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

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.

• Try proving the contrapositive; that is, the product is a left coast for all pairs implies $H$ is normal. Apr 2, 2019 at 16:04
• My previous comment isn't very helpful, considering that you've tried a proof by contradiction. Sorry. Apr 2, 2019 at 16:10

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.
• Could you explain why $\fbox{$1\in lH$}$ ? Apr 2, 2019 at 16:21
• @YadatiKiran it holds as $1\in H$. That is, we are assuming that $ghg^{-1}H\subset lH$ for all $h\in H$, so take $h=1$. Apr 2, 2019 at 16:28
• $gg^{-1}=g1g^{-1}1\in gHg^{-1}H$. Apr 2, 2019 at 16:30
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.