Recent content by CoachBryan

So {ge, gh1, gh2,..., ghk} {eg, h1g, h2g,..., hkg} Right? Looking at that, I am still not making a connection. :/

Let H subgroup in G. Prove H is normal in G iff for all g in G and for all h in H, there exists an h1 and h2 in H such that hg=gh1 and gh=h2g.Completely lost on this. All I know for normal is that gH=Hg. Where else do I go from here? Help a brother out! Thanks :)

Yes it converges to zero

I've been messing with this proof for while and I'm stuck on this. I've started with a(n) converges to 0, let epsilon > 0, then there exists an n0 in N such that for all n >= n0. I'm stuck here thus far. Any help? Thanks for your time.

Hey, thanks a lot. Now i understand it.

if it's possible to find at least one convergent subsequence in ANY sequence. Definition of converge: "A sequence {a(n)} converges to a real number A iff for each epsilon>0 there is a positive integer N such that for all n >= N we have |a(n) - A| < epsilon."

I'm not sure if this is true or not. but from what I can gather, If the set of Natural numbers (divergent sequence) {1, 2, 3, 4, 5,...} is broken up to say {1}, is this a subsequence that converges and therefore this statement is true?

Gotcha. Thanks a lot, now it makes sense. So, not every bounded sequence is cauchy, but it's subsequences can be convergent (cauchy).

I've been very confused with this proof, because if a sequence { 1, 1, 1, 1, ...} is convergent and bounded by 1, would this be considered to be a Cauchy sequence? I'm wondering if this has an accumulation point as well, by using the Bolzanno-Weirstrauss theorem. I really appreciate the help...

Nevermind, it was a typo. The answer was a^-5 = a^95

It's asking me to find a^-5 From what I understood about inverses is that it's the element before you hit the identity would be the inverse. So a^5 = (a^5, a^10, a^15,..., a^90, a^95, a^100 = e), then a^-5 = a^95. The correct answer is actually a^20. Can anyone help me out with understanding...

Thanks guys! On the inverse problem as well. Appreciate it.

Thanks a lot! Now it makes sense. Do you know how to find the inverse?

So I've been sitting here for a while looking at my study guide and I am not sure how to find the product (or even the inverse) of this permutation in S9: (2 5 1 3 6 4) (8 5 6)(1 9) = (1 3 8 5 9)(2 6 4) (Correct answer) I know it starts off with 1 --> 3, then you get (1 3 and then after you...