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You could download GAP or Sage or acquire a copy of Magma. In GAP there's a function called $$\text{"AllSmallGroups(N)"}$$ which gives you a list of all the groups of order $N$ up to isomorphism. For each one you can then use "StructureDescription(G)" for GAP to give you a short description of the group in terms of common names of groups, (semi)direct ...

2

We have \begin{align*}\phi:A\times\operatorname{Mat}_n(K)&\to\operatorname{Mat}_n(A)\\ (a,M)&\mapsto aM\end{align*} a well-defined bilinear homomorphism. By definition of the tensor product this means there is a unique linear map $\varphi$, such that commutes. This map is given by ...

2

Note that $\dim A = n^2 \dim D$, so $A$ is finite dimensional if and only if $D$ is finite dimensional. It shouldn't be hard for you to show that if $A\simeq M_n(D)$ then the center of $A$ (which is $k$) is isomorphic to the center of $M_n(D)$. But it is a rather easy exercise to show that the center or $M_n(R)$ for any ring $R$ is isomorphic to the center ...

2

The expression $S \otimes_A Hom_A(S,S)$ doesn't make sense, because if $S$ is an irreducible right $A$-module, there is usually not any natural left $A$-module structure on $Hom_A(S,S)$. For instance, if $A=M_n(\mathbb{C})$ for some $n>1$ and $S=\mathbb{C}^n$, then $Hom_A(S,S)=\mathbb{C}$ cannot be made into an $A$-module (in any way compatible with the ...

2

You need to set up the correct hom–tensor adjunction. Recall that, for rings $R, S, T$, an $(R, S)$-bimodule $M$, $(S, T)$-bimodule $N$, and $(R, T)$-bimodule $P$, we have: $$\mathrm{Hom}_{(R, T)} (M \otimes_S N, P) \cong \mathrm{Hom}_{(R, S)} (M, \mathrm{Hom}_{(\mathbb{Z}, T)} (N, P))$$ $$\mathrm{Hom}_{(R, T)} (M \otimes_S N, P) \cong \mathrm{Hom}_{(S, T)} ... 2 Bad news It is impossible to prove it with the strategy you chose: in fact R is a perfect ring, and it is known that perfect rings satisfy the DCC on principal ideals. Your descending chain of ideals will therefore have to be more sophisticated. Triangular rings The ideal structure of this type of construction is explained in both of the following ... 1 To answer your first question, a very accessible example of a non-commutative multiplication is concatenation of words. Say you are given an alphabet A (that is, a set of letters). For the sake of making explicit examples, let's pick$$A = \{a,b,\ldots,x,y,z,\_\}, where we will use $\_$ to indicate a space. We define the free group over $A$, denoted by ...

1

How can we find the simple modules of this algebra Denoting this algebra as $R$, the simple $R$ modules are exactly the simple $R/J(R)$ modules, where $J(R)$ is the Jacobson radical you have found. This is isomorphic to the product ring $k^3$. Now you just need to conclude what the isoclasses of simple modules over that ring are. The same strategy was ...

1

Be careful. It's cleanest to describe the tensor-hom adjunction with three different rings instead of one, to make it as hard as possible to accidentally write down the wrong thing, so let $A, B, C$ be three different rings, let $_A M_B$ be an $(A, B)$-bimodule, let $_B N_C$ be a $(B, C)$-bimodule, and let $_A K_C$ be an $(A, C)$-bimodule. Then ...

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