Would there anyone tell me reason for this statement hold and give some hint? $Q)$ Let the group $G$ and its normal subgroups $N$ and $M$. The orders of groups are  $\Vert G \Vert = 100$,  $\Vert M \Vert =25$ and $\Vert N\Vert =20$
Defining the homomorphsim $f$ and $g$ like the below
$f : G \to   G/M$ by  $f(x) = \bar x$ and $g : G \to   G/N$ by  $g(x) = \bar x$
Find the $c\in G$ $s.t. f(c) = \bar a$ and $g(c) = \bar b$ for given $\forall a, 
b(\neq a) \in G$ 

The answer sheet said 
considering the homo, $\phi : G \to G/M \times G/N $ by $\phi(x) = (f(x),g(x))$
The number of the $\beta$ is $\vert ker \phi \vert$ for $\phi^{-1}(\alpha) = \beta \bullet ker\phi$ (Here, $\phi(\beta)=\alpha$ ) [statement(*)]
The number of the $c$ is $\vert ker \phi \vert$ for $ \phi(c) = (\bar a, \bar b)$ 
Plus I did the another method motivated by $ker\phi = M \cap N$
My attempt)  I was motivated by the $ker \phi = M \cap N$. 
So I tried to show the $c$ which are satisfying the $f(c) = \bar a $and $g(c) = \bar b$ would be $c \in M \cap N$
For $c \in G$, $f(a) = \bar a = aN = f(c) \Rightarrow \exists n \in N s.t. c = an$
Likewise $f(b) =  \bar b = bM = f(c)  \Rightarrow \exists m \in M s.t. c = bm$
So All I left to just showing $c  \in M\cap N$. But I've stuck. 
Therefore my question is first, Why does the statement (*) hold? Second,What should I next to show $c  \in M\cap N$ in my trial?
Thanks.

In general, if $f\colon G\to A$ and $h\colon G\to B$ are group homomorphisms, then the corresponding map $G\to A\times B$ given by $g\mapsto (f(g),h(g))$ has kernel $\mathrm{ker}(f)\cap\mathrm{ker}(g)$. 
If I understand correctly, you are trying to show that given any $a,b\in G$, you can find $c$ such that $cM=aM$ and $cN=bN$.
Now, $G/M$ has order $4$, and $G/N$ has order $5$. So $(G/M)\times(G/N)$ has order $20$. Here, the kernel of your map into the product is $M\cap N$, which has order dividing $\gcd(|M|,|N|) = 5$. Thus, the kernel is either trivial or of order $20$. Since the image has order $|G|/|M\cap N|$, which must divide $20$, it follows that $M\cap N$ has order $5$, and hence that the map $G\to (G/M)\times (G/N)$ is surjective. 
Given that the morphism is surjective, given $a,b\in G$, the element $(aM,bN)$ is in the image of the map, and hence there exists $c\in G$ that maps to $(aM,bN)$; that is, such that $cM=aM$ and $cN=bN$, which is what you want to establish.
