Show that $Z$ is an abelian subgroup of $G$ I am trying to solve this question:  The center of a group $G$ (denoted by $Z$) is defined to be the set of elements $$\{z_1, z_2,...\}$$ that commute with all elements of $G$, that is $z_ig=gz_i$ for all $g \in G$. Show that $Z$ is an abelian subgroup of $G$. 
So I know that if it's a subgroup of $G$ and it commutes with the elements of $G$ it must commute with itself, but I'm not sure how to prove that it is a subgroup. Don't I have to know how many elements are respectively in $Z$ and $G$ to check if Lagrange's Theorem applies?
 A: Basically you have to check the following three facts:


*

*Presence of inverse elements    If $zg = g z$ for all $g \in G$ you obtain by multiplication that 
$z^{-1}(zg)z^{-1} = z^{-1}(g z)z^{-1}$
and then
$$gz^{-1} = z^{-1}g$$
and $z^{-1}$ commutes with all $G$ and hence is contained in $Z$.

*Presence of identity  It is obvious that $e$ is contained in $Z$ because it commutes with all elements of $G$.

*Closure by multiplication  You have to show that $z \cdot z'$ is in $Z$ if $z, \, z' \in Z$, but this is easy because $$z z' g = z(z'g)= z (g z')= (zg)z'=(gz)z'=gzz'$$
A: The definition of a subgroup is the following:

Definition: A subset $H \subseteq G$ of a group $G$ is called a subgroup if the following condititons are satisfied:
  
  
*
  
*$1 \in H$, where $1$ denotes the neutral element of $G$,
  
*for all $x, y \in H$ one has $x y \in H$, and
  
*for every $x \in H$ one has $x^{-1} \in H$.
  

To check that
$$
  Z = \{z \in G \mid \text{$zg = gz$ for all $g \in G$}\}
$$
is a subgroup of $G$ you need to check that this subset satisfies all three of the above conditions.

Regarding your proposed use of Lagrange's theorem:

Theorem: (Lagrange)
  If $H \subseteq G$ is a subgroup of a finite group $G$, then $|H|$ divides $|G|$.

First off, $G$ is not assumed to be finite, so we already cannot apply the theorem.
But even if we assume $G$ to be finite, we gain nothing from this:
The theorem only tells us that if we already know that $Z$ is a subgroup, then we can conclude that $|Z|$ divides $|G|$.
The converse of the theorem does not hold, i.e. if we have a subset $H \subseteq G$ such that $|H|$ divides $|G|$, then it does not necessarily follow that $H$ is a subgroup of $G$.
So even if $G$ was finite and $|Z|$ would divide $|G|$, we could not use this to conclude that $Z$ is a subgroup.
The only thing we could conclude from the theorem is that if $G$ was finite and $|Z|$ would not divide $|G|$, then $Z$ would not be subgroup of $G$.
But of course this won’t happen, as it is a subgroup.
