Question: Let $G$ be a group. Prove that $Z(G)$ is a subgroup of $G$.
If I want to show that $Z(G)$ is a subgroup of $G$ that means I have to show that it is closed under group operation?
Here is my attempt.
Let $a,b$ be elements in $Z(G)$ and $x$ be an element in $G$. Then $ax=xa$ which is under group multiplication commutative and under inverse $(a^{-1}) x=x(a^{-1})$. And hence $(a^{-1})bx= (a^{-1})xb=x (a^{-1})b$ which is under group operation so $(a^{-1}),b$ are elements in $Z(G)$ thus a subgroup of $G$.
Really appreciate if anyone can help me by directing me if my attempt is not good.
Thanks in advance.
Best Answer
You need to show that $Z(G)\leq G$.
First of all clearly $1\in Z(G)$ since $1x=x1$ for all $x\in G$.
Let $a,b\in Z(G)$ then $ax=xa$ and $bx=xb$ for all $x\in G$. Then $(ab)x=a(bx)=a(xb)=x(ab)$ so $ab\in Z(G)$.
If $a\in Z(G)$ consider its inverse $a^{-1}$ in $G$. Since $ax=xa$ for all $x\in G$ we have that $a^{-1}(ax)a^{-1}=a^{-1}(xa)a^{-1}$ so $xa^{-1}=a^{-1}x$ for all $x\in G$ namely $a^{-1}\in Z(G)$.