Algebra [Lecture notes] - download pdf or read online

By I. M. Isaacs

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Proof. Let N Sn . Assume that N = 1 and N = Sn . We show that N = An . By the diamond lemma, N ∩ An An , so either N ∩ An = An or N ∩ An = 1 as An is simple. If N ∩ An = An , then An ⊆ N . As |Sn : An | = 2, we must have either An = N or N = Sn . As N = Sn by assumption, we have An = N . So assume that N ∩ An = 1. , we can only have that |N | = 2. So N = {1, x} for some x ∈ Sn . WLOG, assume that x maps 1 to 2. Then xg ⊆ N for all g ∈ Sn . Yet x(2,3) maps 1 to 3, which happens neither in x nor for 1.

Proof. We wish to show that Jl (R) ⊆ J(R). Jl (R) is an ideal and therefore Jl (R) is a two sided ideal. It suffices to show that elements of Jl (R) are rqr. Let u ∈ Jl (R). We know that U is lqr, so there exists an element x ∈ R such that x(1 − u) = 1. Let y = 1 − x. Then x = 1 − y, and we have that x(1 − u) = (1 − y)(1 − u) = 1. So −u − y + uy = 0, which means that y = yu − u. Now u ∈ Jl (R) and yu ∈ Jl (R) as Jl (R) is a left ideal. So yu ∈ Jl (R) and y ∈ Jl (R) so y is lqr. So there exists z ∈ R such that z(1 − y) = 1.

Proof. We must show that if x1 x2 . . xr = y1 y2 . . yr , then xi = yi for all i. So assume that it is false, and choose k minimal such that xk = yk . Then x1 x2 . . xk = y1 y2 . . yk , so moving things around yields: −1 −1 xk yk−1 = x−1 k−1 xk−2 . . x1 y1 y2 . . yk−1 where the result, xk yk−1 lies in Nk as xk , yk ∈ Nk . However, due to the right hand side of the equation above, the result is also in N1 n2 . . Nk−1 . By our hypothesis, we see that we must then have that xk yk−1 = 1, and hence xk = yk .

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Algebra [Lecture notes] by I. M. Isaacs


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