Partitions and Equivalence Relations I have been working on this question for some time and I have looked throughout the forum and internet but have not found any success.

Let H be a subgroup of the group G and let the index of H in G be 2. Show that if the elements a, b ∈ G are not in H, then the product ab ∈ H.

I greatly appreciate any hints as to where to start.
 A: This is a proof by contradiction. To be clear, we are going to assume the opposite of what we want to prove, and show this leads us to a contradiction.
So here is what is given, that we can rely on:


*

*There are only two cosets of $H$. These PARTITION $G$, so it's "one or the other".

*$a,b$ are not in $H$. This means they BOTH lie in the "other coset".


From 2), we have $aH = bH$. Let's see why:
If $aH = H$, we would have $ah \in H$, say $ah = h'$, for some $h' \in H$. Then $a = ae = a(hh^{-1}) = (ah)h^{-1} = h'h^{-1} \in H$, since $H$ is closed under inverses and multiplication. But this is impossible, since $a \not\in H$. So $aH$ is "the other coset".
Similarly, $bH \neq H$, so $bH$ is also the other coset. Since there's only one other coset, it must be $aH$.
So here is where we assume the opposite of what we want to prove: we suppose it is somehow possible that $ab \not\in H$. Since $ab$ is either in $H$, or $aH$ (because $G = H \cup aH$, and $H \cap aH = \emptyset$), if $ab$ is NOT in $H$, it must be in $aH$.
So $ab = ah$, for some element $h \in H$ (we don't even care which one). Thus:
$a^{-1}(ab) = a^{-1}(ah)$, so:
$(a^{-1}a)b = (a^{-1}a)h$ (by associativity), and then:
$b = eb = eh = h$.
But this says $b \in H$, and we know this is not true. This is our contradiction: $b \in H$, and $b \not\in H$ cannot both be true at the same time.
And how did we get to such a sorry state of affairs? We assumed it was possible that $ab \not\in H$. This was our downfall, and to avoid the annihilation of the known universe, we must not make that assumption ever again. But this forces upon us the only OTHER conclusion: $ab \in H$, quod erat demonstrandum.
