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# 数学代写|同调代数代写Homological Algebra代考|MAST90068 Group Representations

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## 数学代写|同调代数代写Homological Algebra代考|Group Representations

By Lemma 1.1.14 a module over a $K$-algebra $A$ is given by an algebra homomorphism $A \longrightarrow E n d_K(M)$. In the case of a group, we want to study representations as well. Here, staying in the philosophy expressed previously, we understand a representation to be a concrete realisation of an abstract group by an action on a “real world object”, in our case a $K$-vector space where $K$ is a field.

Group Algebras and Their Modules
Since group elements have inverses, the same should be true for the representation of a group. So, a representation of a group $G$ is a homomorphism of $G$ to the group of invertible matrices.

Definition 1.2.1 Let $G$ be a group and $K$ be a field. A representation of $G$ over $K$ of degree $n \in \mathbb{N}$ is a group homomorphism $G \longrightarrow G L_n(K)$. If the kernel of this homomorphism is all of $G$, then we say that $G$ acts trivially, and if in addition $n=1$, we call the representation the trivial representation.

There are at least two obvious drawbacks to this concept. First, it would be nice if we could phrase this idea in our setting of modules over an algebra. Second, we see that we would like to use at least some of the additional information the matrices provide. One of the most obvious ones is the fact that $G L_n(K)$ is the unit group of an algebra, the ring $\operatorname{Mat}_{n \times n}(K)$ of degree $n$ square matrices. Just as we can form linear combinations of invertible matrices of the same size to get a (not necessarily invertible) square matrix, we can formally form $K$-linear combinations of group elements and get a $K$-algebra. This is the concept of a group ring.

Definition 1.2.2 Let $G$ be a group and let $R$ be a commutative ring. The group ring $R G$ as an $R$-module is the $R$-module $\oplus_{g \in G} R \cdot e_g$, where $R \cdot e_g=R$ for all $g \in G$. This becomes an $R$-algebra when one sets
$$\left(\sum_{h_1 \in G} r_{h_1} e_{h_1}\right) \cdot\left(\sum_{h_2 \in G} s_{h_2} e_{h_2}\right):=\sum_{g \in G}\left(\sum_{h \in G} r_h s_{h^{-1} g}\right) e_g .$$
Occasionally we shall also call a group ring a group algebra if we want to stress the algebra structure.

We should explicitly mention that if $G$ is infinite, the fact that $R G$ is defined as a direct sum $\coprod_{g \in G} R$ of copies of $R$ as $R$-modules implies that an element $x=$ $\sum_{g \in G} r_g e_g$ has only a finite number of non-zero coefficients $r_g$. In other words, for any $x=\sum_{g \in G} r_g e_g \in R G$ we have
$$\left|\left{g \in G \mid r_g \neq 0\right}\right|<\infty .$$
Now, the group of invertible elements $(R G)^{\times}$of $R G$ contains $G$ as a subgroup. Indeed, the mapping
\begin{aligned} &G \longrightarrow(R G)^{\times} \ &g \mapsto 1 \cdot e_g \end{aligned}
is a group monomorphism. The image of this homomorphism will again be denoted by $G$.

## 数学代写|同调代数代写Homological Algebra代考|Maschke’s Theorem

We now prove one of the most important results in the representation theory of finite groups. The result actually states that if $G$ is a finite group of order $n$ and $K$ is a field with $n \cdot K=K$, then any module of $K G$ is semisimple.

Theorem 1.2.8 (Maschke 1905) Let $G$ be a finite group and let $K$ be a field. If the order of $G$ is invertible in $K$, then $K G$ is semisimple.

Proof We shall first show that for any $K G$-module $M$ and any submodule $N$ of $M$ there is another submodule $L$ of $M$ such that $M \simeq N \oplus L$.

Once this is done we proceed by induction on the dimension of $M$. The statement is clear for one-dimensional modules. Suppose we have shown that any module of dimension up to $n$ is semisimple, and that we have given an $n+1$-dimensional module $M$. If $M$ does not have a proper non-zero submodule, then $M$ is simple. If this is not the case, then there is a proper non-zero submodule $N$ of $M$. By the argument which we are about to develop below, there is another submodule $L$ of $M$ such that $M \simeq L \oplus N$. Since the dimensions of $L$ and of $N$ are both at most $n$, these two modules are semisimple, and hence $M$ is semisimple.

Suppose that $N$ is a proper non-zero submodule of $M$. Then the quotient $\tilde{L}:=$ $M / N$ is a $K G$-module as well. Denote by $\pi: M \longrightarrow \tilde{L}$ the mapping $\pi(m)=m+N$.

