Showing W=ku+kv has a unique module structure over kG

Let $$G= \times $$ where $$|x|=|y|=p$$ and so $$|G|=p^2$$. Let $$k$$ be a field of characteristic $$p$$. Let $$W$$ be the $$k$$-span of $$v$$ and $$u$$. We wish to show the module structure given by

$$x \cdot u = u + \alpha v$$

$$x \cdot v = v$$

$$y \cdot u = u + \beta v$$

$$y \cdot v = v$$

where $$\alpha,\beta \in k$$.

We can represent $$x$$ and $$y$$ by matrices

$$x = \begin{pmatrix} 1 & 0\\ \alpha & 1 \end{pmatrix}$$

$$y = \begin{pmatrix} 1 & 0\\ \beta & 1 \end{pmatrix}$$

If I understand: To show this is structure is unique we are essentially be asked to show that the only commutative set of $$2 \times 2$$ matrices which commute and are of order $$p$$ are of the form given above. However, this seems to miss a lot of what I might already know about representation theory and so I wanted to confirm that really is the approach and how to continue from that idea if that is the case.

• It seems to me I might be on the right track because the upper triangulars similar to the form above would also satisfy – Aaron Zolotor Apr 24 at 13:30
• What you've written provides an appropriate left module structure, and the upper triangulars form an appropriate right module structure. You'll have to pick one or the other. – rschwieb Apr 24 at 14:20

For a $$k$$ algebra $$A$$, to show that "$$M$$ has a unique left $$A$$ module structure" amounts to showing there is exactly one algebra homomorphism $$A\to End(M_k)$$.
Now, we're assisted in the fact that there is a universal property that if $$A$$ is an $$R$$ algebra, then any group homomorphism $$G\to U(A)$$ extends uniquely to an algebra homomorphism $$R[G]\to A$$.
In your case, $$R=k$$ and $$A=M_2(k)$$, and the assignments you provided make a group homomorphism $$G\to U(M_2(k))= GL_2(k)$$. Therefore by the universal property this extends uniquely to an algebra homomorphism $$k[G]\to M_2(k)$$.
• Assuming I didn't know about that property how might I prove this? Seeing as we want to show there is exactly one homomorphism of algebras would it be sufficient to show that the only other possible non-trivial ones must correspond to the subgroups generated by $x$ or $y$? – Aaron Zolotor Apr 24 at 13:49
• @AaronZolotor It seems to me that there is no profitable reason to prove it for the special case and to avoid proving the general case. It is no harder to prove the general case. Just show that given $\phi:G\to U(A)$, the obvious map from $kG\to A$ given by $\psi(\sum \alpha_gg)=\sum \alpha_g\phi(g)$ extends $\phi$ and moreover anything else extending $\phi$ would have to match it. – rschwieb Apr 24 at 14:14