**Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 3.2 Exercise 3.2.6**

Let $G$ be a group, $H \leq G$ a subgroup, and $g \in G$ a fixed element. Prove that if $Hg = kH$ for some element $k \in G$, then $Hg = gH$ and $g \in N_G(H)$.

Solution: Suppose $Hg = kH$ for some $k \in G$. In particular, $k \cdot 1 \in Hg$; then $k = hg$ for some $h \in H$, so that $$Hg = kH= hgH.$$ Left multiplying by $h^{-1}$ we have $h^{-1}Hg = gH$, but since $h \in H$, $h^{-1}H = H$, and we have $Hg = gH$. Therefore, $g^{-1}Hg = H$ so that $g \in N_G(H)$.

## Max

18 Apr 2021Hi. If H is not a normal group in G how can we still left multiply by h^-1?

## Linearity

18 Apr 2021You can left multiply any sets. I mean if $A=B$ (as sets), then $h^{-1}A=h^{-1}B$.