Normal subgroups of GL(n, R)?
822775
Every group has a compact normal subgroup, namely the identity.  \nAs noted, the scalar matrices form the center of your group G.  Therefore the two-element subgroup {plus/minus I} is also normal in G.\n\nAs noted, the group SL_n is normal in G.  Therefore if H is also normal in G, its intersection H' with SL_n is normal in SL_n.  It is known that if H' is a normal subgroup of SL_n, then H' is contained in the scalar matrices, which are trivial if n is odd and {plus/minus I} if n is even.\n\nIf H is not equal to H' it must contain a matrix of determinant not equal to 1.  If this determinant is not plus/minus 1, then H is not compact.  Let us note that the set of matrices with determinant plus/minus 1 is a normal subgroup of G containing SL_n.\nFor that matter, also note that the set of matrices with positive determinant is normal in G.\n\nThis discussion should enable you to classify the normal subgroups of GL_n.  If you mean *Lie* subgroups we are done.  \nIn particular, there are exactly two compact ones, already noted.
984229
D_4 is NOT abelian.\nnow if f:G ->D_4 was onto, then:\ntake a, b in D_4, then there exist g1 and g2 in G such that\nf(g1) = a and f(g2) = b\nnow\nab =f(g1)f(g2) =f(g1g2) =f(g2g1) = f(g2) f(g1) = ba,\nwhich would imply that D_4 is also abelian!!!! contradiction, therefore f cannot be onto\n`
1088813
Here you go...\nhttp://www.mathsisfun.com/platonic_solids.html
1062218
a) Find the point of intersection of DF and GE.  Call that point M.\nShow that the length of DM = MF and the length of GM = ME.\nb) Find the length between each set of points (DE, EF, FG and GD).  If they are all equal, then it is a rhombus.
735897
Let s be in S.  Since f is a homomorphism, f(r)=f(1)*f(r), thus \ns*f(r)=s*f(1)*f(r).  We rearrange and factor to get\n[s-s*f(1)]*f(r)=0.  The hypotheses imply that f(r) is not a zero divisor, so we conclude that s=s*f(1).  Similarly using f(r)=f(r)*f(1) we get s=f(1)*s, thus f(1) is an identity element.
244218
http://mathworld.wolfram.com/GaloisGroup.html\n\nI'm not sure if I can give you the exact answer since I'm trying to understand it myself.  \nWhat you want is the Galois group of Q(r1, r2, r3, r4, r5) over Q where the ri are the roots to the polynomial x^5 + x + 1.  This group is going to be a subgroup of the group of permutations on 5, but I don't know if I can say anything else about it since x^5 + x + 1 is an irreducible polynomial in Q.  \nHope this helps.
1147832
The previous poster has the right method, but use 11 as your prime. You then need a divisor of 2*7*13 which is 1 modulo 11. This shows that there is only one 11-Sylow subgroup, so it is normal.
466194
Consider K_3,3, which is the graph obtained when you draw all lines between 3 points {a,b,c} and three other points {x,y,z}. So there is a line between a and x, but not one between a and b.\n\nAn automorphism, f, will take a to one of the six points. There are two possibilities: either f(a) is in {a,b,c} or f(a) is in {x,y,z}. In the first case, {f(a), f(b), f(c)} is a permutation of {a,b,c} and {f(x),f(y),f(z)} is a permutation of {x,y,z}. In the second case, it goes the other way around. Furthermore, any two permutations in both sets will lead to an automorphism of your graph.\n\nThus, the order of the automorphism group will be 6*6*2=72.\nThere is a subgroup isomorphic to S_3 cross S_3 of index 2. Just checking commutativity tells us that the reflection between {a,b,c} and {x,y,z} is not in the center, and in fact, conjugation by the reflection reverses the order of elements in S_3xS_3. Thus, the automorphism group is the semi-direct product of S_3 cross S_3 with Z_2 where Z_2 reverses the order of the cross product.\n\nI hope this helps!
795193
My opinion is that no one will be able to answer your 'question' because it doesn't make sense in any known language.\n\nHowever, I will try.  Imagine a universe that is divided into two large groups.  \n\nThe members in group A include all 14 of your conditions and some others as well (e.g., 15 blue, 16 tall, etc).\n\nThe members of group B includes everything not included in group A.  \n\nItems 1 to 14 'may' be represented in some subsets of group A.  (Paragraph 1)\n\nNo members of group B will ever be represented in any subset of group A.  (Paragraph 2).
1155862
to many to post.....they are all on this site \n\nhttp://www.geom.uiuc.edu/docs/reference/CRC-formulas/
662149
No. If you look at the group of {1,-1,i,-i,j,-j,k,-k} of quaternions, every subgroup is normal and every quotient group is abelian. I should also point out that every proper subgroup is abelian also.
414419
http://www.math.niu.edu/~beachy/aaol/frames_index.html
209070
http://www.robertrtg.com/g3.html
880717
You are going to run into some difficulties because mathematicians and physicists have rather different emphases and notations in this area. Most of the math books that cover Lie groups assume a lower level abstract algebra course (which covers finite groups). The best treatment for physicists I have found is 'Symmetries in Physics' by J. P. Elliot and P. G. Dawber.\n\nIf you really want a mathematical treament, you should look into books on differential geometry first (to get a good definition of Lie group) and then check out a book on representation theory for Lie groups.
314064
Think of the projective plane as the disk with opposite boundary points identified. Let the base point for the fundamental group be the center of the disk. The loop that goes directly across the disk and comes back through the other side is the generator of the fundamental group.\n\nAnother way to see this is to notice that the 2-sphere is the universal cover and is a double cover. Thus the fundamental group has two elements. There is only one group of order two up to isomorphism.
1026074
1s2, 2s2, 2p6, 3s2, 3p6, 4s2, 3dn (where n=1 to 10)\n\nThere's the first Transition series for you.  It can also be shown as:\n     [Ar],4s2, 3dn
902933
Au - 6s2 4f14 5d9\nNd - 6s2 4f3 5d1\nCl - 3s2 3p5\nununillium - 7s2 5f14 6d9\nZr - 5s2 4d2