Algebra [Lecture notes] by I. M. Isaacs

By I. M. Isaacs

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M Nr = G. Look at two consecutive terms of S; say M ∩ Ni ⊆ M ∩i+1 . Let U = Ni (M ∩ Ni+1 ). This is a very diamond lemma type argument, so I will again leave a blank space for a picture. 38 Since M G, we have that M ∩ Ni+1 Ni+1 . Thus we have by the diamond lemma that U Ni+1 . Likewise, M ∩ Ni Ni+1 as it is the intersection of two normal subgroups of G. By the way the Nj are nested, we then find that M ∩ Ni M ∩ Ni+1 . Since Ni+1 /Ni is X-simple, there are only two possibilities for U ; either U = Ni+1 or U = Ni .

Nr ) = n1 n2 . . nr , which is in G. Note that θ is surjective as G is the internal direct product of the Ni . Also, θ is injective by the uniqueness assumption of the internal direct product. So we must show that θ is in fact a homomorphism. So: θ((x1 , x2 , . . , xr ) · (y1 , y2 , . . , yr )) = θ(x1 y1 , x2 y2 , . . , xr yr ) = x1 y1 x2 y2 . . xr yr Yet x1 y1 x2 y2 . . xr yr = x1 x2 . . xr y1 y2 . . yr as each yi commutes with all xj such that i = j. This is θ(x1 , x2 , . . , xr )θ(y1 , y2 , .

Tk = s1 s2 . . sl where the ti and sj are transpositions, such that k is odd and l is even. We then get that t1 t2 . . tm = 1, where m = k+l, and the ti are all transpositions. Note that this follows as transpositions are their own inverses. We do this with the smallest possible m. We note a clever trick; that if t and s are transpositions, we have that st = tst . This follows as tst = tt−1 st = st. As conjugation preserves cycle structures, this allows us to rearrange the order of multiplication within a product of transpositions without altering the result OR the number of transpositions.

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