1. ## a question about normal subgroups and index.

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.

2. Originally Posted by abhishekkgp
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.
(1) $\displaystyle (HN)/N=H/(H \cap N)$

Since N is a normal subgroup of G and H is a subgroup of G, HN is a subgroup of G. Thus, $\displaystyle |(HN)/N|=|H/(H \cap N)|$ divides [G:N] by (1) and Lagrange's theorem. It follows that $\displaystyle |H/(H \cap N)|$ divides both [G:N] and |H|. Since gcd(|G:N|,|H|)=1, we see that $\displaystyle |H/(H \cap N)|=1$. Thus, $\displaystyle H \cap N=H$. We conclude that H is a subgroup of N.

3. ## discussing the solution.

so your solution uses the fact that if N is a normal subgroup and H is a subgroup then
(H intersection N) is a normal subgroup of H??
is that right?

(why \cap not working?)

4. If $\displaystyle |H/(H \cap N)|=1$, then $\displaystyle H \cap N = H$. If $\displaystyle H \cap N = H$, then H is a subgroup of N.

5. ## discussion

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??

6. It is the second isomorphism theorem. Yes, H intersection N is a normal subgroup of H. Define a surjective homomorphism f:H--->HN/N such that its kernel is (H intersection N). Now the kernel of f has to be the normal subgroup of H.

7. 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.

8. Alternatively one can us the first isomorphism theorem. Note that since $\displaystyle N\unlhd G$ we have that $\displaystyle G/N$ is a group with order $\displaystyle \left(G:N\right)$. Consider then the canonical projection $\displaystyle \pi:G\to G/N$. We see then that the restriction $\displaystyle \pi_{\mid H}:H\to G/N$ is also a homomorphism but since $\displaystyle \left(|H|,\left|G/N\right|\right)=1$ it follows that $\displaystyle \pi_{\mid H}$ is the trivial homomorphism $\displaystyle 1:H\to G/N:h\mapsto N$. In particular it follows that $\displaystyle \displaystyle \pi_{\mid H}\left(H\right)=HN=N$ but this is true if and only if $\displaystyle H\leqslant N$.

P.S. I implicitly used the FIT by using it to prove that if $\displaystyle \phi\in\text{Hom}\left(G,G'\right)$ with $\displaystyle |G|,|G'|<\infty$ then $\displaystyle \left|\phi(G)\right|$ divides both $\displaystyle |G|$ and $\displaystyle |G'|$.

9. ## Re: a question about normal subgroups and index.

Originally Posted by Drexel28
Alternatively one can us the first isomorphism theorem. Note that since $\displaystyle N\unlhd G$ we have that $\displaystyle G/N$ is a group with order $\displaystyle \left(G:N\right)$. Consider then the canonical projection $\displaystyle \pi:G\to G/N$. We see then that the restriction $\displaystyle \pi_{\mid H}:H\to G/N$ is also a homomorphism but since $\displaystyle \left(|H|,\left|G/N\right|\right)=1$ it follows that $\displaystyle \pi_{\mid H}$ is the trivial homomorphism $\displaystyle 1:H\to G/N:h\mapsto N$. In particular it follows that $\displaystyle \displaystyle \pi_{\mid H}\left(H\right)=HN=N$ but this is true if and only if $\displaystyle H\leqslant N$.

P.S. I implicitly used the FIT by using it to prove that if $\displaystyle \phi\in\text{Hom}\left(G,G'\right)$ with $\displaystyle |G|,|G'|<\infty$ then $\displaystyle \left|\phi(G)\right|$ divides both $\displaystyle |G|$ and $\displaystyle |G'|$.
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 $\displaystyle |G:N| = k, |H| = m$.

since $\displaystyle gcd(k,m) = 1$, there are integers $\displaystyle r,s$ with $\displaystyle rk + sm = 1$
If $\displaystyle H$ is trivial then $\displaystyle H \leq N$ and we are done.

so suppose $\displaystyle H$ is not trivial. Let $\displaystyle h \neq e$ be chosen arbitrarily in $\displaystyle H$.

then $\displaystyle Nh = (Nh)^{rk + sm}= (Nh)^{rk}(Nh)^{sm}$

$\displaystyle \Rightarrow Nh= [(Nh)^k]^r[(Nh)^m]^s$

$\displaystyle \Rightarrow Nh= (N^r)(N(h^m))^s = N(Ne)^s = N(N) = N$

so $\displaystyle h$ must be in $\displaystyle N$, thus $\displaystyle H < N$.