Product of non-disjoint k-cycles

Question

I have two $k$-cycles $\alpha=(a \dots c \dots b \dots)$ and $\beta=(a \dots b \dots c \dots)$ and $\alpha \neq \beta^{-1}$. How to show that the product $\alpha \beta$ does not result in a cyclic permutation (just one cycle permutation).

Or, perhaps it has a counter example.

Thanks.

Without further assumptions, yes indeed: $(1,2,3)(1,3,2)=id$. — Jonathan Y., Nov 07 '13 at 23:22
Thanks, Jonathan. Actually, I have forgotten to add the assumption that $\alpha \neq \beta^{-1}$. — LuizG, Nov 07 '13 at 23:26
Well, @N.S. has already provided another example, but that is a superficial restriction. For example $(1,2,3,4)(1,3,2) = (4,1)(1,2,3) = (4,1)id=(4,1)$. — Jonathan Y., Nov 07 '13 at 23:31
Thanks, @Jonathan Y. Dou you think it is possible that product could result in another k-cycle? – — LuizG, Nov 07 '13 at 23:46

score 1 · Accepted Answer · answered Nov 07 '13 at 23:26

1

$$(1,4,2,3)(1,4,3,2)=(1,3,4) \,.$$

To find the example I just picked two 4-cyles in $S_4$, and made sure that 1 doesn't end in a 2-cycle. Note that in $S_4$ the only non-cyclic permutations are the products of two $2$-cycles.

answered Nov 07 '13 at 23:26

N. S.

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Product of non-disjoint k-cycles

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