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Why don't $3$-cycles generate the symmetric group? was asked earlier today. The proof is essentially that $3$-cycles are even permutations, and products of even permutations are even.

So: do the $3$-cycles generate the alternating group? Similarly, do the $k$-cycles generate the alternating group when $k$ is odd?

And do the $k$-cycles generate the symmetric group when $k$ is even? I know that transpositions ($2$-cycles) generate the symmetric group.

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2 Answers 2

If $n\geq5$, then the only normal subgroups of the symmetric group $S_n$ are the trivial group, the alternating group and the symmetric group itself. Since the $k$-cycles form a full conjugacy class, it follows that the subgroup they generate is normal. This determines everything if $n \geq 5$.

More specifically: the $k$-cycles in $S_n$ generate the alternating group if $k$ is odd and $k \ne 1$; they generate the full symmetric group if $k$ is even.

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Nice proof. ---- –  user5262 Nov 27 '11 at 18:00

Yes, k-cycles generate the symmetric group when k is even and alternating group, when k is odd. As you've said, for k=2 you know the answer. Suppose k>2.
(1,2,...,k)(k,...,3,1,2)=(1,3,2)
Similarly you can get any 3-cycle.

Suppose a is even element of Sn. Then, as you know, a is a product
$(k_1,k_2)(k_3,k_4)...(k_{4l-1},k_{4l})$
of 2l transpositions. But product of 2 transpositions can be written using 3-cycles:
$(k_1,k_2)(k_2,k_3)=(k_1,k_2)(k_2,k_3)*(k_2,k_3)(k_3,k_4)$.
Any of two products, separated by * in right hand side, is either 3-cycle or unity.

If a is odd and k is even, than a multiplied by any k-cycle is even, so we can apply the previous algorithm to it.

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