given that N is a normal subgroup of G. also H is given to be a subgroup of G.
also given that G is a finite group with gcd(|G:N|,|H|)=1.
prove that H is a subgroup of N.
i got that part... i am talking about the very first line in your first reply.
there you have written HN/N = H/(H intersection N).
For the RHS of the above equation to be well defined its necessary that
(H intersection N) is a normal subgroup of H. is this correct??
thanks in advance..
thanks a lot. that helped.
also i found another solution which you may find intersesting and more algebraic.
here it is:
suppose that h is in H. and suppose [G:N] = k, |H| = m.
since gcd(k,m) = 1, there are integers r,s with rk + sm = 1.
it may be that H is trivial, in which case H < N.
so suppose not. let h ≠ e be chosen arbitrarily in H.
then Nh = (Nh)^(rk + sm) = (Nh)^(rk)(Nh)^(sm)
= [(Nh)^k]^r[(Nh)^m]^s
= (N^r)(N(h^m))^s = N(Ne)^s = N(N) = N,
so h must be in N, thus H < N.
Alternatively one can us the first isomorphism theorem. Note that since we have that is a group with order . Consider then the canonical projection . We see then that the restriction is also a homomorphism but since it follows that is the trivial homomorphism . In particular it follows that but this is true if and only if .
P.S. I implicitly used the FIT by using it to prove that if with then divides both and .
i had asked this long back but somehow i didn't see it and i checked it out today. brilliant!
i got another solution which is cool too:
Suppose .
since , there are integers with
If is trivial then and we are done.
so suppose is not trivial. Let be chosen arbitrarily in .
then
so must be in , thus .