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Generalizing the case $p=2$ we would like to know if the statement below is true.

Let $p$ the smallest prime dividing the order of $G$. If $H$ is a subgroup of $G$ with index $p$ then $H$ is normal.

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    $\begingroup$ Hint: I think you should try to work with what Alex suggested. It is usually referred to as the "Strong Cayley Theorem". $\endgroup$
    – Ludolila
    Commented Feb 10, 2013 at 10:30
  • $\begingroup$ @Ludolila the link no longer works, can you update it? $\endgroup$ Commented Feb 22, 2020 at 19:16
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    $\begingroup$ @S.SundaraNarasimhan , here is another link: math.stackexchange.com/questions/2714715/… $\endgroup$
    – Ludolila
    Commented Apr 25, 2020 at 11:38
  • $\begingroup$ See also: Lang, Algebra, Chapter I Groups, §7 Sylow Subgroups, Lemma 6.7 (page 36) $\endgroup$
    – hbghlyj
    Commented May 22 at 16:12

9 Answers 9

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This is a standard exercise, and the answer is that the statement is true, but the proof is rather different from the elementary way in which the $p=2$ case can be proven.

Let $H$ be a subgroup of index $p$ where $p$ is the smallest prime that divides $|G|$. Then $G$ acts on the set of left cosets of $H$, $\{gH\mid g\in G\}$ by left multiplication, $x\cdot(gH) = xgH$.

This action induces a homomorphism from $G\to S_p$, whose kernel is contained in $H$. Let $K$ be the kernel. Then $G/K$ is isomorphic to a subgroup of $S_p$, and so has order dividing $p!$. But it must also have order dividing $|G|$, and since $p$ is the smallest prime that divides $|G|$, it follows that $|G/K|=p$. Since $|G/K| = [G:K]=[G:H][H:K] = p[H:K]$, it follows that $[H:K]=1$, so $K=H$. Since $K$ is normal, $H$ was in fact normal.

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    $\begingroup$ @hermes: If $x \in K$ then $xaH = aH$ for every coset $aH$. In particular this is true for the coset $H$ itself: $xH = H$, and so $x \in H$. It's easy to verify that $K = \cap_{g \in G}(gHg^{-1})$, which is the largest normal subgroup of $G$ which is contained in $H$. This normal subgroup $K$ is called the core of $H$ in $G$: en.wikipedia.org/wiki/Core_(group) $\endgroup$
    – user169852
    Commented Nov 4, 2015 at 20:18
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    $\begingroup$ How does it follow that if $p$ is the smallest prime dividing $|G|$ then $|G/K|=p$? $\endgroup$
    – sequence
    Commented Nov 27, 2015 at 6:04
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    $\begingroup$ @sequence: $|G/K|$ has $p$ as a prime factor since $|G/H| = p$ divides $|G/K|$. Also, $|G/K|$ divides $p!$, which does not have $p^2$ as a factor, so $p^2$ is not a factor of $|G/K|$. No prime smaller than $p$ divides $|G/K|$ because no such prime divides $|G|$. No prime larger than $p$ divides $|G/K|$ because no such prime divides $p!$. Conclusion: $|G/K|$ must be exactly $p$. $\endgroup$
    – user169852
    Commented Dec 4, 2015 at 19:07
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    $\begingroup$ @sequence: If $p-1$ is composite, then it can be expressed as a product of prime numbers, each of which will be smaller than $p-1$, then the argument in my previous comment applies. $\endgroup$
    – user169852
    Commented Dec 6, 2015 at 2:46
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    $\begingroup$ @Wlliam: Yes, your reading is correct; no, it is not particularly "powerful", because the condition (i) only applies to a single value for each finite group; and (ii) "most" of the time you don't have such subgroups. The idea of using the action on cosets to get a morphism to a finite group is really the take-away you want to take from this result. $\endgroup$ Commented Jan 10, 2022 at 21:53
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Here is a slightly different way to prove the result:

We will do it by induction on $|G|$. If $G$ has just one subgroup of index $p$ then clearly that subgroup is normal, so let $H_1$ and $H_2$ be distinct subgroups of index $p$. We then have that $|H_1H_2|$ is a multiple of $|H_1|$, but due to the choice of $p$ we must in fact have $H_1H_2 = G$ which means that if we let $K = H_1 \cap H_2$ then $K$ has index $p$ in $H_1$ and $H_2$ so by induction we know that $K$ is normal in $H_1$ and $H_2$ and thus normal in $G$. Now we know that $G/K$ has order $p^2$ so it is abelian. Now since both $H_1$ and $H_2$ contain $K$ they correspond to subgroups of $G/K$ and since this is abelian, they correspond to normal subgroups, which shows that they are normal in $G$ as desired.

