$G$ is a finite group and $\emptyset \ne H \subseteq G$ such that $\forall a,b \in H: ab\in H$. Prove that $H\le G$.
My attempt:
Suppose $a\in H$: $a = a\cdot 1_G \in H \Rightarrow 1_G\in H \Rightarrow a\cdot a^{-1}\in H \Rightarrow a^{-1}\in H.$
Associativity is ok, because $G$ is a group.
This seems wrong to me, because this question is marked as a relatively hard one and I solved it using barely 2 lines. Am I doing it right?
Thank you.
You can't conclude that $1_G\in H$ that easily. The only thing we know is that if we have two elements in $H$, then their product is in $H$. We do not know that if the product of two things are in $H$, then each of the factors are in $H$ (not even if we already know that one of the factors are in $H$).
Instead, to show that $1_G\in H$, you need to consider $a^2, a^3, a^4, \ldots$, show that every one of those is in $H$, and show that $1_G$ is one of them.
Associativity you get for free, as we already know that $G$ is a group, and closure is assumed from the get-go. The only remaining thing is inverses. But that proof is basically the same as the proof for $1_G$, so that shouldn't cause too much of an issue.