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A proof that $A_n$ is simple ($n>4$) begins as follows:

Suppose $H$ is a nontrivial normal subgroup of $A_n$. We first prove that $H$ must contain a $3$-cycle. Let $\sigma \neq e$ a permutation that moves the leat number of integers in $n$, Being an even permutation $\sigma$ cannot be a cycle of even length. Hence, $\sigma$ must be a $3$-cycle or have a decomposition of the form $(a b c \cdots)\cdots$ or $(a b)(c d)\cdots$ , where $a,b,c,d$ are distinct. [CUT]

Why $\sigma$ cannot be, for example, $(abc)(def)$ ?

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In the first type, $(abc\cdots)\cdots$, the possibility that the first cycle is just a $3$-cycle (provided there is some disjoint cycle later) is not excluded. That is, the first description is any element of $A_n$ that has a cycle of length at least $3$, plus perhaps some other stuff. – Arturo Magidin Mar 14 '12 at 2:30
Well. Put it as answer so I can mark it. – Oo3 Apr 9 '12 at 19:39
up vote 2 down vote accepted

You are misinterpreting the notation. $(abc\cdot)\cdots$ means that the first cycle is of length at least $3$, and there are possibly some other cycles afterwards. So $(abc)(def)$ is contemplated in that notation, as is $(abcd)$.

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