a result from one view, and applying it to another.

6. Originally Posted by Deveno
sometimes the extreme abstraction of group theory overwhelms people. it involves a bit of "un-learning", because operations like + or * no longer act in the ways we have been drilled into accepting as "self-evident".

the integers modulo n are somewhat helpful, because they are a concrete representation of cyclic groups, and ordinary things we learned from 3rd grade (or so) are still helpful in guiding our intuition (ok, 3 divides 6, so i believe that {0,3} is a subgroup of {0,1,2,3,4,5}).

although cyclic groups are important (they are the "prime numbers" of the abelian groups), they are still rather specialized. abelian groups are also rather specialized, they are "almost" vector spaces (the technical term here is Z-module). non-abelian groups are, in some sense, the most "typical" (the worst case).

the smallest non-abelian group is S3, which has only a scant 6 elements. it is a worth-while exercise to do the following:

version 1: imagine you have a set of 3 elements {a,b,c}. describe all possible bijections on this set.

version 2: imagine you have an equilateral triangle in R^2, with vertices at (1,0), (-1/2, √3/2), and (-1/2, -√3/2). describe all 2x2 matrices that fix these 3 points.

version 3: imagine you have the free group on 2 letters (formally, these are all strings such as aabbbab*a*a*ba, etc, with the proviso that aa* = bb* = a*a = b*b = [ ], the empty word, the multiplication is "concatenation"), subject to the following rules for "reduction": aaa = bb = [ ], ba = aab. describe all possible words.

now, convince yourself these are all "the same thing". are they all groups? what are the possible subgroups?

if any subgroups exist, explicitly calculate left and right cosets, xH and Hx. for which subgroup(s) are these two cosets always the same? express this in terms of:

1) how many elements are "fixed" by the elements of the subgroup.

2) the "kind" of geometric operation it is.

3) a's and b's.

these examples were not chosen at random. (1) corresponds to an "action" view, (2) corresponds to a "matrix" view (i think you can see the physical significance of this), and (3) corresponds to an "elements" view. in order of abstraction, we have (2), (1), (3). with any group, all 3 views are possible, and sometimes we can gain unique information by getting a result from one view, and applying it to another.
Thanks for the information. My major problems at this point include
1) Learning group theory from a Physicist. We hit the Sylow theorems by chapter 3 and headed directly toward character tables. Useful in decomposing groups, but as we only knew a few groups it never seemed especially useful. (Except when trying to work out spin group decompositions.)

2) I have never actually taken an undergraduate level group theory/algebra course. I don't even have an entry level text in group theory. All I have is a graduate level Algebra text and a graduate level Group Theory text. (And my Physics text is more of a graduate level text itself. It concentrates on crystal groups and we didn't really use it much except to construct character tables.)

The outcome is that I know a few advanced concepts, but little or nothing about where they come from and how to apply them generally. I have been trying to counteract that, but it takes time. I didn't realize how much help I needed until I started to really work at introductory Topology. First I thought it was just my set theory that needed a brush-up, but I'm finding that my knowledge of Group Theory has some H U G E gaps in it. So I picked up the better of my two texts (the Algebra one instead of the Group Theory as Algebra is used in Quantum Field Theory as well) and here we are today.

The only good thing about this is, especially after teaching college for a few years, I really know how to get the best out of a textbook. And having MHF around whilst I edumacate myself is a fantastic resource.

-Dan

7. my guess is, that approach (2) will be most amenable to what you want to know a la physics and chemistry. bijections on a set correspond 1-1 (another bijection!) with "permutation matrices" (you need nxn matrices to permute n elements). cyclic groups are basically "rotation groups". an important class of groups in crystallography are the dihedral groups, which are "half rotation" and "half reflection". in one of those coincidences that makes you feel as if the terminology was well-chosen, the rotation subgroup of a dihedral group is "normal" (geometrically this means you can perform a rotation without ever "leaving the space".

the "daddy-group" of matrix groups is the general linear group, important subgroups are the special linear group, and the unitary(orthogonal, if real) subgroup. as with any group, the intersection of these two subgroups is another group, the special unitary group (these are isometries). in the real case, the orthogonal group of R^2 is the infinite generalization of the dihedral group, it's the full symmetry group of the circle (instead of the n-gon), while the "proper" symmetry group (the rotations), form the special orthogonal group.

the infinite-fold symmetry of these particular groups, makes them ideal tools for investigating situations where symmetry exists. as one might suspect, one can recover finite subgroups, by requiring a finite set of fixed points, or a finite invariant set.

8. o.o why did it double-post?

9. Originally Posted by Deveno
These examples were not chosen at random. (1) corresponds to an "action" view, (2) corresponds to a "matrix" view (i think you can see the physical significance of this), and (3) corresponds to an "elements" view. In order of abstraction, we have (2), (1), (3). With any group, all 3 views are possible, and sometimes we can gain unique information by getting a result from one view, and applying it to another.

This is not true. A group is called Linear if it can be embedded into GL(n, K) for some integer n and some field K. Not all groups are linear. However, all finite groups are.

I believe the Symmetric group on infinitely many points should work (meta proof: it acts on something very' infinite, while linear groups act on finite-dimensional vector spaces).

The next level up' is sometimes taken to be things called `PI-groups'. These are the group which embed into the group of units of some PI-rings. Ever linear group is a PI-group, but not every PI-group is linear (this paper mentions the difference - but I think it just alludes to it).

10. i was thinking of finite groups. that darn "any". although couldn't you embed G in K[G] for an infinite group?

PI-groups sound bizzarre...exterior algebras? lie groups?

cayley's theorem works just fine in the infinite case: for any g in G, the map h-->gh is a bijection, so G is isomorphic to a subset of Sym(G).