Proof that normalizer and center are subgroups

Question

I've seen this proof for the center of a group $G$:

$$C = \{x\in G:xg = gx \ \ \ \forall g \in G\}$$

So, the center is the set of all elements that commute with every $g$ of $G$. This subset of $G$ is nonempty because $eg = ge$ for all $g\in G$. If we have $a,b\in C$, then we know that:

$$\forall g\in G:\qquad ag = ga,\qquad bg=gb,$$

and therefore, we have

$$(ab)g = a(bg) = a(gb) = (ag)b = (ga)b = g(ab),$$

which shows that $ab\in C$. Also, for the inverse, if we have $c\in C$ then for all $g\in G$ we have $gc=cg$ hence also $$c^{-1}gc=c^{-1}cg=g,$$ from which it follows that $gc^{-1}=c^{-1}g$, and hence that $c^{-1}\in C$.

The normalizer of $S\subset G$ is defined as:

$$N = \{g\in G: gS = Sg\}.$$

This subset is nonempty, since $eS = Se$. For all $a,b\in N$ we have $aS=Sa$ and $bS=Sb$, and hence

$$(ab)S = a(bS) = a(Sb) = (aS)b = S(ab),$$

which shows that $ab\in S$. For the inverse; if $c\in N$ then $Sc=cS$ and so $$c^{-1}Sc=c^{-1}cS=S,$$ from wich it follows that $c^{-1}S=Sc^{-1}$, and hence that $c^{-1}\in N$.

Am I right?

Answer 1

You might have noticed that the steps in each of your proofs are quite similar. In fact, they are both special cases of a more general result involving group actions.

Given a set $X$, a group $G$ acts on $X$ if there is a map $G \times X \to X$, denoted $g \cdot x$, such that

• $1 \cdot x = x$ for all $x \in X$, where $1$ is the identity in $G$, and
• $g\cdot(h \cdot x) = (gh)\cdot x$ for all $g,h \in G$ and $x \in X$.

Let $G$ be a group acting on a set $X$. For $x \in X$, we define the stabilizer of $x$ to be $$ \text{Stab}_G(x) = \{g \in G \mid g \cdot x = x\} \, . $$ One can show that the stabilizer of a point is in fact a subgroup of $G$. Try to prove this yourself: you will find that the steps are almost exactly the same as those in your two proofs above.

Now let's see how the centralizer and normalizer are both examples of stabilizers of group actions.

• A group $G$ acts on itself by conjugation, that is, we can define the map \begin{align*} G \times G &\to G\\ (g, x) &\mapsto gxg^{-1} \end{align*} and show that it satisfies the two axioms of a group action. For this action, we find that, given $x \in G$, the stabilizer of $x$ is $$ \text{Stab}_G(x) = \{g \in G \mid gxg^{-1} = x\} = \{g \in G \mid gx = xg\} = C_G(x) $$ the centralizer of $x$ in $G$. Thus the centralizer of any point is a subgroup. The center $C = Z(G) = \bigcap_{x \in G} C_G(x)$ is then the intersection of all the centralizers, hence is also a subgroup.
• Let $\mathscr{P}(G)$ denote the power set of $G$, i.e., the collection of all subsets of $G$. Then $G$ also acts on $\mathscr{P}(G)$ by conjugation, as one can show that the map \begin{align*} G \times \mathscr{P}(G) &\to \mathscr{P}(G)\\ (g, S) &\mapsto gSg^{-1} = \{gsg^{-1} \mid s \in S\} \end{align*} satisfies the two group action axioms. In this case we find that, for $S \in \mathscr{P}(G)$, the stabilizer of $S$ is $$ \text{Stab}_G(S) = \{g \in G \mid gSg^{-1} = S\} = \{g \in G \mid gS = Sg\} = N_G(S) $$ the normalizer of $S$ in $G$.

Answer 2

Yes, both your proofs are right.