= Solution
For the <characteristic-three Specht module for the two-by-two partition>, reduce the <matrices> modulo three. The vector $e_1+e_2$ is fixed by every displayed <transposition>: $(12)$ and $(34)$ send it to $e_1-2e_2=e_1+e_2$, and $(23)$ interchanges its two terms. Thus its span is a proper nonzero \b[trivial <submodule>].
On the quotient impose $e_2=-e_1$. Each generating <transposition> then acts as minus one, so the quotient is the \b[<sign representation>]. Consequently
$$
\boxed{0\longrightarrow\mathbf1\longrightarrow S^{(2,2)}\longrightarrow\operatorname{sgn}\longrightarrow0,}
$$
and the two <composition factors> are trivial and sign, each once. In the $D$ notation they are $D^{(4)}$ and $D^{(2,2)}$ respectively. Indeed the restricted <tabloid> form has <Gram matrix> $\begin{pmatrix}4&2\\2&4\end{pmatrix}$, whose characteristic-three radical is precisely the trivial line, so the simple quotient is $D^{(2,2)}$.
The extension does not split: the minus-one <eigenspace> of $\rho(12)$ is the line through $e_2$, but $\rho(23)$ sends that line to the line through $e_1$. Thus there is no sign <submodule> complementing the trivial line, consistent with the failure of semisimplicity.
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