# 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.

Thanks.

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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

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.

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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
@SquiresMcGee yes – SorcererofDM Mar 17 '15 at 22:30
@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 $\ker \phi$?

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@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.

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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 $q\nmid p!$.

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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.

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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 $G$.

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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