The question is
Let G be a group, such that
let with finite order show that
if
then
i am a little unsure as to how [G:N] = r is relevant to your question, other than saying N is of finite index. i also don't know what you mean by "one subgroup is in the center of the other" in terms of your subsequent question.
if H is in Z(N), of course nh = hn, but then it cannot be the case that H∩N = {e}, unless Z(N) is likewise trivial.
but if H,N are both normal in G, with H∩N = {e}, then for any h in H, and n in N, we have:
since , because N is normal, while
, since because H is normal.
since H∩N = {e}, we see that h and n commute.
thanks
the problem in Abstract algebra when you see the answer you say oooh it is easy how i did not figure that
the hardness is the idea of the solution
I think if I had thought in it in reverse nh = hn and see what will happen
anyway thanks very much