Every Quotient group of a group is a homomorphic image of the group (First theorem of isomorphism?) 
"Every Quotient group of a group is a homomorphic image of the group."
Is the statement equivalent to the first group homomorphism theorem?

I am not able to understand how because according to the theorem only the Quotient group with the kernel as normal subgroup is homomorphic then how does it changes in this particular definition.
 A: This statement is not equivalent to the first isomorphism theorem. As far as I can tell, it is a simple fact that falls right out of the definition of a quotient group.
Here is Wikipedia's statement of the first isomorphism theorem.
I'm reproducing it here below for convenience:

Let $G$ and $H$ be groups, and let $φ: G → H$ be a homomorphism. Then:
  
  
*
  
*The kernel of $φ$ is a normal subgroup of $G$, 
  
*The image of $φ$ is a subgroup of $H$, 
  
*The image of $φ$ is isomorphic to the quotient group
  $G / \text{ker}(φ)$. In particular, if $φ$ is surjective then $H$ is isomorphic to
  $G / \text{ker}(φ)$.

The first isomorphism theorem, however, is not a definition of what a quotient group is.
Wikipedia defines a quotient group as follows:

A quotient group or factor group is a mathematical group obtained by
  aggregating similar elements of a larger group using an equivalence
  relation that preserves some of the group structure (the rest of the
  structure is "factored" out).

So, if we have a group $G = \langle X, \cdot \rangle $ and an equivalence relation $\simeq$, then we can define $X'$ and $\circ$
$$ X' = \{\{ a \mathop| a \in X \land a \simeq b \} \mathop| b \in X \} $$
$$ (a \circ b) = \{ m \cdot n \mathop| m \in a \land n \in b \} $$
If $G' = \langle X', \circ \rangle$ is a group, then it is a quotient group of $G$.
Let's look at the statement you provided.

"Every Quotient group of a group is a homomorphic image of the group."

Restated, that statement is:
For every group $G$ and group $L$, if $L$ is a quotient group of $G$, then there exists a homomorphism $\phi : G \to L$ such that $\text{Im}(\phi) = L$ .
Let's assume we're given a $G$ and an $L$.
If $L$ is equal to $G$, then we're done.
If $L$ is a quotient group of $G$, then we construct a specific $\phi$ as follows:
Let's pick the $\phi$ that sends every $g \in G$ to the equivalence class in $L$ containing that particular $g$ .
$$ \phi(g) = l \iff g \in l $$
This map is a homomorphism because of how the product of $L$, $\circ$, is defined.
$$ (a \cdot b = c) \implies (\phi(a) \circ \phi(b) = \phi(c)) $$
