56
$\begingroup$

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.

$\endgroup$
  • 3
    $\begingroup$ Hint: I think you should try to work with what Alex suggested. It is usually referred to as the "Strong Cayley Theorem". $\endgroup$ – Ludolila Feb 10 '13 at 10:30
90
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Thanks. The part 'it follows that $|G/K|=p$' is crucial. Now everything is done! Bye. $\endgroup$ – Sigur Jun 28 '12 at 17:51
  • 2
    $\begingroup$ Why $K\subset H$? It seems that $K=\{k:k=hg^{-1},\:h\in H\}$. $\endgroup$ – Math Wizard Aug 10 '15 at 6:29
  • 8
    $\begingroup$ @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) $\endgroup$ – Bungo Nov 4 '15 at 20:18
  • 3
    $\begingroup$ @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 $p$. $\endgroup$ – Bungo Dec 4 '15 at 19:07
  • 1
    $\begingroup$ @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. $\endgroup$ – Bungo Dec 6 '15 at 2:46
20
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ How is it clear that if $G$ only have one subgroup of index $p$, it must be normal? $\endgroup$ – leo Jun 3 '13 at 18:54
  • 1
    $\begingroup$ Aha: $[H_1:K] \leq [G:H_2]=p$ but $1<[H_1:K] | G$ so this must be exactly $p$. $\endgroup$ – Lior Silberman Sep 25 '14 at 0:22
  • 3
    $\begingroup$ That is a very neat argument which I have not seen before. $\endgroup$ – Geoff Robinson Apr 12 '16 at 8:54
  • 1
    $\begingroup$ @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. $\endgroup$ – Tobias Kildetoft Apr 12 '16 at 9:09
  • 3
    $\begingroup$ 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). $\endgroup$ – Geoff Robinson Apr 12 '16 at 9:18
15
$\begingroup$

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 $\ker \phi$?

$\endgroup$
  • $\begingroup$ @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 Mar 4 '13 at 12:16
7
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ I corrected some mistakes in your proof. Can you tell me why $|H|/|H\cap H^g| \geq p?$ $\endgroup$ – user370967 May 19 '17 at 11:56
  • $\begingroup$ Since $p$ is the smallest prime dividing the order of $G$. $\endgroup$ – mesel May 19 '17 at 21:49
  • $\begingroup$ Can you elaborate? I have thought about it for a while. $\endgroup$ – user370967 May 19 '17 at 21:50
  • 1
    $\begingroup$ 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 May 19 '17 at 21:53
  • 1
    $\begingroup$ you are welcome :) $\endgroup$ – mesel May 19 '17 at 21:56
5
$\begingroup$

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 $q\nmid p!$.

$\endgroup$
  • $\begingroup$ 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 $q\nmid(p-1)!$. $\endgroup$ – Hagen von Eitzen Feb 10 '13 at 10:28
  • $\begingroup$ @HagenvonEitzen This is practically the same idea--this was just perhaps why my answer was in "hint" form. $\endgroup$ – Alex Youcis Feb 10 '13 at 10:31
  • $\begingroup$ It just looked to me like your hint would silently assume that $p^2\nmid |G|$. $\endgroup$ – Hagen von Eitzen Feb 10 '13 at 10:59
0
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.