The order of the center of a finite group divides the group order 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.
 A: *

*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 $\lvert G\rvert$.
A: Just prove the center is a group, and you get the result.
A: I just want to make sure you know the following terminology:
We say that an integer $m$ divides an integer $n$ whenever $\ell=mk$ for some integer $k$, that is $\dfrac \ell m$ is an integer.
We say that an integer $m$ is divisible by $n$ whenever $m=\ell k$ for some integer $k$, that is $\dfrac m\ell $ is an integer. 
Thus, saying $m$ divides $\ell $ is the same as saying $\ell$ is divisible by $m$.
Thus, the theorem should read:

THEOREM Let $H\leq G$. Then $|H|$ divides $|G|$, or $|G|$ is divisible by $|H|$. In particular >$$|G|=|G:H||H|$$ 

