Recent content by Sprooth

I quickly found this website. It seemed pretty informative and covers what I'd consider a good range of concepts to introduce you to group theory. http://dogschool.tripod.com/index.html Another note: sometimes the "axioms of group theory" are considered to be four or perhaps five in number...

In order to gain a better understanding of groups and to feel more comfortable with them, it is important to understand the idea behind groups and to look at a number of specific examples. The main objective of algebra is solving equations. We have systems of numbers such as integers, rational...

Oh, one more thing. Yeah, abstract algebra is a tough beast. I took a couple abstract algebra courses in college, and they were pretty difficult. For me what made them hard was the lack of an inherent way to visualize the subject matter, unlike analysis or topology. I am now taking my first...

Oh no, at the thought of someone turning away from math, I had to say something! Sorry this is your first experience with math in college. The further you advance in math during college, the more you get into "real math". During my second or third year, I realized the math I was learning...

I'd still like to figure out this problem. It's not for an assignment or anything, rather for an independent project I'm working on, somewhat of an A.I. for a certain strategy game. Would it belong better in the homework section anyway? I'd really appreciate any help someone could offer. I'm...

Hi, I'm trying to figure out a variation of the problem where you determine the number of paths on a discrete grid. For example, for the paths from (0,0) to (a,b), you can consider it to be a rearrangement of a word with a x's and b y's, so the number of paths is (a+b)! / a!b!. The...

Yes I know that definition. H is a collection of elements, and aH is the collection of all those elements left-multiplied by a. But you can't do things like: aH = bH implies a = b, and you can't say that if aH = Ha, then ah = ha, for all h in H. You can say that for h1 in H, h1 left-multiplied...

I understand the elements in the subgroup commute amongst themselves, but saying aH = Ha does not show that they commute with the other elements. You could have a * h1 = h2 * a, for different h1, h2 in H. How do you use this to show that each individual element of H commutes?

The normality of each subgroup gives you aH = Ha for every a in G, which means that the subgroups commute as subgroups, but the individual elements do not necessarily commute. For some element n of a normal subgroup N and an element g in G, gn = gn2 for some n2 in N, but n is not necessarily...

I see that if you have a p-subgroup of order p, then it is cyclic, but what if you have a p-subgroup of order p*p? Then there are some elements of order p and an element of order p*p. Can you still show that the element of order p*p is the generator of the group and that the p-subgroup of order...

Thanks for your help. Yes, I am confused. What you're saying makes sense. I think I almost have it, but there are two things which I'm still not sure about. 1) It is true that if a subgroup H is abelian (i.e. the elements of H commute with one another), then the elements of H also commute...

Are they normal? I know that all subgroups of a cyclic group are cyclic. I already know from Sylow's theorems that the group of order m is normal in the group of order m^2 and the group of order n is normal in the group of order n^2. The groups of order m and order n are the only groups with...

Homework Statement Let G be a group with order (m^2)(n^2), where m and n are primes and m not equal to n. Prove that every subgroup of G is normal if and only if G is abelian. The Attempt at a Solution The <== direction is simple (I think), since all subgroups of an abelian group are...

fantispug: Thanks for pointing out that b^2 = 1 implies either b = 1 or b = c. I hadn't thought of that. Taking that into consideration, I played around with the equations some more, but there was something that confused me a little bit. When considering b with b^2 = 1, when b is 1 or c must be...

I need to prove that if a group contains exactly one element with order 2, then that element is the center of the group. Here is how I formulated the problem: Let A be a group with an element c such that c^2 = 1 (i.e. c = c^-1), and b^2 = 1 implies b=c. Want either: ac = ca, for all a in...