Recent content by sairalouise

I'm trying to show that there is not one sentence (formula) that if a group satisfies this formula it is equivalent to the group being infinite. I can show this in a hap hazard way analogous to the same problem in the empty language , but how do you use the fact that the model is a group and...

How would you show that for every countable order, there is an order embedding function f: (A,<) -> (Q,<) ? Is this similar to the proof of the w-categoricity of the Theory of Dense Linear Orders?

If you assume that ZFC is consistent, then by the main theorem of model theory ZFC has a model, let the model be countable. Since ZFC proves: "there is a set consisting of all real numbers" there is a point a belonging to M such that: M satisfies " a is the set of all real numbers" But since...

Given H,K and general finite subgroups of G, ord(HK) = [(ord(H))(ord(K))] / ord(H intersection K) I know by the first isomorphism theorem that Isomorphic groups have the same order, but the left hand side of the equation is not a group is it? I am struggling to show this.

If a is the only element of order 2 in a group, does it belong to the centre Z(G)?

Hey, now i can!

i just don't see how to get from one side of the equation to the other.

I can show that the orders of an element and its inverse are equal, and have tried supposing that ab and ba have different orders to reach a contradiciton but i can't work the problem though.

I'm looking to show that: Order (ab(a^-1)) = Order b So far... Let x be order of ab(a^-1), so we have: e = (ab(a^-1)^x) = (a^x)(b^x)(a^(x-1)) = (a^x)(b^x)((a^-1)^x) so by associativity we have... = (a^x)((a^-1)^x)(b^x) = (((a)(a^-1))^x)(b^x) = (e^x)(b^x) = b^x Hence...

Hi, I don't understand the step that goes: Using associativity, we get (ab)^{x} = a(ba)^{x-1}b = 1. Could someone elaborate, thanks!