PROPERTY
Let $G$ be an finite group. Then for every group $P\subset G$ the order of $P$ must be a multiple of the order of $G$.
Does the same property hold for the center of the group? I mean, is the order of $Z(G)$ a multiple of the order of $G$?
Where $Z(G)$ is the set of element that commute with all other element of the group.
Best Answer
Because of $x1 = 1x$ for all $x\in G$, $1\in Z(G)$.
Assume $g\in Z(G)$ and let $x\in G$. Then $$g^{-1}x = xg^{-1} \iff g(g^{-1}x) = g(xg^{-1}).$$ This is true, since the left hand side is $g(g^{-1}x) = (gg^{-1})x = 1x = x$ and the right hand side is $g(xg^{-1}) = (gx)g^{-1} = (xg)g^{-1} = x(gg^{-1}) = x1 = x$. So $g^{-1} \in Z(G)$.
Let $g,h\in Z(G)$ and $x\in G$. Then by the group axioms and the center property $$(gh)x = g(hx) = g(xh) = (gx)h = (xg)h = x(gh).$$ So $gh\in Z(G)$
Together, we have shown that $Z(G)$ is a subgroup of $G$ and therefore, $\lvert Z(G)\rvert $ divides $\lvert G\rvert$.