Recent content by OMM!

There is a Theorem that says FG-Modules are equivalent to group representations: "(1) If \rho is a representation of G over F and V = F^{n}, then V becomes an FG-Module if we define multiplication vg by: vg = v(g\rho), for all v in V, g in G. (2) If V is an FG-Module and B a basis of V...

Thanks for your help, I'll give it another shot in the morning. Found a bit on Frobenius Groups which seems quite related, which I think may be my bed-time reading! Thanks again, you're a great help!

And the order of that element divides q? It can't be 1, else the group would be Abelian i.e. yxy^{-1} = x ===> yx = xy So the order must be q, as q is prime?

I'm not entirely sure! I am basically saying that C_{q} is the cyclic group of order q, thus is generated by < y >. In other words: C_{q} = < y | y^{q} = 1 > The elements of which are: {1, y, y^{2}, ... , y^{q-1}}

I'd say yes (With very little conviction! Haha) As I said in the last post, if we have an element y in C_{q} which is obviously of order q, then y is also in C_{p-1}?

So for y in C_{q}, we have y^{q} = 1 And if q divides p-1, then y^{p-1} = 1? Does the order of the element \phi(y) divide p-1? Or does it divide q? (Sorry, probably obvious, but having a mental moment!) Then to find the possible element for \phi(y) for C_{p-1} = < \alpha > \phi(y)...

Hi micromass, thanks for your help once again! So if we say, < y > = C_{q}, then we need to show there is a homomorphism \phi that sends y to \phi(y) which is an element of Aut C_{p} = C_{p-1}. And since p \equiv 1 (mod q), then C_{p-1} isomorphic to C_{q}? And clearly there exists a...

Homework Statement Let p, q be distinct primes s.t. p \equiv 1 (mod q). Prove that there exists a non-Abelian group of order pq and calculate the character table. Homework Equations Semi-Direct Product: Let H = < Y | S > and N = < X | R > be groups and let \phi : H \rightarrow Aut...

So if G = Q8 = <a, b : a^4 = 1, b^2 = a^2, b^{-1}ab = a^{-1}> I'm fine with the notion of the derived subgroup G' = <[g,h] : g, h in G> (Where [g,h] = g^{-1}h^{-1}gh) But I can't see why G' = {1, a^2}, I can only seem to get everything to be 1!? i.e. g = a, h = a^3 ===> a^{-1}a^{-3}aa^3...

Thanks for your help! After a bit of trial and error, I can see we could take c = 1/n and that would satisfy for e_2! So we now have e_1 = 1 and e_2 = (1/n)E For e_3 I assumed we'd take a complex conjugate of "c" or (1/-n), but that obviously won't work as we end up with: e_3^2=-e_3. Or am...

I managed to get a bit more information about part (iii) and I was told to use a specific example of S_3, the symmetric group of order 6. (iii) asks to find 3 solutions to the equation (e_i)^{2} = e_i inside the group algebra C[S_3]. where C = complex numbers. I guess the identity...

Hi it's "Representations & Characters Of Groups" by Liebeck & James. Although the questions aren't from there, they were set for me to help me understand irreducible CG submodules and CG algebras a bit better! As I'm struggling here on this! Aghhh! Thanks, I can see the injectivity solution...

Apologies, the LaTeX thing doesn't seem to be working, so not very clear! I am working through a book on representation theory, but am stuck on these exercises. Homework Statement Let G be a finite group and let E = Sum g (running over all g in G). (i) Prove: Ex = E (forall x in G)...