Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.


share|cite|improve this question
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
up vote 38 down vote accepted

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 index 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 $G\to S_p$, 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.

share|cite|improve this answer
Thanks. The part 'it follows that $|G/K|=p$' is crucial. Now everything is done! Bye. – Sigur Jun 28 '12 at 17:51
Is there a simpler homomorphism that would work for the case $|G|=p^2$? – misi Apr 17 '13 at 7:23
What is $S_p$? The symmetric group? – Squires McGee Mar 17 '15 at 19:32
@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

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 $\ker \phi$?

share|cite|improve this answer
@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

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.

share|cite|improve this answer
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
@leo If a group has only one subgroup of some fixed order, then that subgroup is normal, since any conjugate of a subgroup is a subgroup of the same order. – Tobias Kildetoft Jun 3 '13 at 18:55
That is a very neat argument which I have not seen before. – Geoff Robinson Apr 12 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 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 at 9:18

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 $q\nmid p!$.

share|cite|improve this answer
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

proof: If $H$ is not normal then assume $H\neq H^g$. Then $$|HH^g|=|H|\dfrac{|H|}{|H\cap H^g|}$$.

Notice that $\dfrac{|H|}{|H\cap H^g|}\geq p$. Thus, $|HH^g|\geq G$. We must have $HH^g=G$.

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

share|cite|improve this answer

Since $[G: H]=2$, the group $G$ has only two distinct left cosets and only two distinct right cosets.

Now $H$ itself is a left as well as a right coset in $G$. Let $a \in G$. If $a \in H$, then $$aH=H=Ha$$.

Suppose $a$ does not belong to $H$.Then $aH \neq H$. Hence $G=H \cup aH$ and $H \cap aH= \varnothing$. Then $aH=G-H$. Since $a$ does not belong to $H$ and $G$ has only two right cosets, we find that $G= H \cup Ha$ where $H \cap Ha= \varnothing $. Thus $Ha=G-H$. Hence $Ha=aH$.Thus we find that $aH=Ha$ for all $a \in G$ and so $H$ is a normal subgroup of $G$.

share|cite|improve this answer
The OP is asking for a generalization of the $p=2$ case, not a proof of the $p=2$ case. Also, unfortunately this proof does not generalize for larger primes. – Arturo Magidin Jun 28 '12 at 17:02

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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