Can someone explain and give brief examples of centralizer and normalizer? This is a chapter on group action and I need to better understand its relationship with centralizer and normalizer. Quick examples would be very helpful! Thank you so much.
 A: The centralizer of $H$ in $G$ is the set $x\in G$ such that $xhx^{-1}=h$ for all $h$ in $H$ (that is, the set of elements that commute pointwise with things in $H$.
The normalizer of $H$ in $G$ is the set of all $x\in G$ such that $xhx^{-1}\in H$ for all $h\in H$. (notice the slight difference). This just says while normalizing elements don't necessarily commute with things in $H$, but conjugating by them at least lands back in $H$.
Clearly then, the centralizer is a subset of the normalizer.
As you might have guessed, both of these things are related to the conjugation action of a group $G$ on itself: $g\mapsto hgh^{-1}$ for some fixed $h$. You probably know this is a special automorphism of $G$ called an inner automorphism. If $h$ is in the centralizer of an element $g$ then $g$ remains still after being acted upon by $h$.
Extending that idea a step further, we can say that the action of $h$ does not move anything in a subset $S$ outside of $S$ if $h$ is in the normalizer of $S$. With this in mind, you can see that a normal subgroup is just a subgroup that holds still under the conjugation action of any element $h$ on $G$.
Consider the symmetries of the square: $\{1,\sigma,\sigma^2,\sigma^3,\tau,\sigma\tau,\sigma^2\tau,\sigma^3\tau\}$.
You can try these exercises:


*

*Show that both the normalizer and centralizer of a nonempty subset of $G$ are subgroups of $G$.

*The centralizer and the normalizer of a single element $\{g\}$ are the same set.

*Show that the normalizer of $\{1,\sigma,\sigma^2,\sigma^3\}=G$, but that its centralizer is not all of $G$. 

*Check that the centralizer of $\{1,\sigma^2\}=G$.

*Compute the normalizer of $\{1,\tau\}$
More standard exercises:


*

*Show that $H$ is a normal subgroup of its normalizer.

*Show that the centralizer of $H$ is all of $G$ iff $H$ is contained in the center of $G$.

*Show that the centralizer of $H$ is a normal subgroup of the normalizer of $H$.
A: If you're studying group actions, then you know that for a group $G$ acting on a set $X$, the stabilizer $G_x$ is all the elements of $G$ that fix $x$ under $G$'s action on $X$.
Consider $G$ acting on the set of subsets of $G$ by conjugation.  Then the stabilizer $G_H$ for some $H\leq G$ is $\{g\in G : gHg^{-1}=H\}$.  This particular stabilizer is the normalizer of $H$.
Now consider $G$ acting on $H$ by conjugation.  Here $G_h=\{g:ghg^{-1}=h\}$ is the centralizer of the point $h\in H$.  Taking the intersection of the stabilizers over all $h\in H$ is the centralizer of $H$.
So they're both just special cases of the stabilizer of $G$ acting on different things.
