when a subgroup is in the center of other subgroup

**Amer** · December 14th 2011, 07:03 AM

The question is
Let G be a group, $N \lhd G$ such that $\[G:N\] = r$
let $H \lhd G$ with finite order show that
if
$H \cap N = e$ then $nh = hn \;\; \forall \; h\in H \; , n\in N$

**Deveno** · December 14th 2011, 07:39 AM

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:

$hnh^{-1}n^{-1} \in N$ since $hnh^{-1}, n^{-1} \in N$ , because N is normal, while

$hnh^{-1}n^{-1} \in H$ , since $h, nh^{-1}n^{-1} \in H$ because H is normal.

since H∩N = {e}, we see that h and n commute.

**Amer** · December 14th 2011, 07:52 AM

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

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