Group Theory : What is $Ha \ne Hb$?

Question

As a beginner of Group Theory, I got stuck with the following question:

Suppose that $H$ is a subgroup of $G$ such that whenever $Ha \ne Hb \space ,$ then $aH \ne bH$. $(a,b \in G)$ Prove that $gHg^{-1} \subset H \space\space\forall g\in G$.

My first doubt is what is exactly implied by $Ha \ne Hb$ in the above question? Does it mean $$ whenever \space ha\ne hb , ah\ne bh \space \forall\space h\in H $$ $$OR$$ $$whenever \space h_1a\ne h_2b , ah_1\ne bh_2 \space \forall \space h_1,h_2 \in H \space ?$$

I dont know whether this is a very silly doubt or not ; please help me clarify.

Secondly , it will be very helpful if you give me a hint how to proceed with this problem. Thank you in advance..

If this question is a repetition please give a link , but do not down-vote. I am very low in reputation.

Answer 1

I think your first question has been answered,

Now Hints to the second one:

$Ha\neq Hb\implies aH\neq bH$ means $aH=bH\implies Ha=Hb$ means $b^{-1}a\in H\implies ab^{-1}\in H$.

Now let $g\in G,h\in H;h(g^{-1}g)=he\in H\implies (hg^{-1})g\in H\implies ghg^{-1}\in H$ which holds $\forall h,g$.

Thus $gHg^{-1}\subseteq H$