Two subgroups of the same order, and the normaliser of one in the other

Question

Let $G$ be a group, $H\unlhd G$, and $|H|$ and $|G|/|H|$ are coprime. I want to show that $H$ is a characteristic subgroup of $G$. To do this, I figure that I just need to prove that $H$ is the only normal subgroup of order $|H|$. Looking around in my book, I see a related exercise that could help me do this:

(Exercise, or lemma) Let $G$ be a group, $H,K\le G$, and $|H|=|K|$. $|H|$ and $|G|/|H|$ are coprime. Show that $N_H(K)=N_K(H)=H\cap K$, where $N_H(K)$ is the set of normalisers of $K$ in $H$.

Once I finished this exercise, I could solve the problem at the beginning of this post. However, after thinking for a while, I was unable to make any use of the condition $|H|=|K|$, so I looked at the hints given.

Hint: Let $\pi$ be the set of all prime factors of $|H|$. Try to show that if $H\cap K\subsetneq N_H(K)$, then all prime factors of $|N_H(K)|$ lies in $\pi$. $|N_H(K)|$ divides $G$, contradicting the fact that $|H|$ and $|G|/|H|$ are coprime.

I cannot understand this hint - the first sentence is very obvious, so obvious that it doesn't need proof. The second sentence doesn't make sense - why is it a contradiction?

So, how can I solve this exercise?

Answer 1

Note that $KN_{H}(K)=N_{H}(K)K$ is a subgroup of $G$ containing $K$. By Lagrange we have that $|K|$ divides $|KN_{H}(K)|$ divides $|G|=|K||G:K|$. Hence $|KN_{H}(K)|/|K|$ divides $|G:K|$.

However we also have $|KN_{H}(K)|=|K||N_{H}(K)|/|K\cap N_{H}(K)|$ divides $|K||N_{H}(K)|$, which by Lagrange divides $|K||H|=|K|^2$.

As $|G:K|$ and $|K|^2$ are coprime we have that $|KN_{H}(K)|/|K|=1$, and so $N_{H}(K)\leqslant K$. As $N_{H}(K)\leqslant H$, we get $N_{H}(K)\leqslant K\cap H$. It is clear that $K\cap H\leqslant N_{H}(K)$

Answer 2

Suppose $K$ is a subgroup of $G$ such that $|K|=|H|$. Consider the homomorphism $f: K\to G/H$ given by $f(k)=k+H$ for all $k\in H$. Note that ${\rm Im}(f)$ is a subgroup of $G/H$ and ${\rm Im}(f)\cong K/(K\cap H).$ So $|{\rm Im}(f)|$ divides $|G/H|$ and $|K|=|H|$. Since $|G/H|$ and $|H|$ are coprime, $|{\rm Im}(f)|=1$, that is, ${\rm Im}(f)=\{0\}$. So $K\subseteq H$. Since $|K|=|H|$, we have $K=H$.