If $H\trianglelefteq G$, show that $a\cdot h=aha^{-1}$ (for every $a\in G$ and every $h\in H$) is an action of $G$ on $H$.

Question

If $H\trianglelefteq G$, show that $a\cdot h=aha^{-1}$ (for every $a\in G$ and every $h\in H$) is an action of $G$ on $H$.

I don't know how to use normal group and action.

Answer 1

Normality ensures that this action is well-defined, since normal subgroups contain all conjugates. Now, just check that this is a group action:

$$e\cdot h=ehe^{-1}=h,$$ and $$(ab)\cdot h=(ab)h(ab)^{-1}=abhb^{-1}a^{-1}=a(bhb^{-1})a^{-1}=a\cdot (b\cdot h)$$