Need a counterexample: A product of a cycle with itself will in general be a cycle.

Question

I was thinking something in S_4.

You should mention at least the word "permutation" somewhere in your question's title... — DonAntonio, Apr 05 '14 at 19:53

score 1 · Answer 1 · answered Apr 05 '14 at 19:52

1

In $\;S_4\;$:

$$(1234)^2=(13)(24)$$

answered Apr 05 '14 at 19:52

DonAntonio

1

Note: This suggests that any cycle with order divisible by two might work. A cycle $c$ of order $2n$ generates a cyclic group of order $2n$. $c^2$ generates a cyclic group of order $n$. – Mark Bennet Apr 05 '14 at 20:04
Indeed so, @MarkBennet +1 – DonAntonio Apr 05 '14 at 20:06

1 Answers1