$G$ is a finite group such that $|G| = n$ and $p$ is minimal prime dividing $n$. $H \subset G, [G:H] = p$. Prove $H$ is normal in $G$

Question

Let $G$ be a finite group of order $n$ and $p$ be the minimal prime number dividing $n$. Assume that $H \subset G$ is a subgroup of index $p$. Prove that $H$ is normal: $H \trianglelefteq G$

To do this, I said, lets first take the set $X = \{aH\}$, i.e the set of left cosets of H, where $|X| = p$.

If we then consider the action of $G$ on $X$ by left shifts, we get $(g,aH) \rightarrow (ga, H)$. We then have just one orbit such that $X = Orb(H)$.

Now, lets consider the same action but by group $H$. This gives us $(h,aH) \mapsto (ha,H)$. Here, there is a special orbit consisting of just one element, $H : hH = H \,\,\,\,\forall h \in H$. (How do you get this?)

We then conclude that $X \ \{H\}$ consists of several orbits, where their index is given by $|X \ \{H\}| = p -1$, and the number of elements in an orbit divides $G$, $p$ is the minimal prime dividing $|G| \implies $ All orbits consist of just 1 orbit (What does that mean? How can all orbits have 1 orbit?)

We therefore get, for any $h \in H, a \in G$, we get

$$h(aH) = aH \implies a^{-1}haH = H \iff a^{-1}Ha \in H \iff H \trianglelefteq G$$

Does this proof make sense? Can you help me with the questions in text please? Thank you.

Answer 1

Lets construct a proof for this problem. $G$ is a group and $H\leq G$ be its subgroup. Assume $\Omega=\{Ha\mid a\in G\}$ and define $$(Ha)^x=Hax, \;\; Hax\in\Omega, x\in G$$ This defines an action and you know that. What is the stabilizer of $Ha$? It is $a^{-1}Ha$. Now assume this mapping $\phi:G\to S_{\Omega}$ which takes $g$ to $\bar{g}:\Omega\to\Omega$ with $\bar{g}(Ha)=Hag$. It is easy to see that this map is a group homomorphism. What is its kernel? It is $\cap_{a\in G}a^{-1}Ha$ which is the maximal subgroup in $H$ normal in $G$. Do you know it from somewhere else? check it. It can be proved that if index of $H$ in $G$ is $n$, so there is a normal subgroup of $G$ such $K\subseteq H$ such that $[G:K]<\infty$ and divides $n!$. Now take $n=p$. Doesn't $[G:K]\big||G|$? Doesn't $[G:K]\big|p!$? Isn't $p$ is the smallest prime dividing the order of $G$? What does it mean? It means that $[G:K]\big|p$ and so $K=H$. :)

Answer 2

Let $N=N_G(H)$ (the normalizer of $H$). Then it is clear that $H\subseteq N\subseteq G$. Since $[G:H]=p$, which is the smallest prime dividing $n=|G|$, it must be the case that $H=N$ or $G=N$ (otherwise there is a smaller prime dividing the order of $G$). If $N=G$, we're done, so assume $N=H$. Let $S$ be the set of all conjugates of $H$ and let $G$ act on $S$ by conjugation. Then there exists a homomorphism $\varphi:S\to\text{Perm}(S)$ given by this action. Now, observe that $|S|=[G:N]=[G:H]$, so $\text{Perm}(S)\cong S_p$, so this gives a homomorphism $\varphi^*:S\to S_p$.

Let $K=\ker(\varphi^*)$; then $G/K\hookrightarrow S_p$ (an imbedding). This implies that $|G/K|=[G:K]$ divides $p!$, but $K\subseteq N_G(H)=H$, thus $[G:K]=[G:H][H:K]=p[H:K]$. This says that since $p[[H:K]$ divides $p!$, we have $[H:K]$ divides $(p-1)!$. By Lagrange's theorem, we also have $|H|=|K|[H:K]$ divides $n$, hence $[H:K]=1$, which only happens if $H=K$, but then $H$ is the kernel of a homomorphism, so is normal in $G$, a contradiction to our assumption that $H=N_G(H)$. Therefore, it is the case that $N_G(H)=G$, i.e., $H\lhd G$.

Answer 3

$[G:H]$ is the index of H in G i.e. $[G:H] = \frac{|G|}{|H|}$, not the order of H. It is also the number of left cosets

Answer 4

Consider the homomorphism $\phi:G\rightarrow S_{G:H}$ (Where $S_{G:H}$ is the group of permutations of $G:H$) that sends $x$ to the permutation $T_x$ of $G:H$ that sends $aH$ to $xaH$.

Claim: $Ker \,\phi$ is not trivial implies $H\lhd G$.

Proof: Suppose that $Ker \,\phi$ is not trivial. We find that $HKer \,\phi $ is a subgroup of $G$ and $|H|<|HKer \,\phi|$. since $|G:H|=p$, we deduce using Lagrange's theorem that $HKer \,\phi =G$.It is easy to verify that $Ker \,\phi=\{x\in G|\forall a\in G[axa^{-1}\in H]\}$ . Let $g\in G$, we know that $\exists h\in H\,\exists x\in Ker \,\phi [g=hx]$ . Now let $h'$ be any element of $H$, we get: $$h'gh'^{-1}=h'hxh'^{-1}=h'hh'^{-1}h'xh'^{-1}$$ Since both of $h'hh'^{-1},h'xh'^{-1}$ belong to $H$, therefore $h'gh'^{-1}\in H$. Thus, $H\lhd G $.

Hence assume WLOG that $Ker \,\phi$ is trivial. It follows that $G$ is isomorphic to a subgroup of $S_{G:H}$. Thus, $|G|\ |\ |G:H|$. Hence, $|G|$ divides $p!$. From this and the minimality of $p$ one can deduce that the order of $G$ is exactly $p$. Now that $G$ is cyclic, we get that $H$ is normal.