Finding a form of Representations of algebra $FV_4$, where $F=Z_2$ I am hoping for some help with the following exam question I attempted. It is as follows, broken in to two parts:  

Let $A$ be a finite-dimensional algebra over a field $F$.
  $1.)a)i)$ What is a representation of algebra $A$? What are equivalent representations? Define the $A$-module $V$ corresponding to a representation $\rho$?
  $ii)$ Let $V, W$ be finite dimensional $A$-modules. Define the socle of $V$. Denote the socle of $V$ by $soc(V).$ Let $\phi:V \to W $ be an $A$-module isomorphism. Show that $$\phi(soc(V)) = soc(W) $$

Part $a)i)$ is fine I only included it here for completeness. However I didn't even know the definition of 'Socle' (it's not even in the lecture course...) so I'm not confident about the next part.
After looking the definition up, my answer for $a)ii)$ is as follows:  
$soc(V) = \sum \{N : N\; \text{is a simple submodule of V}\}$
If $\phi :V \to W$ is an $A$-module isomorphism then given a simple submodule $N \subset V$; $\phi(N)$ must be simple (if not a non-zero proper submodule $U \subset \phi(N)$ maps to non-zero proper submodule $\phi^{-1}(U) \subset N$ under $A$-module isomorphism $\phi^{-1}$.
  Hence $$\begin{align} \phi(soc(V)) &=\phi(\sum \{N : N\; \text{is a simple submodule of V}\})\\
&=\sum \{\phi(N) : N\; \text{is a simple submodule of V}\} \\
&\subset soc(W) \end{align}$$
And since $\phi$ is an isomorphism we can do the same thing with $\phi^{-1}$ to get equality.  
Is this correct?? I think it should be having done similar things a million times but given that I am not fully confident of the definition of socle I thought I'd check... But now on to the bulk of the question:  

$b)$ Let $F= \Bbb{Z}_2$ be the field with two elements, and let $V_4$ be the klein-Four-Group ($C_2\times C_2$)- generated by elements $x$ and $y$. For $a, b \in F$, define matrices:
  $$
\begin{array}{cc}
\begin{array}{c|cccc}
\rho_{(a,b)}(x) =\begin{pmatrix}
                 1 & 0 & 1 \\
                 0 & 1 & 0 \\
                 0 & 0 & 1 \\
                 \end{pmatrix}
\end{array}
&
\begin{array}{c|cccc}
\rho_{(a,b)}(y) =\begin{pmatrix}
                 1 & a & 1 \\
                 0 & 1 & b \\
                 0 & 0 & 1 \\
                 \end{pmatrix}
\end{array}
\end{array}  
$$
  i) Determine all tuples $(a,b)$ such that $\rho_{(a,b)}$ is a representation of the group algebra $FV_4$
  ii) Denote by $V_{(a,b)}$ the module corresponding to a representation $\rho_{(a,b)}$. Determine the socle of the module $V_{(a,b)}$ for each pair $(a,b)$ such that $\rho_{(a,b)}$ is a representation.
  iii) Are the representations $\rho_{(0,0)}$ and $\rho_{(0,1)}$ equivalent?
  [Hint: you may use without proof the fact that $FV_4$ has precisely one simple module.]  

