Recent content by nebbish

Does this mean there exists a subgroup H of G such that o(H/(a)) = p^{n-2} and therefore o(H) = p^{n-1} since o(H/(a))= \frac{o(H)}{o(a)}? (Since the n =1 case is trivial I am assuming n > 1. Also note that the H I am using is a subgroup of G whereas the H you were using above is a subgroup of...

According to my Professor, the key to the problem is to recognize that Z(G) must have an element of order p, say element a. Then "lift up" G/(a) to find a larger proper subgroup within G itself as necessary. Anybody have an idea what "lift up" means? I asked him about it. The phrase seems to...

Abstract Algebra -- lifting up a factor group After spending an extended period with my Professor during office hours I must admit I am mystified. He kept on talking about "lifting up" factor groups. I think this has something to do with using a factor group, say G/N, to show that there...

My original non-abelian proof now seems fatally flawed. The normalizer always contains the center, I can't hope to generate a larger subgroup by the product of the center with some subgroup of a normalizer group.

If H happens to have order p^{n-1} then by induction we are done. However, if H had order of just p, the induction really tells us nothing. It is critical that we find a subgroup of order p^{n-1}, not just any order less than p^n. We could attempt to generate a subgroup of order p^{n-1}...

The text has a theorem stating that every finite abelian group is the direct product of cyclic subgroups, and indeed I think I could use this to prove the abelian case. However, the theorem appears later in the book than the Sylow material, and indeed Sylow material is used in the proof.

Still, a hint would be appreciated. A pointed reference to a theorem or equation that gives the answer away would be fine with me. I've spent tremendous time and effort on this problem and the course in general. I'm disappointed you didn't evaluate the work that I did for the non-abelian...

Actually the important thing is that every interval of the reals contains a rational number. Therefore every interval around every irrational point contains a point that is not in the set of irrationals. Therefore the irrationals are not an open set. For the irrationals to be open there...

Abstract Algebra -- no Sylow allowed Please note Sylow's theorem(s) may not be used. Using Theorem 1 as a tool, prove that if o(G)=p^{n}, p a prime number, then G has a subgroup of order p^m for all 0\leq m\leq n. Theorem 1: If o(G)=p^{n}, p a prime number, then Z(G)\neq (e). Theorem 1 uses...

you're right Undoubtedly you are all correct and I hope we cover that material in short order. If not, I should go after it myself. To a large extent I blame myself for not providing you guys with more information about how my textbook author intended the problem to be solved. This would...

nearly a heart attack! I'm not sure what exactly you're saying is false. The Klein 4 group's elements are of order 2 (other than e). My comment referred to an element of order p^2, which is 4 in this case. The Klein 4 group has no such element. To be clear, I said EITHER there is an...

how i did it If I've got this figured correctly, it turns out that either there is an element of order p^2, in which case immediately the group is cyclic and abelian (therefore any subgroup is in the center); or, each element x in G other than e is in a normal subgroup of order p, namely, (x)...

I didn't end up looking anything up (beyond what I already knew when I came across the problem) but in any case I believe I have solved the problem.

Abstract Algebra -- group Show that in a group G of order p^2 any normal subgroup of order p must lie in the center of G. I am pretty sure here that p is supposed to be a prime number, as that is the stipulation in preceding and later problems. However, the problem statement does not...

I have two problems I would like to discuss. 1.For any group G prove that the set of inner automorphisms J(G) is a normal subgroup of the set of automorphisms A(G). Let A be an automorphism of G. Let T_{g} be an inner automorphism, i.e. xT_{g}=g^{-1}xg The problem can be reduced to the...