- #1
Samuelb88
- 162
- 0
Sylow's theorem tells us that there is one 7-Sylow subgroup and either one of seven 3-Sylow subgroups. Call these subgroups H and K respectively. Sylow's theorem also tells that H is normal in G.
I'm not going to write it all out as I don't think it's necessary but in the case when we have seven 3-Sylow subgroups, we conclude that the generators [itex]x[/itex] (of order 7) and [itex]y[/itex] (of order 3) generate the entire group G. Since H is normal, we have know that [itex]yxy^{-1} = x^k[/itex], for some [itex]k[/itex], [itex]0 \leq k \leq 6[/itex]. [itex]k[/itex] cannot equal 0 and 1 because that would imply [itex]x = e[/itex] in the first case and G is abelian in the second case, contrary to assumption in both cases.
Here's where I get lost:
Since [itex]y[/itex] has order 3, and [itex]y^3 x y^{-3} = x^{k^3}[/itex] ...
How did he infer that [itex]y^3 x y^{-3} = x^{k^3}[/itex] from what was given?
I'm not going to write it all out as I don't think it's necessary but in the case when we have seven 3-Sylow subgroups, we conclude that the generators [itex]x[/itex] (of order 7) and [itex]y[/itex] (of order 3) generate the entire group G. Since H is normal, we have know that [itex]yxy^{-1} = x^k[/itex], for some [itex]k[/itex], [itex]0 \leq k \leq 6[/itex]. [itex]k[/itex] cannot equal 0 and 1 because that would imply [itex]x = e[/itex] in the first case and G is abelian in the second case, contrary to assumption in both cases.
Here's where I get lost:
Since [itex]y[/itex] has order 3, and [itex]y^3 x y^{-3} = x^{k^3}[/itex] ...
How did he infer that [itex]y^3 x y^{-3} = x^{k^3}[/itex] from what was given?