advertisement

Review of Two Quasi 2-Groups There are several places where you can improve the techniques of proof. The argument for the existence of 15 elements whose order is a power of 2 in Z5 oφ Z4 is not correct. Consider the element x2 y. You claim that the order of this element is a power of 2. Since 4 xyx−1 = y −1 , we have that x2 y = yx2 and (x2 y) = x8 y 4 = y −1 6= e. Hence, |x2 y| does not divide 4. Therefore, x2 y is not a power of 2, since |Z5 oφ Z4 | = 20. You do not need to count elements to verify your claim that P (G) = G (see below). You will find specific comments about the paper below. NOTE: A positive number is the line number from the top. A negative number is the line number counted from the bottom. Comments Page Line Number Comment 1 -7 The number on the definition should be 2.1 (not 1.1). Usually a definition has one statement in it. This seems to be a cross between a definition and a lemma. In the paper you just focus on proving criterion 2. You have not used the other ‘definitions’ here. 4-5 Do you mean φ(x)(y) = y −1 where x is a generator of Z4 ? -6 There are 5 Sylow 2-subgroups, each with 4 elements. Are you claiming that the intersection of any two Sylow 2-subgroups is trivial? How do you know this? This is not true. You can fix your argument (see below). 1-2 I think this may be the typesetting device you are using. The spacing makes it hard to read. For example: x2 y 2 instead of x2 y 2 . 4 “In particular ... power of 2” You are basing this off of page 2 line -6. 2 3 1 Comments Page Line Number Comment 9 To show that P (G) = G you can use the fact that the orders of x and xy are powers of 2. You do not need to use Sylow here. 2 [(x1 , y1 ), z1 ] is much easier to read than [(x1 , y1 ), z1 ]. 4 There are at least 10 elements in the subgroup generated by the elements of order 2. Now use a theorem of Lagrange. 4 5