battletoad

Thanks for the input. Totally forgot that the centre of any group is non-trivial if the order of the group is some power of a prime (although i proved it an earlier question during the exam!). As things stand, the manner in which I defined K, is it anyway possible to conclude that it's non-trivial without knowing what we're dealing with (apart from the fact that the initial group is of order 6655)?

So i just got back from a Group Theory exam. One question's bugging me. |G|= 6655. Prove that G has nontrivial characteristic subgroups H & K, K a proper subgroup of H. Now i've proved that there exists a unique Sylow 11-subgroup (you guys may confidently assume this to be the case if you don't want to do it yourselves), thus it is normal in G. Let's call this subgroup H. We proceed to show that it is a characteristic subgroup of G. If f is any automorphism in G, the order of H is the same as that of the order of H under f (Hf is a subgroup of G). Since H is the unique Sylow 11-subgroup, H=Hf (if not, because Hf qualifies as a Sylow 11-subgroup of G, H!=Hf contradicts H's uniqueness). Hence H is a characteristic (non-trivial) subgroup of G. Now |H|=11^k where k is some positive integer, k<=3. By Sylow, H has Sylow 11-subgroup M. There may be multiple such Sylow 11-subgroups, denote them M=M1, M2, ... ,Mr for some non-negative integer r, r some power of 11. Let K be the intersection of all these Mi's. Thus K is a subgroup of (possibly equal to, if Mi is the unique Sylow 11-subgroup of H) Mi < H => K<H Any automorphism in G permutes the Mi's among themselves, and since K contains all common elements of these Mi's, K is sent to itself under any such automorphism. Thus K is a characteristic subgroup of G. This last part is what's bugging me. All that is required to show is that K is non-trivial. I did some things involving the order of elements in these Mi subgroups dividing the order of H but didn't find to killer blow to tie it up all together. What i have is that H,K are characteristic subgroups of G, K<H and H nontrivial. Not the K nontrivial part. Appreciated if anyone can add something, even if it is criticism of the proof.

Thanks for the heads up. Ur 'vector spaces where the scalars are now in the ring' (tbh thought that same thing but maybe i needed time to fully understand it all) fits in nicely with what im doing in group theory atm, automorphisms and such. Planning to read up on vector spaces again forgot most of it...would've been great if i didnt sell that textbook...

Your derivations are right for the most part so you definitely on the right track when it comes to these AWFUL (fun for me tho;)) number theory exercises My A1-A9 would look like this...(im going to draw it out so you can get the full understanding behind it all) . Given n is odd. Thus n=2k+1 for some integer k. It follows that (n^2)=4(k^2)+4k+1. After factorisation we get n^2=2([2k^2]+2k)+1. <---Just to show that n^2 can be written in the form of 2t+1 for some integer t (in this case our t=([2k^2]+2k), which is our definition of an odd integer Clearly n^2 is odd as well. Since n^2 is odd, n^2-1 is even, i.e. divisible by 2. As such, (n^2-1)/2 is an integer. We have now shown that m=(n^2-1)/2 is an integer <---AFAIS, not needed for your proof but lets be precise shall we (the part that follows is your derivation of earlier with an added 'twist'...) Given m=(n^2 - 1)/2 2m=n^2-1 2m + 1=n^2 m^2 + 2m + 1=m^2 + n^2. ....Added m^2 both sides, permissible since its an integer (m+1)^2 = m^2 + 2m + 1 = m^2 + n^2 => (m+1)^2 = m^2 + n^2 And there you go. As with all maths, doing plenty of exercises will give you the experience needed to get to know the tricks of the trade and to know when to use them. I advise you to do as much of the exercises at the end of the section and all this will be a breeze at the end of the day Good luck with your course

Thanks. Your proof puts the matter to rest. Strange, im usually the first one to shout 'proof by induction' yet the idea alluded me. Meh, six months of working has made me soft!

Cant seem to fathom what an R-module is. Thing is i can prove if something (elementary structures for now) is an R-module or not, but i dont get what the basic structure 'looks' like. Let me explain... Graphs and rings are fine, in being sets which satisfy certain criteria. So i understand that they are special sets, operative word SETS, a structure i understand. R-modules on the other hand utilise these special sets (abelian groups, rings), uses an endomorphism onto, say, the group 'space', then end up being a structure i vaguely understand (obviously a set itself, but different somehow to me). So for now i view an R-module as a ring (again, satisfying certain axioms) but il ultimately would like to know what it 'looks like'. Hope im making myself understood. All in all, im thinking i should move away from my need to familiarise myself with mathematical structures, ideas, concepts and so forth, only deal with them on a definition basis. Perhaps then il go into middle age with a full head of hair :/

Yip, tripped up over there. Thanks for the reply. But why is |P(S)|=2n? Is there a proof for it? Merged post follows: Consecutive posts mergedIts fine i got it. If anyone's interested its binary basically. For a set with n elements, draw n columns, a column for each element in the set. Then for every combination, add 1 if the element is in ur subset, 0 if not. Clearly there will be 2n combinations

Started off with graph theory with some elementary things. There's one thing however that i cant wrap my head around, the order of the edges. I get that if the vertices of a graph were included in a set S, the number of edges is equal to the order of the power set of S. For the life of me i cant remember the proof if i did it or not. Can someone please provide the proof of |P(S)|=2^(nC2), where P(S) is the power set of set S, |S|=n and C refers to the combination function? Much appreciated

Been searching for the proper place to submit this post, seems to be in this section. Soz if not. So 2010 has come along, with it implications for my future, i.e. which discipline i'm to follow in post grad. Desperately in need of some advice. I've been accepted for mathematics, physics and statistics post grad courses at my uni. Problem is, ideally i'd like to pursue mathematics (number theory, modern algebras, that sort of thing) but i'm not sure if i'd be able to get a job within an industry with these disciplines. Seems as if (in my country, South Africa, at least) companies favour a financial based mathematics qualification, in which i suppose i'm competent enough to pursue, but my interest isn't really there. If i'm to deviate from math, i'd ideally want to go down the physics route, perhaps even an applied math degree with an engineering/physics slant (my lack of interest in financial math would rightly correlate to my lack of interest in statistics-type occupation). Question is, if i'm to stick with my math, which type of jobs would i be setting myself up for? Same question would go for physics (here again physics qualifications would lead more often than not only to research institutions if one's lucky enough). I'd love a job within an industry and not forced to stay within academia/research for employment, although i definitely wouldn't be opposed to academia/research if the opportunity arises. Advice would be much appreciated, the new academic year starts roughly at the end of Jan. PS I have a BSc in math and stat sciences and was unable to get a job in the last six months which explains my current disillusionment i guess. Hopefully with the recession ending sometime soon companies can start hiring people with degrees again like before.