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    $\begingroup$ How is it clear that if $G$ only have one subgroup of index $p$, it must be normal? $\endgroup$
    – leo
    Commented Jun 3, 2013 at 18:54
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    $\begingroup$ That is a very neat argument which I have not seen before. $\endgroup$ Commented Apr 12, 2016 at 8:54
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    $\begingroup$ @GeoffRobinson Yeah, I found it quite appealing as well when I found it, though it does slightly hide some technicalities in the fact that it uses that groups of order $p^2$ are abelian. $\endgroup$ Commented Apr 12, 2016 at 9:09
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    $\begingroup$ That's OK!- but you could finish in other ways too. Since $H_{1}H_{2} = G$, it is clear that $H_{1}$ and $H_{2}$ are not $G$- conjugate (and we may assume that $H_{2}$ does not normalize $H_{1}$). Then $H_{1}$ has $p$ different conjugates, all containing $K$, which forces $H_{2}$ to be normal as it is the only other subgroup of index $p$ containing $K$ ( just by counting). $\endgroup$ Commented Apr 12, 2016 at 9:18
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    $\begingroup$ The conclusion that $K$ is normal in $G$ could use a little clarity. $\endgroup$ Commented Nov 12, 2021 at 16:45
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Hint: Consider the set of cosets $G/H$ of which there are $p$. Then $G$ acts on these cosets by left multiplication. So you have a homomorphism $\phi: G \rightarrow S_p$. If $p$ is the smallest prime dividing $|G|$ then what can you say about $|\mathrm{im} \phi|$ and what does this imply about $\ker \phi$?

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  • $\begingroup$ @J: Is it possible to do it as follows: As $|H| \le |N(H)|$, implies $[G: N(H)] \le [G:H]$, as p is the smallest prime, this means $N(H)=H$ or $N(H)=G$. how do I show that $N(H) \ne H$ $\endgroup$
    – user23238
    Commented Mar 4, 2013 at 12:16
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proof: If $H$ is not normal then assume $H\neq H^g$ for some $g \in G$. But a classical formula says \begin{equation} |HH^g|=\dfrac{|H|\cdot |H^g|}{|H\cap H^g|} . \label{1} \tag{1} \end{equation}

Notice that $H\cap H^g$ is a proper subgroup of $H$ (proper since $H \neq H^g$). Hence, $|H\cap H^g|$ is a proper divisor of $|H|$. Since every prime divisor of $|H|$ is $\geq p$, this leads to $\dfrac{|H|}{|H\cap H^g|}\geq p$. Thus, \eqref{1} yields $|HH^g|\geq p|H^g| = |G|$. Therefore, $HH^g=G$.

Thus, $g=hg^{-1}kg$ for some $h,k\in H$. Therefore, $g=kh\in H$. As a result, $H=H^g$, which is a contradiction.

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  • $\begingroup$ I corrected some mistakes in your proof. Can you tell me why $|H|/|H\cap H^g| \geq p?$ $\endgroup$
    – user370967
    Commented May 19, 2017 at 11:56
  • $\begingroup$ Since $p$ is the smallest prime dividing the order of $G$. $\endgroup$
    – mesel
    Commented May 19, 2017 at 21:49
  • $\begingroup$ Can you elaborate? I have thought about it for a while. $\endgroup$
    – user370967
    Commented May 19, 2017 at 21:50
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    $\begingroup$ Since $H\neq H^g$, $H\cap H^g<H$. Since $1\neq |H:H\cap H^g|$ is a number dividing the order of $G$ then it must be greater than $p$. (if $q$ is prime dividing the |H:H\cap H^g| then $q\geq p$). $\endgroup$
    – mesel
    Commented May 19, 2017 at 21:53
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    $\begingroup$ you are welcome :) $\endgroup$
    – mesel
    Commented May 19, 2017 at 21:56
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Here's yet another, slightly different proof:

Instead of considering the action of $G$ on the left cosets of $H$, let us consider the action of $H$ on left cosets of $H$.

By the orbit-stabilizer theorem, the size of every orbit of cosets divides $|H|$, and hence also $|G|$. Since there are exactly $p$ cosets of $H$ and $p$ is the smallest prime dividing $|G|$, it must be that either there is a single orbit of size $p$ or there are $p$ different orbits, all of size $1$.

The first option, however, is impossible, since for every $h\in H$, $hH=H$, meaning the action fixes the coset corresponding to the identity. Hence, there exists an orbit of size $1$, so they must all be of size $1$.

This means that for every $h \in H, g \in G$ we have: $$hg^{-1}H=g^{-1}H\implies \exists h_1\in H \space s.t.\space hg^{-1}=g^{-1}h_1\implies \\ ghg^{-1}=h_1\in H$$

And we are done.

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Another answer if you know character theory over the complex numbers.

Let $H$ be a subgroup of order $G$ of index $p$, smallest prime dividing $|G|$. Look at the trivial character of $H$ and induce it to $G$. Then $1_H^G=\sum_{\chi \in{\rm Irr}(G)} a_{\chi}\chi$, with $a_{\chi} \in \mathbb{Z}_{\geq 0}$. Since $[1_H^G, 1_G]=[1_H,1_H]=1=a_{1_G}$, it follows that the irreducible constituents of $1_H^G$ have degree $\chi(1) \leq p-1 \lt p$. Since these degrees divide $|G|$ it follows that all the irreducible constituents $\chi$ must be linear, that is $\chi(1)=1$. Hence $H \supset \ker(1_H^G)=\cap\{{\ker(\chi) \in{\rm Irr}(G): a_{\chi} \gt 0}\} \supseteq G'$, whence $H \lhd G$.

