A normal subgroup with index relatively prime to its order

Question

This is an exercise from Lang's Algebra. The theorem and my work on it are below:

Let $G$ be a finite group and $N$ be a normal subgroup such that $N$ and $G/N$ has relatively prime orders. I need to show the following two:

i)Let $H$ be a subgroup of $G$ having the same order as $G/N$. Prove that $G =HN$.

ii)Let $g$ be an automorphism of $G$. Prove that $g(N)=N$.

For the first one, I do the following: It can be shown that $HN$ is a subgroup of $G$. By our assumption, I can say that $|H \times N |= |H||N|=|G|.$ Then I define the following map: $\varphi: H \times N \rightarrow G$ where $\varphi(h,n) = hn$. I tried to show that $\varphi$ is an isomorphism but I could not find a way to show it is bijective or surjective. One is enough to finish the proof.

For the second one, I defined two maps $f_1$ and $f_2$ such that: $f_1 : G \rightarrow G / N$ with $f_1(g) = gN$ and $f_2:G / N \rightarrow G$ with $f_2(gN)=g$. By assumption, I have $g: G \rightarrow G$ which is an isomorphism. We have to show that $g(N)=N$.

(*)By first isomorphism theorem, I can say that $f_2(f_1(x))=g(x)$. Now, $g(N)=f_2(f_1(N))=f_2(N)=N$. Is this part correct with * correct?

Answer 1

For $i)$:

Since $|H|=|G/N|$, $|H|$ and $|N|$ are relatively prime by assumption which means that if $x\in H\cap N$, $x=e$ for the order of $x$ must divide both $|H|$ and $|N|$. Because $$|HN|=\frac{|H||N|}{|H\cap N|}$$ we see that $|HN|=|H||N|=|G|$.

For $ii)$ what you've written is not correct. For one, $f_2$ isn't even well-defined because there are many elements in $gN$ and it's unclear how we ought to choose one. Instead try this:
Because $g$ is an automorphism, $|g(N)|=|N|$ and $g(N)$ is also a subgroup so it is sufficient to show that $N$ is the unique subgroup of order $|N|$. To see this, note that if $H$ is any other subgroup, $$|HN|=\frac{|H||N|}{|H\cap N|}=\frac{|N|^2}{|H\cap N|}$$ but $|HN|$ divides $|G|$ by Lagrange's theorem so $$\frac{|G|}{|N|}\frac{|H\cap N|}{|N|}$$ is an integer and since $\frac{|G|}{|N|}$ and $|N|$ are relatively prime, it follows that $|H\cap N|=|N|$ so $H=N$.

Answer 2

Here is another solution for the second part.

Problem. Let $G$ be a finite group and $N\lhd G$ a normal Hall-subgroup, that is, a subgroup s.t. $\gcd(|G/N|,|N|) = 1$. Prove that $N$ is charachteristic in $G$, meaning that each $\theta \in \operatorname{Aut}(G)$ satisfies $\theta(N)=N$.

Proof. We'll prove that $\theta(N)\subseteq N$, and note that the other inclusion follows from the same argument for the automorphism $\theta^{-1}$.

Denote $n=|N|$ and $m=|G/N|$. They are co-prime, therefore there exists $a,b\in \Bbb{Z}$ with $an+bm=1$. Let $x\in N$. From Lagrange's theorem we have $x^n = e$ and we deduce that $x=x^{an+bm}=x^{bm}$. Every automorphism $\theta \in \operatorname{Aut}(G)$ therefore satisfies $\theta(x)=\theta(x^{bm})=\theta(x)^{bm}$.

Denote $g=\theta(x)$, and we wish to prove that $g\in N$. Since $N$ is normal, $G/N$ is a group. Apply Lagrange's theorem to deduce that $g^m N=(gN)^m = N$. Thus $g^m\in N$, so we must have $g^{bm} \in N$ too. But $g^{bm} = \theta(x)^{bm} = \theta(x)\in N$, so we're done. $\square$