Prove ${ Z }_{ 3 }{ \times Z }_{ 4 }\simeq { Z }_{ 12 }$ directly. $${ Z }_{ 3 }{ \times Z }_{ 4 }\simeq { Z }_{ 12 }$$
Above notations are ideal.
I tried $f(a,b)=4a+3b$
But, I run into a brick wall because of multiplication for homomorphism.
How to prove it?
 A: NB: I could not entirely understand the question as posted, and here provide a proof that the two sides are isomorphic as additive groups.
I suggest that you write elements of $Z_3 \times Z_4$ as pairs $(a, b)$, where $a$ is in $Z_3$ and $b \in Z_4$. Then the simplest (group) isomorphism $f$ from $Z_{12}$ is 
$$
1 \mapsto (1, 1)
$$
You have to then ask whether $f(c) = 0$ for any value of $c$ other than $0$) (i.e., check that it's injective). Once you've done that, you know it's surjective, because the two sets have the same number of elements. 
Let's check injectivity. For an element $c$, we have
$$
f(c) = (c \bmod 3, c \bmod 4)
$$
If that's $(0,0)$, then $c$ is a multiple of both $3$ and $4$; hence must be a multiple of 12 as well, and hence must be 0 in $Z_{12}$. 
N.B.2:as @A.P. points out, this is also a homomorphism of rings, for 
\begin{align}
f(cd) &= (cd \bmod 3, cd \bmod 4) \\
&= ((c \bmod 3) (d \bmod 3), (c \bmod 4) (d \bmod 4))\\
&= (c \bmod 3, c \bmod 4) \cdot (d \bmod 3, d \bmod 4) .
\end{align}
A: How about you use $f : Z_{12} \to Z_4 \times Z_3$ such that $a \to (a\ \text{mod}\ 4, a\ \text{mod}\ 3)$
A: Hint:
Use Bézouts identity $4-3=1$. The map is:
$$(a\bmod 3,b\bmod 4)\mapsto 4a-3b\bmod 12.$$
