{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

mm1 - Math Meth HW1 Solutions Szilard Farkas a 1(i No(a b c...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
Math Meth HW1 Solutions Szilard Farkas a) 1. (i) No. ( a · b ) · c = a / ( bc ), a · ( b · c ) = a / ( b / c ) = ac / b , so the product is not associative. (ii) a = (24)(153) and b = (12)(354). We will see that if two permutations are of the same type, which means that the number of cycles of a given length is the same in both permutations, then they are related to each other by conjugation. Let c be for example the permutation (2 , 4 , 1 , 5 , 3) 7→ (1 , 2 , 3 , 5 , 4), that is, c = (1342). Then b = cac - 1 . (iii) Reflexive since a = eae - 1 , where e is the identity element; transitive because b = gag - 1 and c = hbh - 1 implies that c = hgag - 1 h - 1 = ( hg ) a ( hg ) - 1 ; and symmetric since b = gag - 1 means that a = g - 1 bg = ( g - 1 ) b ( g - 1 ) - 1 . (iv) H is a subgroup if and only if H - 1 = H and HH = H . ( gHg - 1 ) - 1 = gH - 1 g - 1 = gHg - 1 and gHg - 1 gHg - 1 = gHHg - 1 = gHg - 1 , so gHg - 1 is also a subgroup. (v) An isomorphism preserves the order of any element. Z 4 has a fourth order element, whereas in Z 2 × Z 2 the order of any element is at most two. 2. I 2 = J 2 = K 2 = - 1 means that I - 1 = - I etc. Inverting both sides in IJ = K , we get J - 1 I - 1 = K - 1 , or JI = - K , so the product is not commutative. The group, which is usually denoted by Q , has eight elements { 1 , I , J , K , - I , - J , - K } . The quaternion group Q and the dihedral group D 4 are the only eight-element nonabelian groups. Q is the smallest noncommutative group which is not a proper semidirect product. (In order for a group to be a proper semidirect product, it has to have two nontrivial subgroups whose only common element is the identity. But - 1 is the element of any nontrivial subgroup of Q .) 3. (i)-(ii) We have to show that two permutations are of the same type b) if and only if they are conjugate to each other. Consider the permutations of a set H . Let a and c be two arbitrary permutations, and define the permutation b by b ( c ( i )) B c ( a ( i )), where a ( i ) etc is the element of H that the permutation a assigns to i H . Since c is a bijection on H , b is indeed a well-defined bijection on H , so it is a permutation. The type of this permutation is the same as the type of a , since c just relabels the elements of H . The definition bc ( i ) = ca ( i ) of b can also be written as b = cac - 1 . So far we have shown that conjugation does not change the type. Now let
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Page1 / 3

mm1 - Math Meth HW1 Solutions Szilard Farkas a 1(i No(a b c...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon bookmark
Ask a homework question - tutors are online