So, I can write down the definition of the socle of a module but I'm having trouble actually putting it in my brain. To understand things I'm trying to write down the radical and socle series of the lower triangular matrices. In general, if $A$ is a finite dimensional $k$-algebra is it the case that $soc^2 A$ is the stuff annihilated by $rad^2 A$ or is there more?
Let $T_n(k)$ be the lower triangular matrices over a field $k$. It is easily seen that
$$rad T_n(k) = \begin{pmatrix}0 & 0 & 0\\ a & 0 & 0\\ b & c & 0 \end{pmatrix}$$
$$rad^2 T_n(k) = \begin{pmatrix}0 & 0 & 0\\ 0 & 0 & 0\\ b & 0 & 0 \end{pmatrix}$$
and $rad^3 T_n(k)=0$. Now, the socle is the stuff annihilated by the radical so
$$soc T_n(k) = \begin{pmatrix} 0 & 0 & 0\\ 0 & 0 & 0 \\ a & b & c\end{pmatrix} $$
the stuff annihilated by $rad^2(u)$ is $$ \begin{pmatrix}a & 0 & 0\\ b & c & 0\\ 0 & x & y \end{pmatrix}$$ and
$$T_n(k)/soc T_n(k) = \begin{pmatrix}a & 0 & 0\\ b & c & 0\\ 0 & 0 & 0 \end{pmatrix}$$
and the socle of this is the sum of the simple submodules of that quotient (which I'm having trouble determining) but it seems these things don't line up? What's going on?
As mentioned in a comment, check your computations again. The radical and socle series of the lower triangular matrices (considered as a right module over itself) should look like this:
radical series: $$ \begin{bmatrix}\cdot&0&0\\\cdot&\cdot&0\\\cdot&\cdot&\cdot\\\end{bmatrix}\supseteq \begin{bmatrix}0&0&0\\\cdot&0&0\\\cdot&\cdot&0\\\end{bmatrix}\supseteq\begin{bmatrix}0&0&0\\0&0&0\\\cdot&0&0\\\end{bmatrix}\supseteq\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\\\end{bmatrix} $$
socle series: $$ \begin{bmatrix}0&0&0\\0&0&0\\0&0&0\\\end{bmatrix}\subseteq \begin{bmatrix}0&0&0\\0&0&0\\\cdot&\cdot&\cdot\\\end{bmatrix}\subseteq \begin{bmatrix}0&0&0\\\cdot&\cdot&0\\\cdot&\cdot&\cdot\\\end{bmatrix}\subseteq \begin{bmatrix}\cdot&0&0\\\cdot&\cdot&0\\\cdot&\cdot&\cdot\\\end{bmatrix}$$
(The obvious generalization to $n\times n$ lower triangular matrices is hopefully self-evident, although inconvenient to write out.)