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!
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.