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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 $C(a):=\{x \in G:ax=xa\}$.

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    $\begingroup$ 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 $$Z(Q_8)=\{ ±1\}\subset C(i)=\{±1,±i\}$$ $\endgroup$
    – Fotis
    Jul 21, 2023 at 12:38
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    $\begingroup$ @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? $\endgroup$
    – Afzal
    Jul 21, 2023 at 12:41
  • $\begingroup$ 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 $a$). $\endgroup$
    – citadel
    Jul 21, 2023 at 12:44
  • $\begingroup$ Oh, now I understand your problem, I posted an answer on where your mistake was on the proof, followed by a correct proof $\endgroup$
    – Fotis
    Jul 21, 2023 at 12:49
  • $\begingroup$ @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? $\endgroup$
    – Afzal
    Jul 21, 2023 at 12:49

1 Answer 1

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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.

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    $\begingroup$ I understood. I misunderstood whole stuff just because of my incorrect proof. $\endgroup$
    – Afzal
    Jul 21, 2023 at 12:56

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