Recent content by math_nerd

First of all, I used to be the kid that hated math back in high school. But did okay, besides in PreCalc. I go to UCSD now, it's my junior year, and I'm now a Math-Computer Science major. I only declared CSE recently. I got a D in my modern algebra class. I studied super hard too, but didn't...

* H is a subgroup. And how is this true? I'm pretty lost on how to show this.

Suppose that H is a subset of S_4, and that H contains (12) and (234). Prove that H=S_4. The order of S_4 is 24. (12)(234) = (1342) shows that the finite subset of group H is closed under the group operation, so it is a subgroup of H. Then, using Lagrange's Theorem, |S_4:H|=24/2 = 12. so...

Hmm...okay. So here's a problem in the book that I think applies this concept. Prove that group order of 12 must have an element of order 2. This problem uses the converse, because 12|1,2,3,4,6,12. But by Lagrange we can say that order of 2 is definitely there. But to prove that order 3 is not a...

This is not a homework problem. In Gallian, there is an example given: The group A_4 of order 12 has no subgroups of order 6. I can't seem to understand what this means in terms of how this is the "converse" of Lagrange's Theorem.

Suppose that H is a subset of S_4, and that H contains (12) and (234). Prove that H=S_4. The order of S_4 is 24. (12)(234) = (1342) shows that the finite subset of group H is closed under the group operation, so it is a subgroup of H. Then, using Lagrange's Theorem, |S_4:H|=24/2 = 12...

Suppose that H is a subset of S_4, and that H contains (12) and (234). Prove that H=S_4. The order of S_4 is 24. The order of H is 2. (12)(234) = (1342) shows that the finite subset of group H is closed under the group operation, so it is a subgroup of H. Then, using Lagrange's...

Okay thanks!

I want to know if the way I have choose g'=c-di and g=a+bi to be two elements in G is written with the correct logic? For the part I said I did the arithmetic, I did incorrectly. I want to prove the mapping from G to G is onto. So I need to show that phi(a+bi)= c+di. How?

So I have to show group G maps to itself.

So I need to show that the mapping Phi(a+bi)=a-bi is an automorphism of the group of complex numbers under addition. The way I have proved it is to show that it is a mapping (we are given this piece of information), the mapping is 1-1 and onto, and the operation is preserved. Also, I assumed...

So for part 2, I can say that a-bi=c-di, so a=c and b=d must be true for a-bi=c-di. This then implies a+bi=c+di is true. Also, is rest of the proof right?

Show that the mapping Phi(a+bi)=a-bi is an automorphism of the group of complex numbers under addition. I have this as of now: Let (c+di) and (a+bi) be elements in group G. 1) phi(a+bi) is function phi from G to G, by assumption. Therefore the function is a mapping. 2) 1-1: assume...

Let me give another example, like to prove that f3n is an even integer, we can apply the formula, and it works. But in the problem stated above, it does not seem to work the same way. It only works when you add F2n+1 to both sides of the equation.

For n= n + 1, I get: ...+ f2n-1 + f2n+1 = f2n+2. So we can state that f2n + f2n+1 = f2n+2...which is true, but my issue is why does this not work when you use the fibonacci formula?