Let $S\_n$ be the finite group given as $n \times n$ permutation matrices.


Define for a given field $K$ the algebra $B\_n$ as the subalgebra of $M\_n(K)$ generated by all permutation matrices of $S\_n$. (more generally we can do this for any subgroup of $S\_n$ to associate to a finite group such a subalgebra. Often we will just obtain the group algebra, but not always.)



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> Question 1: What is the subalgebra $B\_n$? Has it been studied before and does it have a name? Can we determine quiver and relations of the basic algebras of the blocks of $B\_n$ (over a splitting field of $K$)?
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Probably I should know this but I only feel really familiar with quiver algebras.


Here some things that I found out:


-$B\_n$ has vector space dimension $(n-1)^2+1$ and is semisimple over a field of characteristic 0. The center of $B\_n$ is two dimensional. This implies that the algebra is $M\_{n-1}(K) \times K$ since we know two simples that are split over $K$, see the answer of Sasha.


-$B\_2$ over a field of characteristic two is isomorphic to $K[x]/(x^2)$ (so it is a non-semisimple Frobenius algebra in this case).


-$B\_3$ over a field of characteristic 3 is isomorphic to the the Nakayama with Kupisch series [2,3] (in particular it can have finite non-zero global dimension and thus is not a Frobenius algebra in general).



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> Question 2: Is it true that any field $K$ is a splitting field for this algebra? What is the number of simple modules?
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Note that if any $K$ is really a splitting field, then the number of simples of the algebra $B$ is given by the vector space dimension of
$B/(rad(B)+[B,B])$, which seems to be 2-dimensional(?).


It would be interesting to see what the algebra for $S\_p$ is for the primes $p=5$ and $p=7$ (up to Morita equivalence), but the algebra is not baisc anymore.



A finite dimensional algebra associated to the symmetric group
Let $K$ be a field. The literal answer has already been given by several people but let me try and get at the algebraic structure and provide a quiver with relations (***see the addition***). Let $J$ be the $n\times n$ all ones matrix. Then $J$ centralizes $S\_n$ and the centralizer of $J$ consists of all matrices whose rows and columns all sum to some fixed field element $k\in K$. This has dimension $(n-1)^2+1$ and several people have proved the subalgebra spanned by $S\_n$ has this dimension so $B\_n$ is the centralizer of $J$.


If the characteristic of $K$ is zero or does not divide $n$, then we can change the basis to consists of $(1,\cdots,1)$ and $e\_1-e\_i$ with $2\leq i\leq n$ with $e\_i$ standard basis vectors. Then $J$ becomes the elementary matrix $E\_{11}$ in this basis and its centralizer is clearly $K\times M\_{n-1}(K)$, which can also be seen by representation theory. This is a split semisimple algebra with two simple modules and there is nothing much to say.


If $K$ has characteristic $p$ dividing $n$, then $J^2=0$ and it is easy to see the Jordan canonical form of $J$ is a $2\times 2$ block and $n-2$ blocks that are $1\times 1$. In fact the basis $e\_1,(1,\ldots, 1), e\_i-e\_1$ for $2\leq i\leq n-1$ gives the Jordan form with $e\_1,(1,\ldots,1)$ giving the $2\times 2$-block. Thus we can identify $B\_n$ with the centralizer of $$J'=N\_2\oplus 0\_{n-2}$$ where $N\_2= \begin{bmatrix} 0 & 0\\ 1&0\end{bmatrix}$. A more ring theoretic view is the following.


Let $R=K[x]/(x^2)$; its a self-injective local $K$-algebra and $K$ is the unique simple where $x$ acts by $0$. Then the centralizer of $J'$ is $\mathrm{End}\_R(R\times K^{n-2})$. I claim the the semisimple quotient here is $K\times M\_{n-2}(K)$ and hence there are two simple modules and the algebra $B\_n$ is split over $K$. This is not to difficult to check directly since $R$ is a projective indecomposable $R$-module with simple quotient $K$ and simple socle $K$. Also note that the radical of $R$ is its socle. So the radical of $\mathrm{End}\_R(R\times K^{n-2})$ is the direct sum of the $1$-dimensional radical of $R$ (viewed as endomorphisms of the first summand $R$), the $n-2$-dimensional space of $R$-module homomorphisms $R\to K^{n-2}$ and the $n-2$-dimensional space of $R$-module homomorphisms $K^{n-2}\to R$ (which all land in the one-dimensional socle=radical). From this description it is easy to see that the radical squares to $0$ which means that as soon as you know the quiver, you know the basic algebra. The semismple quotient is a copy of $K$ coming from $R/\mathrm{rad}(R)$ and the endomorphisms of the semisimple module $K^{n-2}$, which is $M\_{n-2}(K)$. Note that any composition $K^{n-2}\to R\to K^{n-2}$ is zero since the socle of $R$ is its radical. But the compositions $R\to K^{n-2}\to R$ give the elements of $\mathrm{rad}(R)$ and so the radical cubed is zero.


If we view $B\_n$ as the centralizer of $J$, then the radical (which has dimension $2n-3$) is spanned by those matrices where each row is a constant vector and each column sums to $0$ and those matrices where each column is a constant vector and each row sums to $0$. Hopefully a representation theorist can compute the basic algebra from this description.


Let me add that when $n=2=p$, this description recovers that $B\_2\cong K[x]/(x^2)$ and when $n=3=p$, it recovers that $B\_3$ is a basic algebra with semisimple quotient $K\times K$. I would guess in general that the basic algebra is $\mathrm{End}\_R(R\times K)$ but I am not 100% sure.


**Added and edited based on corrections by the OP and @JeremyRickard.** If I am not mistaken, the quiver of $B\_n$ when $p\mid n$ and $n>2$ has two vertices $v,w$. There is one edge from $v$ to $w$, one edge from $w$ to $v$.and one loop at $v$. Since its a radical-squared zero split algebra, the quiver presentation for the basic algebra then says that all paths of length $2$ are $0$. But an expert should check this. There is a relation saying the path of length two from $v$ to $v$ is zero. Indeed view $B\_n$ as $A=\mathrm{End}\_R(R\times K^{n-2})$. A complete set of orthogonal primitive idempotents are the projection $e$ to $R$ and the $n-2$ projections to $K$, but the latter $n-2$ all give isomorphic projective indecomposables so we just need one of them, say the projection $f$ to the first factor, for the quiver. Then since the radical squared is zero, to compute the quiver we have that $f\mathrm{rad}(A)e=fAe$ is one dimensional and so is $e\mathrm{rad}(A)f=eAf$ since there is a one-dimensional space of homomorphisms $R$ to $K$ and $K$ to $R$. Also $eAe\cong R$ but $e\mathrm{rad}(R$ is the radical squared as noted above and so $e(\mathrm{rad}(A)/\mathrm{rad}^2(A))e\cong 0$. On the other hand, $fAf\cong K$ and so $f\mathrm{rad}(A)f=0$. This gives my description of the quiver. That the compositions $K\to R\to K$ are zero but $R\to K\to R$ are not all zero gives the relation.


