If $G$ is a finite group, then $\operatorname{ord}S(a)=\operatorname{ord}(G)/\operatorname{Ord}(C(a))$ $\DeclareMathOperator{\ord}{ord}\DeclareMathOperator{\Ord}{Ord}$$C(a)=${$x\in G \mid xa=ax$} and $S(a)=${$xax^{-1} \mid x\in G$}. If $G$ is a finite group, then $\ord(S(a))=\ord(G)/\Ord(C(a))$. Also $\ord(g)=\sum \ord(G)/\Ord(C(a_i))$ for finitely may distinct $a_i$.
Attempts: I think I can understand what order of a group which mean that the cardinality of the group. I have showed that $C(a) \le S(a)$ but not sure how that helps to solve it.
 A: First of all note that for any $a \in G$, $C(a)$ is a subgroup of $G$; we have, for $x, y \in C(a)$, 
$(xy)a = x(ya) = x(ay) = (xa)y = (ax)y = a(xy), \tag{1}$
showing $xy \in C(a)$ as well.  Since $G$ is finite, we conclude $C(a)$ is a subgroup of $G$ since it is a multiplicatively closed subset of a finite group.  Next, scrutinize the set $S(a) = \{xax^{-1} \mid x \in G\}$; it is in fact the set of distinct conjugates of $a$.  I claim that $xax^{-1} = yay^{-1}$ if and only if $xC(a) = yC(a)$ as cosets of $C(a)$ in $G$.  For if $xax^{-1} = yay^{-1}$, then $y^{-1}xa = ay^{-1}x$, showing $y^{-1}x \in C(a)$; this implies $xC(a) = yC(a)$, as is well-known.  Similarly, $xC(a) = yC(a)$ implies $y^{-1}x \in C(a)$, or $y^{-1}xa = ay^{-1}x$ whence  $xax^{-1} = yay^{-1}$.  This shows that the distinct conjugates of $a$ are in one-to-one correspondence with the distinct cosets of $C(a)$; indeed if we define the map $\theta_a:xax^{-1} \to xC(a)$, our argument shows it is well-defined and depends only on the conjugate $xax^{-1}$ itself, independently of the particular conjugating element or coset representative $x$.  It follows that the number of conjugates $xax^{-1}$, which is $\text{ord} (S(a))$, the cardinality of $S(a)$, is equal to the number of cosets of $C(a)$, and this is $[G:C(a)]$, the index of $C(a)$ in $G$.  By Lagrange's theorem, $[G:C(a)] = \text{ord}(G) / \text{ord}(C(a))$;
so we see that $\text{ord} (S(a)) = \text{ord}(G) / \text{ord}(C(a))$.
As for the second question, I am going to assume that the OP's equation
$\text{ord}(g)=\sum \text{ord}(G) /\text{ord}(C(a_i) \tag{2}$
actually contains a minor typographical error and that the "$g$" occurring on the left-hand side should in fact be capitalized, so the equation reads
$\text{ord}(G)=\sum \text{ord}(G) /\text{ord}(C(a_i) \tag{3};$
I came to the conclusion that such an assumption is warranted after assuming that the little "$g$" occurring in (2) referred to the cyclic subgroup $\langle g \rangle$ generated by some $g \in G$.  After several attempts to establish (2) based on such a hypothesis, having learned some interesting things but being no closer to a solution, I reverted to the assumption that "$g$" is a typographical error, and that "$G$" is what was intended.  This interpretation is certainly contextually consistent, in any event.  So we will prove (3), not (2).  To do so, observe that for any $a, b \in G$ the sets $S(a)$, $S(b)$ are either disjoint or identical.  For if $c \in S(a) \cap S(b)$, then there exist $x, y \in G$ with $xax^{-1} = c = yby^{-1}$; but then $y^{-1}xax^{-1}y = (y^{-1}x)a(y^{-1}x)^{-1} = b$, showing $b \in S(a)$; but then $S(b) \subset S(a)$ since for any $z \in G$, $zbz^{-1} =  z(y^{-1}x)a(y^{-1}x)^{-1}z^{-1} = (zy^{-1}x)a(zy^{-1}x)^{-1}$; likewise we have $S(a) \subset S(b)$, so in fact $S(a) = S(b)$ if $S(a) \cap S(b) \ne \phi$.  Note that $a = eae^{-1} \in S(a)$ for every $a \in G$, where $e \in G$ is the identity element.  This shows that
$\text{ord}(G) = \sum \text{ord}(S(a_i)), \tag{4}$
where the sum is taken over any collection of $a_i \in G$ such that $S(a_i) \cap S(a_j) = \phi$ and $\bigcup_i S(a_i) = G$.  We have seen that $\text{ord} (S(a_i)) = \text{ord}(G) / \text{ord}(C(a_i))$ so (4) becomes
$\text{ord}(G) = \sum \text{ord}(G) / \text{ord}(C(a_i)), \tag{5}$
as was required.  QED.
It is worth noting at this point that for any $z \in Z(G)$, the center of $G$, $S(z)$ is the singleton set $\{ z \}$ since $xzx^{-1} = zxx^{-1} = ze = z$ for any $x \in G$.  Of course in such a case $C(z) = G$, so the sum in (5), so there is a summand $1$ occurring in (5) for each $z \in Z(G)$; this means (5) can be written
$\text{ord}(G) = \text{ord}(Z(G)) + \sum \text{ord}(G) / \text{ord}(C(a_i)), \tag{6}$
where in (6) the $a_i$ range over the non-central elements out of the collection occurring in $5$.  
What a surprise! It appears that in performing this little exercise I have re-discovered the world-famous class equation!
Hope this helps.  Cheerio, 
and as always,
Fiat Lux!!!
A: $G$ works on itself by conjugation. Your $S(a)$ is the orbit of $a$ under this action, usually called the conjugacy class. $C(a)$ is the stabilizer of $a$. Now go and look for a book that treats the Orbit-Stabilizer Theorem, and in addition realize that $G$ is the union of its distinct conjugacy classes.
