# 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.

• You're stating it wrongly. The theorem is "The order of $G$ is divisible by the order of any subgroup of $G$". – Pedro Tamaroff Mar 26 '13 at 23:10
• To elaborate on Peter's comment, let $G$ be any group. If $e\in G$ is the identity, certainly $\{e\}\subseteq G$ is a subgroup, but $\left|\{e\}\right| = 1$ is not a multiple of $\left|G\right|$ unless $\left|G\right| = 1$ (if $G$ has more than two elements, then $\left|G\right| > \left|\{e\}\right| = 1$, so $1$ can't be an integer multiple of $\left|G\right|$). – Stahl Mar 26 '13 at 23:20
• @PeterTamaroff: Thanks for pointing this out, maybe a language problem. I took the liberty to modify the question to something more meaningful. – azimut Mar 26 '13 at 23:20
• @azimut Oh, but please wait untill the OP sees his/her mistake! Else it might go unnoticed! – Pedro Tamaroff Mar 26 '13 at 23:22

• 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 $\lvert G\rvert$.

• The OP has written a wrong statement above. Did you realize? – Pedro Tamaroff Mar 26 '13 at 23:12
• @PeterTamaroff: Maybe a language problem confusing 'divisible' with 'divides'. – azimut Mar 26 '13 at 23:14
• It seems so. =/ – Pedro Tamaroff Mar 26 '13 at 23:17
• @azimut Is $C(a)$ (centralizer of a $\in G$ ) is also an subgroup. – TLE Mar 27 '13 at 9:07
• @stranger001: Yes, just check it in a similar way as above. – azimut Mar 27 '13 at 10:44

Just prove the center is a group, and you get the result.

• you mean I have to prove center is subgroup – TLE Mar 26 '13 at 22:54
• Any Hints how to prove that??,Don't tell me answer. – TLE Mar 26 '13 at 23:02
• @stranger001, Show 1 is in the center. Let a,b be in the center and show ab is in the center and show that a^{-1} is in the center. – user58512 Mar 26 '13 at 23:05
• The OP seems to mean the otherwise. He says "order of $G$ is divisible by order of $Z_G$", meaning wether $$|G|\;\mid\; |Z_G|$$ is true or not. I pointed it out in the comments above. – Pedro Tamaroff Mar 26 '13 at 23:08
• @caveman Please check my solution $property\ 1:$$e commutes to every element \therefore e must be in Z(G). property\ 2:suppose a,b \in G \therefore ax=xa \implies ex=a^{-1}xa \implies xa^{-1}=a^{-1}x. That means a^{-1} \in Z(G) property\ 3:Given that a \in Z(G) \therefore ax=xa \implies bax=xab \implies (ab)x=x(ab) \therefore \ ab commutes to every element of G, ab must be in Z(G). – TLE Mar 26 '13 at 23:20 I just want to make sure you know the following terminology: We say that an integer m divides an integer n whenever \ell=mk for some integer k, that is \dfrac \ell m is an integer. We say that an integer m is divisible by n whenever m=\ell k for some integer k, that is \dfrac m\ell is an integer. Thus, saying m divides \ell is the same as saying \ell is divisible by m. Thus, the theorem should read: THEOREM Let H\leq G. Then |H| divides |G|, or |G| is divisible by |H|. In particular >$$|G|=|G:H||H|$\$