What's the significance of the union of all non-identity centralizers in a group?

Question

Let's assume we have a finite group G, where G contains ${e, g_1, g_2, g_3, ..., g_n}$, what would $C(g_1)\cup C(g_2)\cup C(g_3)\cup ...\cup C(g_n)$ be?

Well, first, the centralizer of a group is defined as $C_G(S) = (g \in G$ | $gs = sg$ for all $s \in S)$.

Going from there, I've found that the answer is the subgroup all elements in the group, assuming every element is its own centralizer and e is always a centralizer too.

There's also the intersection of the previous statement, i.e. $C(g_1)\cap C(g_2)\cap C(g_3)\cap ...\cap C(g_n)$ - which I got to be the subgroup of all elements in the group, since every element is its own centralizer and so is $e$.

I'm assuming I did something wrong here, because it doesn't make sense that they're important yet I get the same result from both. Any ideas on how to go about solving this?

Answer 1

I think $C_G(S)=\{g\in{G}\;|\; gs=sg \text{ for all } s\in{S}\}$ would make much more sense: all the elements of $G$ that commute with every element of $S$. So we get:

$C(g_1)\cap...\cap C(g_n)$ is the set of all elements of $G$ that commute with every single $g_i$.

$C(g_1)\cup...\cup C(g_n)$ is the set of all elements of $G$ that commute with at least one of the $g_i$.

Answer 2

If $G = \{ e, g_1, \ldots, g_n \}$, then the union of centralizers $C(g_1) \cup \ldots \cup C(g_n)$ is the whole $G$, as $g_i \in C(g_i)$.