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?
Best Answer
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 $$z z' g = z(z'g)= z (g z')= (zg)z'=(gz)z'=gzz'$$