Index of intersection of two normal subgroups

Question

Question:

Let $G$ be a finite group and $H$ and $K$ are normal subgroup of $G$. If $[G:H]=2$ and $[G:K]=3$, determine $[G:H∩K]$.

My attempt: I think $[G:H∩K] = 6$. Since $G$, $H$, $K$ are finite, let $|G| = 6n$, $|H| = 3n$, $|K| = 2n$.

By Lagrange theorem, $|H∩K|$ divides $n $ ($H∩K$ is a subgroup of $H$ and $K$).

If $n$ divides $|H∩K|$, (or $n$ is lower bound of $|H∩K|$) then $|H∩K|=n$, it leads to $[G:H∩K] = 6$.

But I don't know how to show $n$ divides $|H∩K|$.

Normality of $H$ and $K$ is not used yet and I have a trouble with using this fact.

Need help! Thanks.

Answer 1

By the second isomorphism theorem, we have $$\dfrac{HK}{H}\cong \dfrac{K}{H\cap K}$$ and thus $$|H\cap K|=\frac{|H|\cdot|K| }{|HK|}\geq \frac{|H| \cdot |K|}{|G|}=\frac{6n^2}{6n}=n.$$

Answer 2

Hint:

Note that $$[G:H\cap K]= [G:H][H:H\cap K]=2[H:H\cap K]$$ and $$[G:H\cap K]= [G:K][K:H\cap K]=3[H:H\cap K].$$ Since both $2$ and $3$ divides $[G:H\cap K],$ $6$ does as well. Since $|H\cap K|$ divides both $|H|, |K|$ and $|G|$, it follows that there is an integer $n$ such that $|G|=6n$. Can you continue?

Answer 3

In general, if $M, N \unlhd G$, with gcd$(|G:M|,|G:N|)=1$, then $G=MN$, and $|G:M \cap N|=|G:M| \cdot |G:N|$. Proof: the index of the subgroup $MN$ divides both indices of $M$ and $N$. Now apply the second isomorphism theorem.

The result is still true when you drop normality on $M$ or $N$.