Return to Homework Problems, Homework 5

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, $Z(G)\leq G$.

$\hspace{200pt} \spadesuit$