Prove that $Z(G)$ which is the center of $G$ is a subgroup of $G$

Question: Let $$G$$ be a group. Prove that $$Z(G)$$ is a subgroup of $$G$$.

If I want to show that $$Z(G)$$ is a subgroup of $$G$$ that means I have to show that it is closed under group operation?

Here is my attempt.

Let $$a,b$$ be elements in $$Z(G)$$ and $$x$$ be an element in $$G$$. Then $$ax=xa$$ which is under group multiplication commutative and under inverse $$(a^{-1}) x=x(a^{-1})$$. And hence $$(a^{-1})bx= (a^{-1})xb=x (a^{-1})b$$ which is under group operation so $$(a^{-1}),b$$ are elements in $$Z(G)$$ thus a subgroup of $$G$$.

Really appreciate if anyone can help me by directing me if my attempt is not good. Thanks in advance.

• You are trying the use the fact that for $\varnothing \ne S \subset G$, $S$ is a subgroup of $G$ iff $a, b \in S \Rightarrow a^{-1}b \in S$, right? Looks like you got it to me, save for the typo " . . . so $(a^{-1}),b$ are elements i $Z(G)$ . . . "; shouldn't it read "$(a^{-1})b$" instead of "$(a^{-1}),b$"? (No comma after "$(a^{-1})$".) May 2, 2014 at 5:39

2 Answers

You need to show that $$Z(G)\leq G$$.

1. First of all clearly $$1\in Z(G)$$ since $$1x=x1$$ for all $$x\in G$$.

2. Let $$a,b\in Z(G)$$ then $$ax=xa$$ and $$bx=xb$$ for all $$x\in G$$. Then $$(ab)x=a(bx)=a(xb)=x(ab)$$ so $$ab\in Z(G)$$.

3. If $$a\in Z(G)$$ consider its inverse $$a^{-1}$$ in $$G$$. Since $$ax=xa$$ for all $$x\in G$$ we have that $$a^{-1}(ax)a^{-1}=a^{-1}(xa)a^{-1}$$ so $$xa^{-1}=a^{-1}x$$ for all $$x\in G$$ namely $$a^{-1}\in Z(G)$$.

• just a question, will it be necessary if the definition of a "center of the group" be considered? that is: Z(G)={g element of G | xg=gx for all x element of G} May 2, 2014 at 6:23

While you can do this easily by checking the conditions for a subgroup, there is a somewhat easier approach using some general facts. The set of permutations of$$~G$$ (that is bijections $$G\to G$$) is a group under function composition (this is true for any set), call it $$P(G)$$. Define a map $$\def\ad{\operatorname{ad}}\ad:G\to P(G)$$ by $$\ad(g)=(h\in G\mapsto ghg^{-1})$$. To see that actually $$\ad(g)\in P(G)$$ for all$$~g\in G$$, that is that $$\ad(g)$$ is always bijective, it suffice to check that $$\ad(g^{-1})$$ is the inverse map of $$\ad(g)$$ by a computation similar to what follows. Now show that $$\ad$$ is a morphism of groups, that is $$\ad(g_1g_2)=\ad(g_1)\circ\ad(g_2)$$ for all $$g_1g_2$$. One has $$\ad(g_1g_2)=(h\in G\mapsto g_1g_2h(g_1g_2)^{-1})$$, whose final expression can be written as $$g_1g_2hg_2^{-1}g_1^{-1}$$, which is easily checked to be equal to $$\ad(g_1)(\ad(g_2)(h))$$, as required.

Now to conclude, check that $$\def\id{\operatorname{id}}\ker\ad=\{\, g\in G\mid \ad(g)=\id\,\}=\{\, g\in G\mid \forall h\in G:\ad(g)(h)=h\,\}$$ is equal to $$\{\, g\in G\mid \forall h\in G:ghg^{-1}=h\,\}=Z(G)$$. But then $$Z(G)$$ like any kernel, is a subgroup, even better a normal subgroup. (It is actually even better, namely a characteristic subgroup, but that does not follow as easily from the argument given.)