Download Algebra through Practice: A Collection of Problems in by Byth T.S., Robertson E.F. PDF

xy is injective (by the cancellation hw) we deduce that IxXI = IXI and hence, since X is finite, that xX = X. Consequently, x = xe for some e E X. The cancellation hw gives e = 1, and so we have that 1 E X. We now observe from 1 E xX that 1 = xy for some y CC X, which gives X-I = Y E X. It now follows from the fact that XX <;;; X that X is a subgroup of G.

13 ~ Qg with A ~ Z(H) n H' then H Let G be the group with presentation Prove that (xy)4 = 1 and y8 = 1. 25 ~ Qg. 14 Groups Let G be the group with presentation G = (a, b I a7 = (a'2W = (a 3 b? = (ab 5 )2 = 1). Prove that G may also be presented as (x,y I x 2 = y3 = (xy)7 = ((y- 1xyx)4 y -l x )'2 = 1). 15 Let G be the group with presentation ( a, b, c, d, e I ab = c, be = d, cd == e, de = a, ea = b). By eliminating c, d, e show that (1) a == babab'2 ab, (2) b == ab'2 aba. Replacing the final a in (2) by the expression given by (1), show that b5 = a-'2.

Then we have g E HX a for some a E A and so g = hX a for some (unique) h E H. But h E KY{3 for some f3 E B and so we have that g = kY{3x a for some k E K. Thus we see that every element of G belongs to a coset K Y{3x a for some f3 E B and some a E A. The result now follows from the fact that if KY{3x a = KY{3'x a , then, since the left hand side is contained in the coset H X a and the right hand side is contained in the coset H X a ', we have necessarily X a = X a ', which gives KY{3 = Kyf3' and hence Y{3 = Y{3" Now observe that (HnK)x = HxnKx for all subgroups Hand K of G.

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