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OMM!
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Homework Statement
Let p, q be distinct primes s.t. p [itex]\equiv[/itex] 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 [itex]\phi[/itex] : H [itex]\rightarrow[/itex] Aut (N) be a homomorphism. Then the SDP has the presentation:
N x H = < X, Y | R, S, [itex]y^{-1}xyw_{x,y}^{-1}[/itex] >
With x in X, y in Y, [itex]w_{x,y}[/itex] = ([itex]\phi[/itex](y(x))) (i.e. [itex]\phi[/itex] of y and then [itex]\phi[/itex] of y of x etc.) in < Y > = N
The Attempt at a Solution
I've done a specific example where p=7 and q=3 and found a group of order 21, using the semi-direct product of 2 cyclic groups [itex]C_{7}[/itex] and [itex]C_{3}[/itex].
But can't see how to prove that there exists a group more generally. So far I have.
If N = [itex]C_{p}[/itex] = < x > and H = [itex]C_{q}[/itex] = < y >, then [itex]\phi[/itex] : [itex]C_{q}[/itex] [itex]\rightarrow[/itex] Aut ([itex]C_{p}[/itex]) = [itex]C_{p-1}[/itex] = < [itex]\alpha[/itex] >, then we have y [itex]\mapsto[/itex] [itex]\phi[/itex](y) too.
When p=7 and q=3
I had that [itex]\phi[/itex](y) = {1, [itex]\alpha^{2}[/itex], [itex]\alpha^{4}[/itex]}
If [itex]\phi[/itex](y) = 1, then we'd have that the SDP of [itex]C_{7}[/itex] and [itex]C_{3}[/itex] would be ismorphic to [itex]C_{7}[/itex] x [itex]C_{3}[/itex].
If [itex]\phi[/itex](y) = [itex]\alpha^{2}[/itex], then the SDP of [itex]C_{7}[/itex] and [itex]C_{3}[/itex] would be equal to < x, y >
Then by the relation in the SDP, we'd have: yx[itex]y^{-1}[/itex] = [itex]\alpha^{2}[/itex](x) = x^{t}
And we have that: t = {1, 2, 3, 4, 5, 6}, with p=7 and q=3, then: t^{3} [itex]\equiv[/itex] 1 (mod 7), and so: t = 1 and t = 2 and t = 4, thus we have groups:
G = < X, Y | [itex]x^{7} = y^{3} = 1[/itex], [itex]yxy^{-1} = x[/itex] > = [itex]C_{7}[/itex] x [itex]C_{3}[/itex]
G = < X, Y | [itex]x^{7} = y^{3} = 1[/itex], [itex]yxy^{-1} = x^{2}[/itex] >
G = < X, Y | [itex]x^{7} = y^{3} = 1[/itex], [itex]yxy^{-1} = x^{4}[/itex] >
But how would I get to this stage more generally?! Thanks for any help!