1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ Are you saying that $e,g_1,\ldots g_n$ are all the elements? $\endgroup$ – rschwieb Sep 29 '16 at 16:19
  • $\begingroup$ @rschwieb Yeah, all the elements in the group G. $\endgroup$ – Andrew Raleigh Sep 29 '16 at 16:25
  • 2
    $\begingroup$ In general, $\bigcup_{e\neq g\in G}C_G(g)=G$ and $\bigcap_{e\neq g\in G}C_G(g)=Z(G)$, the center of the group. I'm just saying it has nothing to do with finiteness of the group or anything. $\endgroup$ – rschwieb Sep 29 '16 at 17:39
2
$\begingroup$

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 $g_i$.

$\endgroup$
  • $\begingroup$ Yeah you're right, I fixed it. Can I ask how you reached the conclusion? $\endgroup$ – Andrew Raleigh Sep 29 '16 at 16:26
  • $\begingroup$ Well, usually one denotes $a\in{A},b\in{B}...$ etc. Apart from that, if it really were $s\in{G}$ in the definition of the set, then the $S$ wouldn't have any impact on that definition. $\endgroup$ – Janik Sep 29 '16 at 16:30
2
$\begingroup$

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 $g_i \in C(g_i)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.