Show that $Z(G) = \cap_{a \in G} C(a)$

Question

Show that $Z(G) = \cap_{a \in G} C(a)$.

Let $a \in Z(G)$. Then $ax=xa$ for all $x$ in $G$. In particular we can say that $ax_1=x_1a$ and $ax_2=x_2a$ and $ax_3=x_3a$ and so on ($x_i$ are elements of $G$). This is nothing but intersection of all subgroups of form $C(a)$. However I doubt my way of doing this question. Please guide me

Thanks

Answer 1

Maybe this is the shortest: $$\begin{align} x\in Z(G)&\iff \forall a\in G,\;ax=xa\\ &\iff\forall a\in G,\;x\in C(a)\\ &\iff x\in\bigcap_{a\in G}C(a). \end{align}$$

Answer 2

$$Z(G) = \{g \in G \mid \forall a \in G: ga = ag\} = \bigcap_{a \in G} \{g \in G \mid ga = ag\} = \bigcap_{a \in G} C(a)$$

Answer 3

I tried to write a proof in the style you're going for.

Suppose $x\in Z(G)$. Then, $xa=ax$ for all $a\in G$, and hence, $x\in C(a)$ for all $a\in G$. Thus, $x\in\bigcap_{a\in G}C(a)$. Conversely, suppose $x\in\bigcap_{a\in G}C(a)$ for all $a\in G$. Then, $xa=ax$ for all $a\in G$, so that $x\in Z(G)$.

By mutual inclusion, $Z(G)=\bigcap_{a\in G}C(a)$.