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# Algebra: A computational introduction by John Scherk

By John Scherk

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Extra resources for Algebra: A computational introduction

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PERMUTATION GROUPS In[14]:= Order[G] Out[14]= 10 We can verify by computation that (1 2) and (1 2 3 4 5 6) generate S6 : In[15]:= Order[ Group[ P[1,2], P[1,2,3,4,5,6] ] ] Out[15]= 720 Since 6! = 720, they do generate S6 . Let's look for generators of A4 , which is also not cyclic. We know that the even permutations of degree 4 are the 3-cycles, the products of two disjoint transpositions and the identity. /2 (see exercise 11). Therefore the two 3-cycles do generate A4 . How about a 3-cycle and a product of two disjoint transpositions?

6 Introduction to Software The main purpose of this section is to give you a chance to practice using Mathematica. It has several functions which are relevant to this section. For making computations in the integers modulo n there is a built-in function Mod. Thus In[1]:= Mod[25+87,13] Out[1]= 8 and In[2]:= Mod[2^12,7] Out[2]= 1 A more efficient way of computing powers mod n is to use the function PowerMod: In[3]:= PowerMod[2,12,7] Out[3]= 1 For any real number a the function N[a, m] will compute the first m digits of a.

Sn itself is a permutation group, called the full permutation group (of degree n) or symmetric group (of degree n). Another example is V ′ = {(1 2), (3 4), (1 2)(3 4), (1)} ⊂ S4 . We see that (1 2)(3 4) · (1 2) = (1 2) · (1 2)(3 4) = (3 4) (1 2)(3 4) · (3 4) = (3 4) · (1 2)(3 4) = (1 2) (3 4) · (1 2) = (1 2) · (3 4) = (1 2)(3 4) ( )2 (1 2)2 = (3 4)2 = (1 2)(3 4) = (1) ( )−1 = (1 2)(3 4) , (1 2)−1 = (1 2) , (3 4)−1 = (3 4) . 2. CYCLIC GROUPS So V ′ is a permutation group. With appropriate software it is easy to check whether a set of permutations is a permutation group.