# prove : if $G$ is a finite group of order $n$ and $p$ is the smallest prime dividing $|G|$ then any subgroup of index $p$ is normal

Prove:

If $G$ is a finite group of order $n$ and $p$ is the smallest prime dividing $|G|$ then any subgroup of index $p$ is normal where $|G|$ is the order of $G$

This is a result in Abstract Algebra by Dummit and Foote at page 120 .

The proof is produced but there is some points which is not obivous for me !

First, in page 121 in the proof, it says, all divisors of $(p-1)!$ are less than $p$. why this is true ? can any one explain?

Second, why does "every prime divisor of $k$ is greater than or equal to $p$ "force that k=1 ??

Why does $k = 1$ under this condition??

Third, if $k=1$ Then, the order of $K$ = the order of $H$
Why does this mean that $K=H$ in this case??

Can you help in explaining these three things?

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At first question, all prime divisors of $(p-1)!$ are less than $p$. (Because if $q$ is prime and $q|(p-1)!$ then $q|n$ for some $1\le n <p$, by definition of prime number.) But there exists divisors of $(p-1)!$ greater than $p$ - for example, $(p-1)(p-2)$. – Hanul Jeon Jan 28 '13 at 22:36
@tetori If $p=3$ then $(p-1)(p-2)=2<p$. – Jaakko Seppälä Jan 28 '13 at 22:41
@Jaakko Seppälä Oh, it is my mistake. But if $p>3$, then $p<(p-1)(p-2)$ – Hanul Jeon Jan 28 '13 at 22:43
If $k|(p-1)!$ and $q$ is a prime that divides $k$ then $q < p$. But if $q|k$ then $q||G|$. Since $p$ is the smallest prime dividing $|G|$, this is a contradiction, hence NO prime divides $k$, which forces $k = 1$. If $K$ is of index $1$ in $H$, $H = K$. – David Wheeler Jan 28 '13 at 22:49
@tetori , this doesn't prove the statement that q must be less than p ! can you explain this in more detail ?! – Maths Lover Jan 28 '13 at 23:20

For the second question, notice that in the proof $[H:K]=k$, so $k\mid |H|\mid |G|$ since $H\leq G$. So any prime divisor of $k$ is also a divisor of $|G|$. But every prime divisor of $k$ divides $(p-1)!$, hence is less than $p$. Since $p$ is the least prime dividing $|G|$, $k$ cannot have any prime divisors, else we would have a prime divisor of $|G|$ strictly less than $p$. But the only positive integer with no prime factors is $1$, so $k=1$.

But then $[H:K]=1$, and since $H$ and $K$ are finite groups, that means $|H|=|K|$, so necessarily $H=K$.

Added: For the first question, recall that if $q$ is a prime, and $q\mid ab$, then either $q\mid a$ or $q\mid b$. This is Euclid's Lemma. So if $q$ is a prime factor of $(p-1)!=1\cdot 2\cdot 3\cdots p-1$, we must have $q\mid j$ for some $j=1,2,\dots,p-1$. So necessarily $q<p$.

If you're not familiar with that property of primes, another way to see this is to observe that the prime factorization of $(p-1)!$ is the product of the prime factorizations of $1,2,\dots,p-1$. But for each $1\leq j\leq p-1$, the prime factorization of $j$ must not have any prime factors greater or equal to $p$. So the prime factorization of $(p-1)!$ consists only of primes less than $p$.

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Great ! but what about the first question ? why are all prime divisors of $(p-1)!$ are less than $p$ ? – Maths Lover Jan 28 '13 at 23:08
@MathsLover I added something to address that. I hope it is clear. – Ben West Jan 29 '13 at 4:07
thanks :) @BDub – Maths Lover Jan 29 '13 at 19:07

Consider the prime factorizations of $p-1, p-2,$ etc. all the way down to $1$. None of these can be divisible by $p$ because, as $p$ is a prime, the lowest number divisible by $p$ is $p$. So since $p$ does not divide any of those numbers, it does not divide their product either, i.e. $p\not\mid (p-1)!$. The same goes for all primes $q$ greater than $p$, we have that $q\not\mid p-1$, $q\not\mid p-2$, etc. and thus $q\not \mid (p-1)!$. So each prime divisor of $(p-1)!$ is less than $p$.

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3 , 4 , 5 are 3 numbers , 6 can't divide any of them but 6 divide their product which is 60 , doesn't it ?!! so your argument is not right in my opinion according to what i had mentioned ! – Maths Lover Jan 29 '13 at 0:43
Maths Lover: Don't you know that 6 is not prime? Pay attention to the hypotheses. Your so-called counterexample is not relevant, because $p$ is prime. – KCd Jan 29 '13 at 4:38
@SamuelHandwich , thanks ! I though about it and found that it's true :)) thanks so much for all – Maths Lover Jan 29 '13 at 19:04
@KCd , yes , you are right :) – Maths Lover Jan 29 '13 at 19:05

We can use the Cayley Theorem to show that:

Theorem: Let the group $G$ has the subgroup $H$ of index $n$, then there is a normal subgroup $K$ of $G$ such that $$K\subseteq H,~~ m=[G:K]<\infty,~~m\mid n!$$

Now let $H\leq G$ and $[G:H]=p$ which $p$ has the property the question noted. So according to the above theorem there is a normal subgroup $K$ of $G$ contained in $H$ such that $[G:K]\big|p!$. But $[G:H]\big ||G|$ also, so since $p$ is the smallest dividing number then we have necessarily have $$[G:K]\big|p$$ and this means thta $K=H$.

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I like this! +1 – amWhy Jan 29 '13 at 15:58