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Possible Duplicate:
Homomorphism between $A_5$ and $A_6$

Why is it true that every element of the image of the function $f: A_5\longrightarrow A_6$ (alternating groups) defined by $f(x)=(123)(456) x (654)(321)$ does not leave any element of $\{1,2,3,4,5,6\}$ fixed (except the identity)?

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    oops! ${}{}{}{}$2012-10-15

2 Answers 2

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I think you are asking why the following is true:

Every element of $A_6$ of the form $(123)(456)x(654)(321)$, with $x$ a nontrivial element of $A_5$, leaves no element of $\{1,2,3,4,5,6\}$ fixed.

This is actually false: let $x = (12)(34)$. Then $f(x)$ fixes either $5$ or $6$, depending on how you define composition of permutations.

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Whilst actually false, it is false because it leaves the $4$th element of the permutation fixed; like the function $f: x \to x$ leaves the $6$th element of the permutation fixed. (Dion didn't see that, he is using the geometrical proof)

I think in this problem, you have to use the geometrical proof, unless you find the permutation which $(123)$ will return yourself, which I've used Mathematica to determine.