# Uniserial modules are determined up to isomorphism by their composition series?

I have a question on uniserial modules. Here, every modules are finitely generated modules over a finite dimensional algebra $$A$$ over a field $$K$$.
In the book Elements of the Representation Theory of Associative Algebras: Volume 1 , there is a remark as below. (p.164)

We also notice that uniserial modules are determined up to isomorphism by their composition series, that is, if M and N are uniserial modules and have the same composition factors in the same order, then they are isomorphic. An isomorphism is constructed by an obvious induction on the common composition length of M and N.

But I cannot figure out how to make an induction step. Obviously, we have an isomorphism $$M/radM \cong N/radN$$ and by induction hypothesis, have an isomorphism $$radM\cong radN$$, but I don't know how to make an isomorphism between $$M$$ and $$N$$ from these informations. How can I show that those two are isomorphic?

## 1 Answer

This is not true in general. For example, the Klein four group has infinitely many $$2$$-dimensional uniserial modules over a field of characteristic $$2$$. I'm not quite sure what the authors mean in this section. It doesn't look like they are considering specific classes of algebras, where this might well be true.

An explicit example for $$G=\langle x,y\rangle$$ Klein four is to map $$x\mapsto \begin{pmatrix}1&1\\0&1\end{pmatrix}$$ and $$y\mapsto \begin{pmatrix}1&\lambda\\0&1\end{pmatrix}$$ for any $$\lambda$$ in the field. It is an easy exercise to see that there is no invertible matrix that centralizes $$x$$ and maps $$y$$ from one such matrix to any other, hence these representations are all inequivalent.