1
$\begingroup$

I have a question i am trying to prove that if $H<G$ and $\dfrac{|G|}{|H|}$ is a prime number then H is a maximal subgroup.

I prove this by contradiction, thus i assume that $\exists K : H<K<G$ and $K\neq H \neq G$.

I use Langrage's theorem to show that:

$\exists a \in \mathbb{N}$ : $|K|=a|H|$

$\exists b \in \mathbb{N}$ : $|G|=b|K|$

Thus $|G|=ab|H|\Leftrightarrow ab=\dfrac{|G|}{|H|}$ so $ab$ has to be prime.

Now I say that $a=1$ and $b=2$ but then $|K|=|H|$ and we knew that $H<K$ thus $K=H$ and this gives a contradiction is this correct?

$\endgroup$
3
  • 2
    $\begingroup$ Of course and it has nothing to do with groups. If $H\subset G$ and $|H|=|G|<\infty$ then $H=G$. $\endgroup$
    – user223391
    Oct 9, 2016 at 15:43
  • $\begingroup$ Yes i am indeed assuming that G is finite i forgot to mention it in my question. $\endgroup$ Oct 9, 2016 at 15:47
  • $\begingroup$ "Now I say that a=1 and b=2" is not the right line of reasoning. If a product of two numbers is a prime, what can you say about these numbers? $\endgroup$ Mar 3, 2017 at 12:20

1 Answer 1

1
$\begingroup$

Your solution is more or less right. You can streamline it by remarking that $$|G:H|=|G:K| \cdot |H:K|$$ So if $|G:H|$ is prime, then either one of the factors equals $1$, that is $G=K$ or $K=H$.

$\endgroup$
2
  • 1
    $\begingroup$ "Your solution is right" Really? Included when they decide for no reason that a=1 and b=2? $\endgroup$
    – Did
    Mar 3, 2017 at 9:56
  • $\begingroup$ Yes your comment is right. All primes are equal to 2 apparently. :-) The end is wrong, but I meant the intent was OK. Will adapt. $\endgroup$ Mar 3, 2017 at 12:19

This site is temporarily in read-only mode and not accepting new answers.

Not the answer you're looking for? Browse other questions tagged .