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| + | =[[HW2_MA453Fall2008walther|HW2]], Chapter 5 problem 43, Discussion, [[MA453]], [[user:walther|Prof. Walther]]= | ||
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| + | ==Problem Statement:== | ||
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Show that: | Show that: | ||
i. <math>A_5</math> has 24 elements of order 5, | i. <math>A_5</math> has 24 elements of order 5, | ||
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iii. and 15 elements of order 2. | iii. and 15 elements of order 2. | ||
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| + | ==Discussion== | ||
Use the theorem that the order of an element written in disjoint cycle form is the least common multiple of the cycle lengths, and count how many elements of each order we have. | Use the theorem that the order of an element written in disjoint cycle form is the least common multiple of the cycle lengths, and count how many elements of each order we have. | ||
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iii. Elements of order 2 are those of the form (ab)(cd). There are <math>5*4*3*2</math> strings of this form, drawing from five symbols, but each two-cycle can be written in two equivalent ways, and the ordering of the two-cycles in the notation does not matter, since they are mutually disjoint. Hence, there are <math>\frac{5*4*3*2}{2*2*2} = 15</math> elements of order 2. | iii. Elements of order 2 are those of the form (ab)(cd). There are <math>5*4*3*2</math> strings of this form, drawing from five symbols, but each two-cycle can be written in two equivalent ways, and the ordering of the two-cycles in the notation does not matter, since they are mutually disjoint. Hence, there are <math>\frac{5*4*3*2}{2*2*2} = 15</math> elements of order 2. | ||
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| + | [[HW2_MA453Fall2008walther|Back to HW2]] | ||
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| + | [[Main_Page_MA453Fall2008walther|Back to MA453 Fall 2008 Prof. Walther]] | ||
Latest revision as of 15:38, 22 October 2010
HW2, Chapter 5 problem 43, Discussion, MA453, Prof. Walther
Problem Statement:
Show that: i. $ A_5 $ has 24 elements of order 5, ii. 20 elements of order 3, iii. and 15 elements of order 2.
Discussion
Use the theorem that the order of an element written in disjoint cycle form is the least common multiple of the cycle lengths, and count how many elements of each order we have.
i. The elements of order 5 are of the form (abcde), so I can just count the number of strings of this form, and then perform a little correction to account for the fact that rotations of a given string are considered equivalent in the context of cycle notation.
So we have 5! elements of the form (abcde), and for every element, it is within a set of five equivalent rotations. This gives $ \frac{5!}{5} = 24 $ elements of order 5.
ii. The same counting strategy can be used here too.
The elements of order 3 are those of the form (abc) with the singleton cycles not shown. There are $ 5*4*3 $ ways of choosing a string of this form five characters, but every choice belongs to a set of three equivalent choices. Thus, there are $ \frac{5*4*3}{3} = 20 $ elements of order 3.
iii. Elements of order 2 are those of the form (ab)(cd). There are $ 5*4*3*2 $ strings of this form, drawing from five symbols, but each two-cycle can be written in two equivalent ways, and the ordering of the two-cycles in the notation does not matter, since they are mutually disjoint. Hence, there are $ \frac{5*4*3*2}{2*2*2} = 15 $ elements of order 2.
