# How center of group is subset of centralizer of an element $a$ of $G$? [duplicate]

In the book "contemporary abstract algebra by gallian" author writes:

Notice that for every element $$a$$ of a group $$G$$, $$Z(G) \subseteq C(a)$$.

I proved above statement like this:
Proof: Suppose $$a \in Z(G)$$ then $$ax=xa$$ for every $$x \in G$$. By definition of centralizer of $$a$$ in $$G$$ we deduce that $$a \in C(a)$$.

I felt the problem when I used the proof steps on an example. I supposed the group $$Q_8:= \{\pm 1,\pm i \pm j,\pm k\}$$ Under multiplication. Then $$Z(Q_8)=\{ \pm 1\}$$ and $$C(i)=\{\pm 1,\pm i\}$$. Here i stuck because in the proof above I started by letting $$a \in Z(G)$$, but here in the example $$i \notin Z(Q_8)$$. Where is the mistake? Is my proof incorrect or am I making some mistake in example?

Note: $$Z(G):=\{g \in G:xg=gx$$ for all $$x \in G \}$$ and $$C(a):=\{x \in G:ax=xa\}$$.

• I don't understand your problem, your little proposition says that the centralizer of any element contains the center of the group, and that's exactly what you showed in your example, since $$Z(Q_8)=\{ ±1\}\subset C(i)=\{±1,±i\}$$ Jul 21, 2023 at 12:38
• @Fotis Actually my question is: If $a ∈ C(a)$, then it is not necessary $a$ is in $Z(G)$. So if i want to prove proposition for the case when $a ∉ Z(G)$, then how will the step goes? Jul 21, 2023 at 12:41
• That's just by the two definitions: if an element commutes with every element of the group (nemely if it lays in the center), then it commutes in particular with any specific $a$ (namely it lays in the centralizer of such $a$). Jul 21, 2023 at 12:44
• Oh, now I understand your problem, I posted an answer on where your mistake was on the proof, followed by a correct proof Jul 21, 2023 at 12:49
• @citadel Oh I understood you point. But suppose $a$ is an element (lays in centralizer but not center) s.t $a$ does commute with all elements of group. So $a ∉ Z(G)$ right, so my proof is incorrect? Jul 21, 2023 at 12:49

So, in your proof, you wrongly assumed that $$a\in Z(G)$$, as this is not the general case (e.g. the example you gave).

A correct proof would be:

Let $$a\in G$$ and consider $$C(a)$$.
Now, for any $$g\in Z(G)$$, we have that $$gx=xg$$ for every $$x\in G$$, by the definition of $$Z(G)$$.

In particular, for $$x=a$$, we have $$ga=ag$$.
Now by the definition of $$C(a)$$, we conclude that $$g\in C(a)$$, and since $$g$$ was an arbitrary element of $$Z(G)$$, we get: $$Z(G)⊆ C(a)$$

With the above proof, we never make the wrong assumption that the arbitrary element $$a\in G$$ is a central element of the group.

• I understood. I misunderstood whole stuff just because of my incorrect proof. Jul 21, 2023 at 12:56