For i) I noted that I required for a group representation of $V_4$ we need $\rho_{(a,b)}(x)^2 = Id = \rho_{(a,b)}(y)^2$ and we need $\rho_{(a,b)}(x)\rho_{(a,b)}(y) = \rho_{(a,b)}(y)\rho_{(a,b)}(x)$.
And if we have these the enforcing linearity over the field gives a well-defined algebra representation.
I found that over $\Bbb{Z}_2$; $\rho_{(a,b)}(x)^2 = Id$ is always true and $\rho_{(a,b)}(y)^2 = Id$ forces $ab = 0$ so at least one of $a$ or $b$ is $0$.
Then commutativity of these matrices always happens given this condition.
Is there a better way to do this?? I doubt there will be but I thought I'd check. 
For ii) $V_{(a,b)} = \Bbb{Z}_2^3$ as a module under the $\rho_{(a,b)}$ action. I first check for $1d$ subspaces of $V_{(a,b)}$, so I check for eigenvectors. and I find that: $$\rho_{(0,0)}\; \text{and}\; \rho_{(0,1)}\; \text{have 1d submodules}:  
  \left\langle\begin{pmatrix}
               1 \\
               1 \\
               0 \\
              \end{pmatrix}\right\rangle, 
  \left\langle\begin{pmatrix}
               0 \\
               1 \\
               0 \\
              \end{pmatrix}\right\rangle,
  \left\langle\begin{pmatrix}
               1 \\
               0 \\
               0 \\
              \end{pmatrix}\right\rangle$$
Whilst $$\rho_{(0,0)}\; \text{only has 1d submodule:}\; 
  \left\langle\begin{pmatrix}
               1 \\
               0 \\
               0 \\
              \end{pmatrix}\right\rangle$$  
These must be simple as they are $1$ dimensional. I then look for $2d$ submodules case by case and checking if we could have a particular subspace using the $8$ vectors in $\Bbb{Z}_2^3$ and I find none.
Do I need to check possible $2d$ subspaces?? Can I use the hint here?? How come I have found $3$ simple submodules when the hint says there should only be one?? Are they the same module?? In general how does one check for $2d$ subspaces?? 
For iii)If $\rho_{(0,0)}$ and $\rho_{(0,1)}$ were equivalent this means that in each case the matrices $\rho_{(0,0)}(x)$ and $\rho_{(0,1)}(x)$ are similar and  $\rho_{(0,0)}(y)$ and $\rho_{(0,1)}(y)$ would be similar under the same change of basis matrix right??
I just literally used brute force with a matrix $$A =\begin{pmatrix}
                    a & b & c \\
                    d & e & f \\
                    g & h & i \\
                    \end{pmatrix}$$
and try $\rho_{(0,0)}(x) = A^{-1}\rho_{(0,1)}(x)A$ and $\rho_{(0,0)}(y) = A^{-1}\rho_{(0,1)}(y)A$ to show that this can't happen.  
I have forgotten some of my linear algebra and cannot remember if there is a simple way of showing if two matrices are similar in general. Is there?? How else could we do this without brute force?? Am I meant to have gotten that $soc(V) \not \cong soc(W)$ so that I can use the earlier part, or is this to make a point that have the same socle doesn't necessarily make you isomorphic?? 
Apologies for the really really long question. Massive thanks to anyone that can answer any of these questions!
 A: All your work is correct, though in a few places you've made things harder than they need to be.
In (b)(ii), when it says that $FV_4$ has only one simple module, that means that there is only one simple $FV_4$-module up to isomorphism. So there is nothing wrong with finding multiple simple submodules of your representations: it's just that these simple submodules are all isomorphic to each other.
In particular, since you've found a 1-dimensional simple module (any of your simple submodules), all simple modules must be isomorphic to that 1-dimensional simple module and hence are also 1-dimensional.  This means you can skip your work of searching for 2-dimensional simple submodules, since no such thing can exist.  (Note that if you don't assume this, you should really also check for 3-dimensional simple submodules, which amounts to verifying that the entire module is not simple.)
In (b)(iii), you don't need to use brute force!  Just notice $\rho_{(0,0)}(x)=\rho_{(0,0)}(y)$ but $\rho_{(0,1)}(x)\neq\rho_{(0,1)}(y)$, and a change of basis can't change whether two linear maps are equal.
I agree that it seems weird that the work on socles is irrelevant to solving (b)(iii) (and that there is a much easier way to solve (b)(iii) as I noted above).  Perhaps there was a typo and the problem was meant to ask you to compare $\rho_{(1,0)}$ and $\rho_{(0,1)}$ instead.