We may choose a $K$-basis $\mathcal{N}$ of $N$ which we may complete by vectors $\mathcal{L}$ to a $K$ basis $\mathcal{M}=\mathcal{N} \cup \mathcal{L}$ of $M$. Then ${l+N \mid l \in \mathcal{L}}$ is a $K$-basis of $\tilde{L}$. Denote by $\tilde{\rho}$ the $K$-linear mapping $\tilde{L} \longrightarrow M$ defined by $\tilde{\rho}(l+N):=l$ for any $l \in \mathcal{L}$. By definition, $\pi \circ \tilde{\rho}=i d_{\tilde{L}}$. Moreover, $\pi$ is a $K G$-module homomorphism, but there is however no reason why $\tilde{\rho}$ should be a $K G$-module homomorphism.
Define
\begin{aligned} \rho: \tilde{L} & \longrightarrow M \ l+N & \mapsto \frac{1}{|G|} \sum_{g \in G} e_g^{-1} \tilde{\rho}\left(e_g \cdot(l+N)\right) \end{aligned}
Remark 1.2.9 Observe that we assumed that $|G|$ is invertible in $K$, and so the term $\frac{1}{|G|} \in K$ makes sense. This is the only occasion when we shall use this hypothesis.
Since $\tilde{\rho}$ is a well-defined homomorphism of $K$-vector spaces, this is true for $\rho$ as well. Moreover, for any $l+N \in \tilde{L}$ one has
\begin{aligned} \pi \circ \rho(l+N) &=\pi\left(\frac{1}{|G|} \sum_{g \in G} e_g^{-1} \tilde{\rho}\left(e_g \cdot(l+N)\right)\right) \ &=\frac{1}{|G|} \sum_{g \in G} e_g^{-1} \pi\left(\tilde{\rho}\left(e_g \cdot(l+N)\right)\right) \ &=\frac{1}{|G|} \sum_{g \in G} e_g^{-1} e_g \cdot(l+N) \ &=l+N \end{aligned}
and so, $\pi \circ \rho=i d_{\tilde{L}}$.

$$\left(\sum_{h_1 \in G} r_{h_1} e_{h_1}\right) \cdot\left(\sum_{h_2 \in G} s_{h_2} e_{h_2}\right):=\sum_{g \in G}\left(\sum_{h \in G} r_h s_{h^{-1} g}\right) e_g .$$有时，如果我们想强调代数结构，我们也会称一个组环为组代数 我们应该明确地提到，如果$G$是无限的，$R G$被定义为$R$副本的直接和$\coprod_{g \in G} R$为$R$ -modules，这意味着元素x=$$\sum_{g \in G} r_g e_g只有有限的非零系数r_g。换句话说，对于任何x=\sum_{g \in G} r_g e_g \in R G，我们有$$ \left|\left{g \in G \mid r_g \neq 0\right}\right|<\infty . $$。现在，R G的可逆元素组(R G)^{\times}包含G作为一个子组。实际上，映射$$ \begin{aligned} &G \longrightarrow(R G)^{\times} \ &g \mapsto 1 \cdot e_g \end{aligned} $$是组单态。这个同态的像将再次用G . 表示 ## 数学代写|同调代数代写同构代数代考|马士基定理 我们现在证明有限群表示理论中一个最重要的结果。结果实际上表明，如果G是一个有限的n顺序组，K是一个带n \cdot K=K的字段，那么K G的任何模块都是半简单的 定理1.2.8 (Maschke 1905)设G为有限群，设K为域。如果G的顺序在K中是可逆的，那么K G是半简单的 我们首先要证明，对于任何K G -模块M和M的任何子模块N，都有一个M的子模块L，这样M \simeq N \oplus L . 完成后，我们继续对M维度进行归纳。对于一维模块，这条语句很清楚。假设我们已经展示了在n之前的任何维度的模块都是半简单的，并且我们给出了n+1 -dimensional模块M。如果M没有合适的非零子模块，那么M很简单。如果不是这样，那么M的N有一个适当的非零子模块。根据我们将要在下面开发的论证，M的另一个子模块L，例如M \simeq L \oplus N。由于L和N的维数都不超过n，所以这两个模块是半简的，因此M是半简的 假设N是M的一个适当的非零子模块。然后，商\tilde{L}:=$$M / N也是一个$K G$ -模块。通过$\pi: M \longrightarrow \tilde{L}$表示映射$\pi(m)=m+N$ .

\begin{aligned} \rho: \tilde{L} & \longrightarrow M \ l+N & \mapsto \frac{1}{|G|} \sum_{g \in G} e_g^{-1} \tilde{\rho}\left(e_g \cdot(l+N)\right) \end{aligned}

\begin{aligned} \pi \circ \rho(l+N) &=\pi\left(\frac{1}{|G|} \sum_{g \in G} e_g^{-1} \tilde{\rho}\left(e_g \cdot(l+N)\right)\right) \ &=\frac{1}{|G|} \sum_{g \in G} e_g^{-1} \pi\left(\tilde{\rho}\left(e_g \cdot(l+N)\right)\right) \ &=\frac{1}{|G|} \sum_{g \in G} e_g^{-1} e_g \cdot(l+N) \ &=l+N \end{aligned}
，因此$\pi \circ \rho=i d_{\tilde{L}}$ .

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