By A. I. Kostrikin, I. R. Shafarevich

ISBN-10: 0387170065

ISBN-13: 9780387170060

From the stories: "... this can be one of many few mathematical books, the reviewer has learn from disguise to hide ...The major benefit is that just about on each web page you can find a few unforeseen insights... " Zentralblatt für Mathematik "... There are few proofs in complete, yet there's an exciting blend of sureness of foot and lightness of contact within the exposition... which transports the reader without problems around the entire spectrum of algebra...Shafarevich's ebook - which reads as conveniently as a longer essay - breathes lifestyles into the skeleton and should be of curiosity to many periods of readers; definitely starting postgraduate scholars might achieve a most dear viewpoint from it but... either the adventurous undergraduate and the proven specialist mathematician will discover a lot to enjoy..."

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**Additional info for Algebra I. Basic notions of algebra**

**Example text**

If ϕ1 and ϕ2 are non-isomorphic, then χ1 , χ2 = 1 χ1 (g)χ2 (g) = 0. |G| g∈G If ϕ1 and ϕ2 are isomorphic, then χ1 , χ2 = χ1 , χ1 = 1. In other words, the set of irreducible representations of G forms an orthonormal set in the space of class functions. Proof. Let dim V = m and dim W = n, and define eij to be the n × m matrix with a 1 in the (i, j)th position and 0’s everywhere else. First, suppose that V and W are non-isomorphic. 2, the matrix fij = |G| g∈G ϕ1 (g)eij ϕ2 (g ) corresponds to an intertwiner V → W .

Now, suppose that V is a module over a group algebra k[G]. Then, V is a vector space over k ⊂ k[G] (since modules over fields are vector spaces), and each element g of G acts on the vector space V by a k-linear map. Denote this k-linear map by ϕ(g) : V → V . Then, for V to be a module, we must have ϕ(g)ϕ(h) = ϕ(gh) for all g and h in G. e. the group of invertible linear maps from V to itself. Notice that the module structure of V is completely characterized by ϕ and makes no reference to group algebras.

Example. Let G = C∞ , and let V be a 2-dimensional vector space over a field k. Pick a generator g of G and a basis {v1 , v2 } of V . Then, to define a representation of G, we just have to define a homomorphism G → GL(V ). It suffices to pick where we send g in this map; so, we define a representation by sending g → ( 10 11 ). Then, we have gv1 = v1 , so that kv1 ⊆ V is a k[G]-submodule of V which is fixed by the action of G, hence it is a subrepresentation. However, its complement in the vector space is kv2 , which is not fixed by G.

### Algebra I. Basic notions of algebra by A. I. Kostrikin, I. R. Shafarevich

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