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Hint: Let $G$ act on $G/H$ by left multiplication. This gives you a homomorphism $G\to S_p$. Try to show that $H$ is the kernel of this map--note that if $q$ is a prime larger than $p$ then $q\nmid p!$.

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  • $\begingroup$ Better: Note that $H$ leaves $H$ fixed, hence we get $H\to S_{p-1}$ - and that $q\ge p$ (we can't rule out $q=p$) implies $q\nmid(p-1)!$. $\endgroup$ Commented Feb 10, 2013 at 10:28
  • $\begingroup$ @HagenvonEitzen This is practically the same idea--this was just perhaps why my answer was in "hint" form. $\endgroup$ Commented Feb 10, 2013 at 10:31
  • $\begingroup$ It just looked to me like your hint would silently assume that $p^2\nmid |G|$. $\endgroup$ Commented Feb 10, 2013 at 10:59
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In addition to the things answered here, I just want to add this one.

Proposition Let $G$ be a finite group and $H$ a subgroup of prime index $p$, with gcd$(|G|,p-1)=1$. Then $G' \subseteq H$.

Note that this implies that $H \unlhd G$, and that it is in fact sufficient to prove that $H$ is normal, since then $G/H \cong C_p$ is abelian.

Proof Firstly, we may assume by induction on $|G|$, that $H$ is core-free, that is core$_G(H)=\bigcap_{g \in G}H^g=1$. This means that $G$ can be homomorphically embedded in $S_p$. Let $P \in Syl_p(G)$ and note that because $|S_p|=p \cdot (p-1) \cdots \cdot 1$, $|P|=p$. By the $N/C$-Theorem, $N_G(P)/C_G(P)$ embeds in Aut$(P) \cong C_{p-1}$. By the assumption gcd$(|G|,p-1)=1$, we get that $N_G(P)=C_G(P)$. Since $P$ is abelian we have $P \subseteq C_G(P)$, whence $P \subseteq Z(N_G(P))$. We now can apply Burnside's Normal $p$-complement Theorem, which implies that $P$ has a normal complement $N$, that is $G=PN$ and $P \cap N=1$. Note that $|G/N|=p$.

Look at the image of $H$ in $G/N$. Then $G=HN$, or $HN=N$. In the latter case $H \subseteq N$, and $|G:H|=|G:N|=p$, whence $H=N$ and we are done if we can refute the first case. If $G=HN$, then $|G:H \cap N|=|G:N|\cdot|N:H \cap N|=|G:N|\cdot |G:H|=p \cdot p=p^2$, contradicting the fact that $|G| \mid |S_p|$. The proof is now complete.

Corollary 1 Let $G$ be a finite group and let $H$ be a subgroup with $|G:H|=p$, the smallest prime dividing the order of $G$. Then $G' \subseteq H$. In particular, $H$ is normal.

Corollary 2 Let $G$ be a finite group and let $H$ be a subgroup with $|G:H|=p$ and gcd$(|H|,p-1)=1$. Then $H$ is normal.

Observe that this last result renders a well-known result for $p=2$! Finally for fun:

Corollary 3 Let $G$ be a finite group of odd order and $H$ a subgroup with $|G:H|=65537$. Then $H$ is normal.

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If $X$ is a (left) $G$-set, $x$ in $X$ with stabilizer $G_x$, then the stabilizer of $g\cdot x$ is $G_{gx} = g G_x g^{-1}$.

Now consider the transitive action of $G$ on $G/H$ by left translations. The stabilizer of $e \cdot H$ is $H$, and the stabilizer of $g \cdot H$ is $gHg^{-1}$. Therefore, the kernel of the map $$G \to \operatorname{Sym}(G/H)$$ is $$C_G(H) \colon =\bigcap_{g\in G} g H g^{-1}$$ which is the largest normal subgroup of $G$ contained in $H$ ( the normal core of $H$), so we have an embedding of groups

$$G/C_G(H) \hookrightarrow \operatorname{Sym}(G/H)$$

With Lagrange we get the divisibility

$$|G/C_G(H)| = [G\colon H] \cdot [H\colon C_G(H)]\, \mid \, |\operatorname{Sym}(G/H)|$$

so

$$[H\colon C_G(H)]\, \mid \, \frac{|\operatorname{Sym}(G/H)|}{[G\colon H]}$$

But from the hypothesis, no prime dividing $[H\colon C_G(H)]$ can divide $\frac{|\operatorname{Sym}(G/H)|}{[G\colon H]}$, so there are none, and
$[H\colon C_G(H)]=1$ , $H = C_G(H)$ is normal.

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