If $H$ is not normal subgroup of $G$ then there are two left cosets of $H$ which multiplication is not a left coset of $H$

Question

Let $G$ be a group and $H$ be a not normal subgroup of $G$. I need to prove that there are two left cosets of $H$: $g_1H,g_2H$ such that $g_1Hg_2H$ is not a left coset of $H$.

I tried to assume the contrary and get to that $H$ is normal, which will mean a contradiction. I tried just using definitions or to find an homomorphism which I can apply the first isomorphism theorem with maybe. However, it didn't work out for me.

Any help would be appreciated.

Answer 1

Suppose that the product of every left cosets is a left coset. This implies that for every $g$, $gHg^{-1}H=lH$, we deduce that $1=gg^{-1}\in lH$. This implies that $l\in H$ and $gHg^{-1}H=H$, we deduce that for every $h\in H$, $ghg^{-1}=ghg^{-1}1\in gHg^{-1}H=H$. This implies that $gHg^{-1}=H$ and $H$ is normal.

Answer 2

Aliter:

Assuming the negation of the conclusion, it follows that $(aH)(bH) = (ab)H$ (since cosets partition $G$ and $ab\in (aH)(bH)$). It follows that $H(aH) = aH$. If $h\in H$, then $hah\in H(aH) = aH$ so that $ha = ah'$ for some $h'\in H$. Hence $Ha\subseteq Ha$. Now since both left cosets and right cosets partition $G$, no inclusion can be proper. Hence, $Ha = aH$ for every $a\in G$ so that $H$ is normal.