I want to exhibit a counterexample of the following claim
Let $G$ be a group and $p$ a prime which divides $|G|$. Then, there exists a subgroup $H$ of $G$, such that $[G:H] = p$.
I know that $\mathbb{A}_5$ is simple, and therefore with $p = 2$ this cannot hold, because $[\mathbb{A}_5:H] = 2$ would imply that $H$ is normal. I'm looking for more elementary examples which do not rely on $\mathbb{A}_5$ not being simple. Any ideas?
The smallest example is $A_4$, of order 12, which also has no subgroup of index 2. You can see this because such a subgroup would be normal, since it has index 2, and thus a union of conjugacy classes, and the classes in $A_4$ have sizes 1, 3, 4, and 4. No combination of those adds up to 6. Or you could simply notice that any subgroup of order 6 would need to contain an element of order 2 and an element of order 3. But in $A_4$ it's pretty easy to see that any element of order 2 and any element of order 3 generate all of $A_4$.