Recent content by cmj1988

So all diagonal matrices?

What is the center of SL(n,C)? I understand that the center of a group is where all elements commute with the group G. So I figure that I should come up with a case in which matricies commute. I remember a few facts from Linear Algebra: Fact 1: Simultaneously diagonalizeanle matricies...

I don't mean to dig this up, but isn't there a theorem (not sure if it lagrange) that states that the minimum order of an element to belong in a subgroup is a divisor of the group order. Couldn't one arrive at this by creating a bijection and then invoking the fact that |G:N|=|G|/|N|. <= That...

I'm just wondering if there is some sort of relationship between right coset and orbit of x. We just got to cosets, and it seems like the properties of cosets are eerily similar to orbits.

Alright, so here is my new proof of closure: Suppose a \phi1(g1)=g2 and \phi2(g2)=g3. Composing and such, it is still closed. I feel like I don't necessarily have to prove closure to demonstrate its a group just because binary operations are assumed to be closed. Another thing is how do...

Also, demonstrating its a homomorphism. Doesn't that immediately follow since it is a function through composition, composing a function with its inverse is kosher.

How would I demonstrate closure? Since its an isomorphism from G-> G won't it just keep sending elements to itself?

Given a group G. J = {\phi: G -> G: \phi is an isormophism}. Prove J is a group (not a subgroup!). Well we know the operation is function composition. To demonstrate J a group we need to satisfy four properties: (i) Identity: (I'm not sure what to do with this) (ii) Inverses: Suppose a...

Let G be a group and let \phi be an isomorphism from G to G. Let H be a subgroup. Hint: These subgroups should already be familiar to you. Let H={z in C:\phi(z)=z} This would be the subgroup of {-1,1}, this would be the group {-1,1} under multiplication. Let H={z in C: \phi(z)=-z}...

I understand that, I'm just having problems counting them. I haven't taken a combinatorics course or anything.

I would say 2!3!

I guess I'm not sure how to treat this stuff once I have different types of cycles

My rationale for pointing out that dividing by 3 gets the answer follows from this: gap (3 cycle) gap (3 cycle) gap But another thing is 3/3!, that yields 3 on the bottom

I realize that I can divide by 3 and get the right answer, though I'm sure that's not the right reason.

Well my next question is, how do I treat this one: (1,2)(3,4)(5,6)(7,8,9)(10,11,12) I wanted to treat it as such: \binom{12}{3}\binom{9}{3}\binom{6}{2}\binom{4}{2}\frac{1}{5!} However this came out wrong, I believe it undercounted.