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One knows that every finite group is isomorphic to a subgroup of $\operatorname{GL}(n)$ for some $n$ large enough.

Can every finite ring be represented by a ring of matrices, i.e., is every ring isomorphic to a matrix ring?

As an example, let $G = S_3$, the symmetric group on $3$ points. Let $g_1 = (1,2,3)$ and $g_2 = (1,2)$ the generator of $G$. Then the matrix group with gnerators $$ M_1 = \begin{bmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix}$$ $$ M_2 = \begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix}$$ is isomorphic to $G$.

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  • $\begingroup$ I think this is trivially true in the sense that every ring $R$ is isomorphic to the ring of $1\times 1$ matrices over $R$. $\endgroup$ – Daniel Franke Jul 28 '13 at 14:15
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Not every group is isomorphic to a matrix group. See for example this MO post for counterexamples. It is true, however, that every finite group is isomorphic to a matrix group over arbitrary field.

If you meant to ask if every finite ring is isomorphic to a matrix ring over some field, then this is not true: a matrix ring over $k$ is a $k$-algebra, in particular its characteristic is equal to that of $k$, so any ring with nonprime characteristic (like ${\bf Z}/6{\bf Z}$) can't be represented as a matrix ring over a field.

On the other hand, every ring is trivially isomorphic to the ring of $1\times 1$ matrices over itself, or more generally $n\times n$ scalar matrices over itself.

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At least the following is true: If $R$ is some $k$-algebra, then it embeds into $\mathrm{End}_k(U(R))$, where $U(R)$ is the $k$-module underlying $R$. The embedding is just $r \mapsto (m \mapsto rm)$, and can also be seen as a special case of the Yoneda embedding. When $k$ is a field, this module is free, hence the endomorphism ring is a ring of (column-finite) matrices. In particular, every finite-dimensional algebra over a field embeds into some finite-dimensional matrix algebra.

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