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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.

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marked as duplicate by YACP, Erick Wong, Brandon Carter, Clive Newstead, Amzoti Feb 10 '13 at 17:43

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1  
$X=\{aH\mid a\in G\}$. –  Babak S. Dec 31 '12 at 16:26
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The title says $|H| = p,$ but the statement of the question states $[G:H] = p.$ The "proof" seems to mix the two up. It is not true in general that a subgroup of order $p$ is normal in $G$ when $p$ is the smallest prime divisor of $|G|$ (it is true if $[G:H] = p,$ and $p$ is the smallest prime divisor of $|G|$). –  Geoff Robinson Dec 31 '12 at 16:30
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@GeoffRobinson Ohhh, so does index mean the number of left cosets? Not the order? So what bit does the proof mess up, the bit about concluding we have several orbits which give us an order (?) of p - 1? –  Kaish Dec 31 '12 at 16:34
    
And generalization math.stackexchange.com/questions/33051/… –  user641 Dec 31 '12 at 20:28

4 Answers 4

up vote 2 down vote accepted

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 $K=H$. :)

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What's $S_{\Omega}$, the group with elements $\Omega$? Also, does the kernel bit mean it is all the intersections of the stabiliser? Why is this? I don't understand the "take $n = p$ bit and how you know it divides $n!$. –  Kaish Dec 31 '12 at 16:57
    
@Kaish: $S_{\Omega}$ is all bijections form omega to omega. Yes it is. –  Babak S. Dec 31 '12 at 17:06
    
It need not be true in general that $\cap_{a \in G} a^{-1}Ha$ is a maximal subgroup of $H,$ but that is not needed for your argument. –  Geoff Robinson Dec 31 '12 at 18:35
    
@GeoffRobinson: Thanks for noting that. I tried to fix it. I wanted to emphasize the OP indirectly that $K$ is $\cap_{a\in G}a^{-1}Ha$. In fact, it would be too long for the problem if I wanted to indicate what SteveD commented. I thik that link is enough for us to know everything. Thanks again, Geoff and Happy new year to you. $\ddot\smile$ –  Babak S. Jan 1 '13 at 7:48
    
Nice job of "walking through" a proof! +1 –  amWhy Feb 26 '13 at 14:31

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., $H\lhd G$.

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$[G:H]$ 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

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So where does my proof mess up? –  Kaish Dec 31 '12 at 16:47

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.

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