$K$-basis of cyclic $KG$-module determined by group elements

Question

Let $G$ be a finite group and $K$ a field with $\mathrm{char}(K) \nmid |G|$, and let $M$ be a cyclic $KG$-module.

For a cyclic generator $v \in M$ we can pick group elements $g_1, \dots, g_d \in G$ such that $\{ g_1v, \dots, g_d v \}$ is a $K$-basis of $M$. Does that automatically mean that $\{ g_1w, \dots, g_d w \}$ is a $K$-basis for all generators $w$ of $M$ as well? If not, can we choose those $g_i$ in a special way such that the answer becomes positive?

Thank you in advance!

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What I know so far: The set of all elements $w \in M$ such that $\{g_1 w, \dots, g_d w\}$ is a $K$-basis of $M$ is Zariski-open. To see this, we can assume w.l.o.g. $M = K^d$ and $G \leq \mathrm{GL}(d,K)$. Then the set of those elements $w$ coincides with the non-vanishing set of the polynomial $\det(g_1 X, \dots, g_d X)$, where $X = (X_1, \dots, X_d) \in K[X_1, \dots, X_d]^d$.

Therefore “almost all” generators $w$ form a $K$-basis of $M$ in this way. What about the remaining ones?

Answer 1

Let $G$ be the (dihedral of order $8$) subgroup of $\operatorname{GL}(2,\mathbb{R})$ generated by $g:=\begin{pmatrix}-1&0\\0&1\end{pmatrix}$ and $h:=\begin{pmatrix}0&1\\-1&0\end{pmatrix}$.

Then every non-zero element of $\mathbb{R}^2$ generates $\mathbb{R}^2$ as an $\mathbb{R}G$-module. If $v=\begin{pmatrix}1\\1\end{pmatrix}$, then $\{v,gv\}$ is a basis, but if $w=\begin{pmatrix}0\\1\end{pmatrix}$ then $gw=w$, so $\{w,gw\}$ is not a basis.

In this example $\{w,hw\}$ is a basis for every non-zero vector $w$, so you can choose the group elements so that your question has a positive answer. But if you take $G$ to be the group of rotational symmetries of a cube acting on $\mathbb{R}^3$ in the obvious way, then again every non-zero vector is a generator, but for any three distinct elements $g_1,g_2,g_3$, there is some non-zero $u\in\mathbb{R}^3$ fixed by $g_2g_1^{-1}$ and so for $w=g_1^{-1}u$, we have $g_1w=g_2w$ and so $\{g_1w,g_2w,g_3w\}$ is not a basis.

Answer 2

I think your trial will be not success. Because of the point of view "decomposition to the irreducible representations";

$K[G] \cong \oplus_{i=1}^h V_i$ as $K[G]$-module for some sub $K[G]$-modules $\{V_i\}_i$. (Giving $K[G]$-module and giving representation of $G$ are equivalent. And it is possible to take $\{V_i\}$ irreducible $K[G]$-modules, so it gives the decomposition to irreducible representation).

Of course $K[G]$ is a cyclic $K[G]$-algebra, but if you take element $w \in V_i$ for some $i$, $K[G]w \subset V_i$.