Re: The Center of a group

Let's prove:

$\displaystyle Z(G) = \bigcap_{g \in G} C_G(g)$.

First of all, suppose $\displaystyle z \in Z(G)$. Then $\displaystyle zg = gz$ for all $\displaystyle g \in G$, so for any particular g, we have zg = gz, that is $\displaystyle z \in C_G(g)$ for every $\displaystyle C_G(g)$. Well that part was easy.

On the other hand, if z is in EVERY $\displaystyle C_G(g)$, it commutes with every g in G, so $\displaystyle z \in Z(G)$. Done deal.

(one small niggling point remains: one ought to show that the intersection is non-empty, so that it actually defines a group. The identity, however, comes to our rescue, here).

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As a practical measure, it is more efficient to find elements that don't commute. For example, with $\displaystyle S_5$ we have that $\displaystyle (i\ j)(j\ k) = (i\ j\ k)$, whereas $\displaystyle (j\ k)(i\ j) = (i\ k\ j)$ so the center doesn't contain any transpositions.

This means that $\displaystyle Z(S_5)$ can have at most 110 elements, so (by Lagrange) at most 60 elements.

In fact, the centralizers of transpositions are a good place to start. Can you prove that $\displaystyle C_{S_5}((1\ 2))$ has 12 elements? This would show that the center can have no more than 12 elements, a considerable improvement over 60.

Can you show that$\displaystyle C_{S_5}((1\ 2)) \cap C_{S_5}((4\ 5))$ has order 4? Together with the first result above this shows the center has at most order 2. Then you just need to find ONE permutation that doesn't commute with (1 2)(4 5)....

Re: The Center of a group

Good reading, as always! Thank you.

-Dan