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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.

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1  
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

3 Answers 3

up vote 1 down vote accepted
  • 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 $\lvert G\rvert$.

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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.

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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 >$$|G|=|G:H||H|$$

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