Product of two elements not in a subgroup is in a subgroup? So I have seen this post, but my question is a bit different.
Suppose $b_1,b_2\in G$ and $b_1\neq b_2^{-1}$. And suppose you have a subgroup $H\leq G$ and neither $b_1$ nor $b_2$ belong to $H$. Can $b_1b_2\in H$?
So far I have just gone round in s circle and been rearranging without getting anywhere. I have tried proof by contradiction but again I get nowhere. It seems like either the above is true  or I am missing some obvious step.
 A: Sure. Take $G=(\mathbb{Q},+)$ and $H=\mathbb Z$. Then $\frac12+\frac12=1\in\mathbb{Z}$, but $\frac12\notin\mathbb Z$.
A: We can say more.  Given $H$ a proper subgroup of $G$, and an element $b_1 \in G$ that is not in $H$, there are $|H|$ choices for $b_2$, none of them in $H$, such that $b_1b_2 \in H$.  Note that $b_1^{-1} \not \in H$ and $b_1b_1^{-1}=e \in H$.  Then for each $h \in H$ we have $b_1^{-1}h \not \in H$ and $b_1(b_1^{-1}h)\in H$
A: If $g_1, g_2 \not\in H$, then there's nothing we can really say about whether $g_1 g_2 \in H$, without more information. Sometimes the product is in $H$, other times it's not.
An example where $g_1g_2 \in H$: Pick two odd permutations in the symmetric group $S_n$. Then neither are in the alternating group $A_n$, yet their product is.
An example where $g_1g_2 \not\in H$: Look at the symmetric group $S_4$, and its normal Klein $4$-subgroup $V = \{1, (12)(23), (13)(24), (14)(23)\}$. Now just pick an even permutation not in $V$, and an odd permutation (which must not be in $V$). Their product must be odd, hence not in $V$.
A: This certainly can be true. If $H$ is a subgroup of index $2$, then it is necessarily true for any $b_1,b_2\not\in H$. (Example: in $\mathbb{Z}$, odd + odd = even ) Even if $H$ has index greater than $2$, it can be true. You only need that $b_1$ and $b_2$ belong to cosets of $H$ that are inverses in the quotient group.
