Use group homomorphism to prove that for systems of linear equations, general=particular+homogeneous 
Suppose that $x$ is a particular solution to a system of linear
  equations and that $S$ is the entire solution set of the corresponding
  homogeneous system of linear equations. Using the fact that "if
  $\phi(g)=g'$, then $\phi^{-1}(g')=\{x\in G:\phi(x)=g'\}=g\text{Ker}(\phi)$",
  show why $x+S$ is the entire solution set of the nonhomogeneous system.
  In particular, describe the relevant groups and the homomorphism
  between them.

Looking at the statement, it's clear to me that $\text{Ker}(\phi)$ should be related to the null space of the system which is the entire solution set of the homogeneous system, but I am not sure what the mapping of $\phi$ should be and what $g$ is in this case.
 A: The system of linear equations seems to be 
$$
A x = g' \quad (*)
$$
where 
$$
A : K^n=(G,+) \to K^m=(G',+)
$$
is a vector homomorphism and thus a group homomorphism too. Further $x \in G$ and $g' \in G'$. Then 
$$
S = \text{ker}\, A = \{ x \in G \mid A x = 0' \}
$$
is the solution space of the homogenous system $A x = 0'$.
The first isomorphism theorem states


*

*$S$ is a normal subgroup of $G$, indeed the kernel of $A$ is a commutative subspace of $(G,+)$

*$A(G)$ is a subgroup of $G'$, indeed the image of $A$ is a subspace of $(G',+)$

*$A(G)$ is isomorphic to the quotient group $G/S = \{ g + S \mid g \in G \} \quad (**)$


The fundamental theorem on homomorphisms considers a "natural" homomorphism $f : G \to G/S$, with $f(g) = g + S$,  and states the existence of a unique homomorphism $\phi : G/S \to G'$ with $A = \phi \circ f$.

So $A x = g'$ means the particular solution $x$ is mapped by $A$ to the inhomogenity $g'$. 
Plus we have $A x =\phi(f(x)) = \phi(x+S)$, by the above, because $f(x) = x + S$.
Together this gives
$$
g' = A x = \phi(f(x)) = \phi(x + S)
$$
So $x + S$ is mapped by $\phi$ to $g'$. The inverse mapping is $\phi^{-1}(g') = x + S$.
$x + S$ is a set of solutions for $(*)$ because for every $h \in x + S$ we have a $s \in S$ and $h = x + s$, then 
$$
A(h) = A(x + s) = A(x) + A(s) = g' + 0' = g'
$$
so $h$ is a solution of $(*)$.
If there is a $y$ which is a solution of $(*)$, it would mean
$$
A y = g'
$$
but this would mean $f(y) = y + S$ and 
$g' = A y = \phi(f(y)) = \phi(y + S)$ thus
$$
\phi(y + S) = g' = \phi(x + S)
$$
because $\phi$ is an isomorphism, see $(**)$, it is injective and we have $y + S = x + S$, so $y$ is already in $x + S$. So $x + S$ contains the entire solution space.
A: Write your linear system as ${\bf y}' = A(t){\bf y}+{\bf g}(t)$. Then consider $D[{\bf y}]=\dfrac{d}{dt}\Big[{\bf y}\Big]-A(t){\bf y}$. 
Your system is equivalent to $D[{\bf y}]={\bf g}(t)$. 
Notice that $D$ is a linear operator: $D[{\bf y}_1+c{\bf y}_2] = D[{\bf y}_1]+cD[{\bf y}_2]$. This means $D$ is (in particular) a group homomorphism (ignore the scalar multiplication).
Let $S = \mathrm{Ker}(D) = \{ {\bf y} \;|\; D[{\bf y}]=0 \} = \{ {\bf y} \;|\; {\bf y}' = A(t){\bf y} \}$. So $S$ is the homogeneous solution set.
We have that ${\bf w} + \mathrm{Ker}(D)$ is then the set of all solutions of $D[{\bf y}] = D[{\bf w}]$ (just as $g\mathrm{Ker}(\varphi)$ is the same as the set of group elements mapping to $\varphi(g)$).
So if ${\bf x}$ is a particular solution of ${\bf y}' = A(t){\bf y}+{\bf g}(t)$, then ${\bf x}'=A(t){\bf x}+{\bf g}(t)$. In other words, $D[{\bf x}]={\bf g}(t)$.
Therefore, ${\bf x}+S = {\bf x}+\mathrm{Ker}(D) = \{ {\bf y} \;|\; D[{\bf y}]=D[{\bf x}] \} = \{ {\bf y} \;|\; D[{\bf y}]={\bf g}(t) \}$ is the solution set of the non-homogeneous system.
