What are all the elements of $A_5?$
$$\{1,(1,2,3),(1,3,2),(1,2,4),(1,4,2),(1,2,5),(1,5,2),(1,3,4),(1,4,3),(1,3,5),(1,5,3),(1,4,5),(1,5,4),(2,3,4),(2,4,3),(2,3,5),(2,5,3),(2,4,5),(2,5,4)(3,4,5),(3,5,4),(1,2) (3,4),(1,3)(2,4),(1,4)(2,3), (2,3)(4,5),(2,4)(3,5), (2,5)(3,4),(1,3)(4,5),(1,4)(3,5),(1,5)(3,4),(1,2)(4,5),(1,4)(2,5),(1,5)(2,4),(1,2)(3,5),(1,3)(2,5),(1,5)(2,3) \} $$
I guess the remaining ones are: $$(1,2,3,4,5),(1,2,3,5,4),(1,2,4,3,5),(1,2,4,5,3),(1,2,5,3,4),(1,2,5,4,3),(1,3,2,4,5),(1,3,2,5,4),(1,3,4,2,5),(1,3,4,5,2),(1,3,5,2,4),(1,3,5,4,2),(1,4,2,3,5),(1,4,2,5,3),(1,4,3,2,5),(1,4,3,5,2),(1,4,5,2,3),(1,4,5,3,2),(1,5,2,3,4),(1,5,2,4,3),(1,5,3,2,4),(1,5,3,4,2),(1,5,4,2,3),(1,5,4,3,2) $$
Are these correct?
$\endgroup$ 21 Answer
$\begingroup$This doesn't answer your question but I think it's much more fun. The elements of $A_5$ are the even permutations of $S_5$. Therefore, it contains:
- the identity
- 3-cycles
- 5-cycles
- disjoint 2-cycles
Now let's count how many of each we have. For 2. the result is given by $$\binom{5}{3}2!=20$$ For 3. we have $$4!=24$$ For 4. $$\binom{5}{2}\binom{3}{2}/2=15$$
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