Check: Prove that $T_b \circ T_c=T_{bc} \forall b,c \in G$ Let $G$ be a group, $H$ a subgroup of $G$, and $S=\\{Ha|a\in G\\}$. For $b\in G$, define $T_b:S \to S$ by $T_b(Ha)=Hab^{-1}$. Prove that $T_b \circ T_c=T_{bc} \forall b,c \in G$. **Proof:** $$T_{bc}(Ha)=Ha(bc)^{-1}=Hab^{-1}c^{-1}=Hab^{-1}Hac^{-1}=T_b(Ha)T_c(Ha)$$ **edit:** $$T_{bc}(Ha)=Ha(bc)^{-1}=Hac^{-1}b^{-1}=Hac^{-1}Hab^{-1}=T_c(Ha)T_b(Ha)=T_b(Ha) \circ T_c(Ha)$$ Is that it?

> $$T_{bc}(Ha)=Ha(bc)^{-1}=\underbrace{Hac^{-1}}_{T_c(Ha)} b^{-1}= T_c(Ha) b^{-1} = T_b\left(T_c(Ha)\right)=(T_b\circ T_c)(Ha)$$

You should be careful with the meaning of **composition**. That's is why your attempt is not correct.