By I.M. Gel'Fand; Translator A. Shenitzer
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With respect to this basis of kG the matrix representation of the group element g has 1 in the intersection of the i-th row and the j-th column, and has zeroes in all other positions. Note that each nonidentity element of G induces a nonidentity permutation on the basis of kG. So the left regular representation is always faithful. Analogously one can deﬁne the right regular representation of kG. 3. Consider the symmetric group S3 which has the following matrix twodimensional representation based on the correspondence with planar symmetry operations of an equilateral triangle: ϕ(e) = ϕ(c) = 1 0 , 0 1 1 ϕ(a) = 2 1 √ − 3 −1 0 , 0 1 1 ϕ(d) = 2 −1 √ 3 √ − 3 , −1 √ − 3 , −1 1 ϕ(b) = 2 1 ϕ(f ) = 2 √ 3 1 √ 3 −1, −1 √ − 3 √ 3 −1.
1. This follows from the observation that since H ⊆ G, by deﬁnition of commutator subgroups, [H, H] ⊆ [G, G], that is, H (1) ⊆ G(1) . Then, by induction, H (i) ⊆ G(i) for all i ≥ 0. In particular, if G(n) = 1 for some n, then also H (n) = 1. ALGEBRAS, RINGS AND MODULES 24 2. Note that, by deﬁnition of commutators, ϕ([x, y]) = [ϕ(x), ϕ(y)], so, by induction, ϕ(G(i) ) ⊆ K (i) . Since ϕ is surjective, every commutator in K is the image of a commutator in G. Hence again, by induction, we obtain equality for all i.
Consider n as an element of the ﬁeld k. It is not zero in k by hypothesis. Deﬁne 1 π= g −1 π0 g. n g∈G Since π is a scalar multiple of a sum of linear transformations from M to X, it is also a linear transformation from M to X. If x ∈ X, then π(x) = 1 n 1 (x + ... , π is also a vector space projection of M onto X. We now show that π is a kG-module homomorphism. For any h ∈ G we have π(hm) = = 1 n = 1 n g −1 π0 g(hm) = g∈G h(h−1 g −1 ) · π0 ((gh)m) = g∈G 1 n h(r−1 π0 r(m)) = h( r=gh,g∈G 1 n 1 n 1 n g −1 · π0 (g · hm) = g∈G h(r−1 · π0 (rm)) = r=gh,g∈G r−1 π0 r(m)) = hπ(m), r∈G because as g runs over all elements of G, so does r = gh.