This is the question I'm stuck on.
Let $H$ be a sugroup of $G$. If for all $a, b \in G$, there exists a $k \in G$ such that $(aH)(bH)=kH$, then show that $k=ab$.
That means that if $(aH)(bH)$ is still somehow a left coset of $H$ for all $a,b \in H$, then this left coset must be $abH$. $(aH)(bH)$ is defined as $\{ah_1 bh_2 \mid h_1, h_2 \in H\}$. I understand that such an operation is $\textbf{well defined}$ only if $H$ is a normal subgroup, but the question doesn't impose any constraints on $H$. It only asks that $k$ must be equal to $ab$. I'm not sure how to proceed.
So far I've tried using the fact that different cosets are disjoint, and writing out an element of $(aH)(bH)$ as some $ah_1 bh_2 = kh_3$, but it doesn't seem to lead anywhere. Even if I let $a$ be identity and vary $b$, it doesn't show that $k=b$ because I get $b= kh_3 h_2^{-1}$, which isn't strong enough to imply that $b= k$.
EDIT: Saying that $k=ab$ is wrong, I think the original intent was to show that the coset of $(aH)(bH)$ is the same as the coset $abH$ without any presumption of $H$ being normal.
Best Answer
Suppose $G$ is a group and $H$ is a subgroup. Suppose $a,b\in G$ are such that $(aH)(bH)$ is a left coset of $H$. Then we claim $(aH)(bH)=abH$. Indeed, since $H$ contains the identity, we have $$ ab\in (aH)(bH) \cap abH $$ So $(aH)(bH)$ and $abH$ are two left cosets of $H$ that intersect nontrivially. So they must be equal since the left cosets of $H$ form a partition.
Remark: This proof works for just one pair $(a,b)$ satisfying the assumptions. In other words, you don't need to make the assumption for all $a,b\in G$.