# If we have that H<G and |H|=|G| does this imply that H=G?

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?

• Of course and it has nothing to do with groups. If $H\subset G$ and $|H|=|G|<\infty$ then $H=G$.
– user223391
Oct 9, 2016 at 15:43
• Yes i am indeed assuming that G is finite i forgot to mention it in my question. Oct 9, 2016 at 15:47
• "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? Mar 3, 2017 at 12:20

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 $K=H$.