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Been studying Group Rings and their applications to the point that I can represent an element $a \in RG$ as a $n\text{x}n$ matrix where $n=\vert G \vert$. Besides being succesful, I've found others descriptions in literature regarding the matrix representation on $RG$ and I'd like to expose and compare my approach with a paper's description on the same topic.

To obtain the matrix representing $M_a$ first we must define the product of two elements $a\cdot b \in RG$. Remember that $a=(a_1,\cdots,a_n)$ and $\vert RG \vert = \vert R\vert ^{\vert G \vert}$:

$$a\cdot b = \sum_{i=1}^{n} \sum_{j=1}^{n}(a_i\cdot b_j)g_i\cdot g_j$$

By inspection we find that the summation on every coordinate point yields the matrix $M_a$ that affords the obtention of the product $a\cdot b \in RG$ as $M_a \cdot b \in GL(n, F_p)$

$$a\cdot b = M_a \cdot b = \sum_{i=1}^{n} \sum_{j=1}^{n}(a_{i\cdot j^{-1}}\cdot b_j)g_i$$

where $g_i$ represents the position $i$ on the column vector resulting from the product. Finally, I present the desired matrix matrix $M_a$ obtained by the morphism $RG \to GL(n,F)$

$$M_a = \begin{pmatrix} a_{g_1\cdot g_1^{-1}} \cdots a_{g_1\cdot g_n^{-1}} \\ a_{g_2\cdot g_1^{-1}} \cdots a_{g_2\cdot g_n^{-1}} \\ \cdots \\ a_{g_n\cdot g_1^{-1}} \cdots a_{g_n\cdot g_n^{-1}} \end{pmatrix}$$

For example, let $a=(1,0,3,0,1,2,3) \in F_5S_3$. The ordering of the basis is $( e,(23),(12),(123),(132),(13))$. The matrix $M_a$ would be $$M_a = \left( \begin{array}{cccccc} 1 & 0 & 3 & 1 & 0 & 2 \\ 0 & 1 & 0 & 2 & 3 & 1 \\ 3 & 1 & 1 & 0 & 2 & 0 \\ 0 & 2 & 0 & 1 & 1 & 3 \\ 1 & 3 & 2 & 0 & 1 & 0 \\ 2 & 0 & 1 & 3 & 0 & 1 \\ \end{array} \right)$$

More generally:

$$M_{(a_1,a_2,a_3,a_4,a_5,a_6)}=\left( \begin{array}{cccccc} a_1 & a_2 & a_3 & a_5 & a_4 & a_6 \\ a_2 & a_1 & a_4 & a_6 & a_3 & a_5 \\ a_3 & a_5 & a_1 & a_2 & a_6 & a_4 \\ a_4 & a_6 & a_2 & a_1 & a_5 & a_3 \\ a_5 & a_3 & a_6 & a_4 & a_1 & a_2 \\ a_6 & a_4 & a_5 & a_3 & a_2 & a_1 \\ \end{array} \right)$$


Conclusion

In Literature they call this matrix as $M(RG,a)$ but it is not entirely the same as mine. Mine is the transpose of that one plus the subindices are in reverse order (as if elements in $G$ would commute), check https://www.researchgate.net/publication/228928727_Group_rings_and_rings_of_matrices Page 5 Section 2.1

I've read that two representations are isomorphic if both matrices are similar by a permutation matrix $P$ such that $M_a = PM_a'P^{-1}$. Haven't checked this fact yet but it could result that my case and the paper's are related by this description.

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    $\begingroup$ Have you tried writing out some examples for small groups? It’s hard to understand your notation. $\endgroup$ – Joppy Sep 2 at 13:18
  • $\begingroup$ Yes I did. For example $Z_5S_3$. The fact is that when you multiply $M_a \cdot M_b$ you obtain the product $a.b$ in the first column of $M_{ab}$. That's why I prefer doing $M_a \cdot b$ as the result is a $n\text{x}1$ vector. If this matrix $M_a$ has full rank then it's analogue in $RG$ is an unit. In $M_a$ we want row $i$ to sum on the $i-th$ coordinate when multiplying by the coordinate vector $b$. I could provide examples at any point extracting the TeX from Mathematica. $\endgroup$ – kub0x Sep 2 at 13:30
  • $\begingroup$ Okay I introduced an example into my question. I can even provide the Mathematica's code I used but I think is not well suited here! $\endgroup$ – kub0x Sep 2 at 13:53
  • $\begingroup$ @Joppy: Any ideas regarding my scheme? I think the example that I introduced helps a lot in the understanding of how to obtain such a matrix representation of an element $a\in RG$ $\endgroup$ – kub0x Sep 3 at 13:40
  • $\begingroup$ It seems what you are doing is taking a group ring $RG$, fixing a group element $g \in G$, and then writing down a matrix for the operator $L_g: RG \to RG$ corresponding to left multiplication by $g$, i.e. $L_g(h) = gh$. Understanding left multiplication by an element is helpful, but people don't usually write down matrices for it, since it doesn't really give more insight than just being able to multiply group elements. What exactly are you trying to do here, and what question do you want answered? $\endgroup$ – Joppy Sep 8 at 3:45

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