Recent content by Adorno

Homework Statement (A somewhat similar question to my last one). Let J be the ideal of the polynomial ring \mathbb{Q}[x] generated by x^2 + x + 3. Find the multiplicative inverse of (3x^3 + 3x^2 + 2x -1) + J in \mathbb{Q}[x]/JHomework Equations The Attempt at a Solution I think I need to apply...

So, it's essentially just trial and error?

Homework Statement Factorize x^2 + x + 8 in \mathbb{Z}_{10}[x] in two different waysHomework Equations The Attempt at a Solution I can see that x = 8 = -2 and x = 1 = -9 are roots of the polynomial, so one factorization is (x + 2)(x + 9). Is there a systematic way to find all the factorizations?

So we define the map phi : G -> S_3 by phi(g) = (left multiplication by g). I can see why this is a homomorphism. (Since left multiplication by g_1g_2 is the same as left multiplication by g_2, then left multiplication by g_1). Is this right?

So does that action look like (g, n) |-> gn where n is in N_G(H)? Ok, so how about the homomorphism? Do you map an element of g to the action of left multiplication by g on N_G(H)?

I'm not too good at group actions. What would the action be? And where would the homomorphism come from? If the kernel is normal, then it must be trivial, otherwise there would be a non-trivial normal subgroup of G.

Homework Statement Prove that no group of order 96 is simple. Homework Equations The sylow theorems The Attempt at a Solution 96 = 2^5*3. Using the third Sylow theorem, I know that n_2 = 1 or 3 and n_3 = 1 or 16. I need to show that either n_2 = 1 or n_3 = 1, but I am unsure how to do...

Cool, thanks for your help! I think I've worked out the other parts now.

Ok, well my definition of an open set is a set in which every point is an interior point. I think I see now. If there is a nonempty open set, it contains an interior point, and thus there is an open ball within the set, by definition of interior point. So if there are no open balls there can...

Ok, I'll see how it goes. My/your/our reasoning proves i -> iii doesn't it? The only problem is the question talks about "open sets", but we were talking about (open) balls. Or does the fact that there are no balls mean that there are no open sets? I'm a bit confused about that

So it contains points of A, and therefore can't be disjoint from A. I see. But shouldn't I be proving that ii -> iii and iii -> i?

Right. So if you take an open ball disjoint from A, all the points in the ball are limit points... right?

Ok, so a limit point of A is a point x in X such that every open ball centred at x contains another point of A, right? I'm not sure where this is going. :-p

Yes, that's the definition of dense I'm using. The definition of closure of A, as far as I know, is the union of A with its limit points.

Homework Statement Let X be a metric space and A a subset of X. Prove that the following are equivalent: i. A is dense in X ii. The only closed set containing A is X iii. The only open set disjoint from A is the empty set Homework Equations N/A The Attempt at a Solution I can...