# Normal subgroup of prime index

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.

• Hint: I think you should try to work with what Alex suggested. It is usually referred to as the "Strong Cayley Theorem". – Ludolila Feb 10 '13 at 10:30
• @Ludolila the link no longer works, can you update it? – S.Sundara Narasimhan Feb 22 at 19:16

## 8 Answers

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.

• Why $K\subset H$? It seems that $K=\{k:k=hg^{-1},\:h\in H\}$. – hermes Aug 10 '15 at 6:29
• @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) – Bungo Nov 4 '15 at 20:18
• How does it follow that if $p$ is the smallest prime dividing $|G|$ then $|G/K|=p$? – sequence Nov 27 '15 at 6:04
• @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 $p$. – Bungo Dec 4 '15 at 19:07
• @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. – Bungo Dec 6 '15 at 2:46

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.

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

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 $\ker \phi$?

• @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 $N(H) \ne H$ – user23238 Mar 4 '13 at 12:16

proof: If $H$ is not normal then assume $H\neq H^g$ for some $g \in G$. But a classical formula says $$|HH^g|=\dfrac{|H|\cdot |H^g|}{|H\cap H^g|} . \label{1} \tag{1}$$

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.

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

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 $q\nmid p!$.

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

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.

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 $$1_H^G=\sum_{\chi \in Irr(G)} a_{\chi}\chi$$, 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 $$|G|$$ is follows that all the irreducible constituents $$\chi$$ must be linear, that is $$\chi(1)=1$$. Hence $$H \supset ker(1_H^G)=\cap\{{ker(\chi) \in Irr(G): a_{\chi} \gt 0}\} \supseteq G'$$, whence $$H \lhd G$$.

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.