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List all the elements of the alternating group $A_3$ written in cyclic notation.

I come up with

Identity $(1)$ Obviously $(123)$

3 Answers 3

7

(1) (123) (132)

got it

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Hint: If a group is cyclic, then find an element that is not the identity and find the group that is generated by that element. That is a subset of your group (and even more so a subgroup, albeit unrelated here). And if the group generated is the size of your desired group (you should know the size of $A_3$), then you have all the elements within your group. If you are unfamiliar with cyclic notation please reference the Wikipedia article.

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    (1) (123) (132)2010-10-22
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The element of A3 = {(1), (1,2,3), (1,3,2)} The formula for finding how many elements A3 has is (3!/2)=6/2=3.