Hw5 Problem 10

If $G$ is a group, the center of G is defined to be $Z(G)=\{ x\in G \mid x*a=a*x$ for all $a\in G \}$. Show that $Z(G)$ is a subgroup of $G$.

Solution:

By the way that $Z(G)$ is defined, all elements in $Z(G)$ must be in $G$, so $Z(G)$ is a subset of $G$.

Let $a,b \in Z(G)$.
Then for every $c\in G, a*c=c*a$ and $b*c=c*b$.
Now, since $G$ is a group, $*$ is associative on G.
Thus, $(b*a)*c=(b*c)*a$.
Consequently, since $b*c=c*b$, $(b*a)*c=(c*b)*a=c*(b*a)$.
Hence, $(b*a)\in Z(G)$ for all $a,b \in G$.
Therefore, $Z(G)$ is closed under $*$.
$\hspace{200pt} \clubsuit$

Let $a,b,c \in Z(G)$.
Then $a,b,c \in G$.
Now, since $G$ is associative, $(a*b)*c=a*(b*c)$.
Therefore, $*$ is associative on $Z(G)$.
$\hspace{200pt} \clubsuit$

Since G is a group, $e\in G$ such that for all $a\in G, e*a=a*e=a$.
Therefore, by definition of $Z(G), e \in Z(G)$.
$\hspace{200pt} \clubsuit$

Let $a\in Z(G)$.
Then $a \in G$ and for all $b\in G, a*b=b*a$
Now, since $G$ is a group, there must exist $a'\in G$ such that $a*a'=a'*a=e$.
Hence, $a'*a*b*a'=a'*b*a*a'$.
Consequently, $e*b*a'=a'*b*e$.
Thus, $b*a'=a'*b$.
Therefore, $a' \in Z(G)$.
$\hspace{200pt} \clubsuit$

Therefore, since $Z(G)$ is a subset of $G$ closed under $*$ and satisfies all requirements for being a group, $Z(G)\leq G$.
$\hspace{200pt} \spadesuit$