Show that $Z(G)$ is contained in $C(a)$. Let $G$ be a group and $a \in G$. $C(a)$ is the centralizer of $a$ in $G$. Show that $Z(G)$ is contained in $C(a)$.
Wont $C(a)$ be contained in $Z(G)$? 
 A: In fact $Z(G) = \bigcap_{a \in G} C_G(a)$.
A: Not necessarily.
If $x \in C(a)$, then $x$ commutes with $a$, it doesn't mean $x$ commutes with every element of $G$. 
On the other hand, if $x \in Z(G)$, then $x$ commutes with every element of $G$ and hence $x$ commutes with $a$, so $x \in C(a)$.
A: Here, is a "concrete" example, to help you remember.
Let $G = \langle a,b: a^4 = b^2 = e; ba = a^3b\rangle$, so that:
$G = \{e,a,a^2,a^3,b,ab,a^2b,a^3b\}$.
It is clear that all of $\langle a\rangle = \{e,a,a^2,a^3\}$ commutes with $a$, so:
$\langle a\rangle \subseteq C(a)$.
Since $ba = a^3b \neq ab$, it is clear that $b$ does not commute with $a$. This shows that $C(a) \neq G$, and since $C(a)$ has order at least $4$, and there are no divisors of $8 = |G|$ between $4$ and $8$, it must be that: $C(a) = \langle a\rangle$.
Now let's look at $Z(G)$. We know $G$ is non-abelian, so $Z(G)$ has order of at most $4$. We also know $a \not \in Z(G)$ (because $a$ does not commute with $b$). By the same logic $b \not\in Z(G)$. Let's test for two more elements "at the same time", $a^3$ and $a^3b$.
$(a^3)(a^3b) = a^6b = a^2b$
$(a^3b)(a^3) = a^3(ba)a^2 = a^3(a^3b)a^2 = a^6(ba)a = a^2(ba)a =  a^2(a^3b)a = a^5(ba) = a^5(a^3b) = a^8b = b$
So neither $a^3$ nor $a^3b$ lie in $Z(G)$. So we know $Z(G) \subseteq \{e,a^2,ab,a^2b\}$ (ths is just a set).
This $4$-element subset is not a subgroup, since $a^2(ab) = a^3b$ is not in this set. So $Z(G)$ must be a subgroup of less than order $4$. 
Now $a^2b = a^6b = a^3(a^3b) = a^3(ba) = (a^3b)a =  (ba)a = ba^2$, so we see $a^2$ commutes with both $a$ and $b$, thus with all of $G$. Hence $Z(G) = \{e,a\}$.
Here, then, it is clear that $Z(G) \subseteq C(a)$.
A: No, because for $g\in C(a)$ it only has to commute with $a$, while $g\in Z(G)$ if and only if $g$ commutes will all elements of $G$.
A: If an element, $x$, lies in the center of $G$, then $x$ commutes with every element of $G$.  So $xa = ax$, which means that $x \in C(a)$, i.e. $x$ is an element that commutes with $a$.  So $Z(G)$ is contained in $C(a)$.  
Your confusion might stem from the idea that since $a \in G$, you'd think $C(a)$ would be contained in $Z(G)$.  However, membership in $Z(G)$ is a stronger condition than membership in $C(a)$.  To be in $C(a)$, you just need to commute with $a$.  To be in $Z(G)$, you need to commute with every element of $G$, including $a$.
A: The condition $g\in Z(G)$ means $g$ commutes with all elements of $G$, in particular that $g\in C(a)$, for all $a\in G$.
Remember that $C(a)=\{g\in G:ga=ag\}$ (for a fixed $a$), while
$$Z(G)=\{g\in G:ga=ag,\text{ for all }a\in G\}$$
so $Z(G)\subseteq C(a)$, for all $a\in G$.
The reverse inclusion does not hold in general: indeed we can have $Z(G)=\{1\}$, for instance when $G=S_3$ (verify it). However, $\langle a\rangle\subseteq C(a)$, so you can take any non identity element of $S_3$ and find the required counterexample.
