## Sequence of permutations without a fixed point

### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)

I need to find $m$ permutations $A_1,..,A_m$ where each $A_i$ is a permutation on $n$ objects such that any of the compositions $A_jA_{j+1}..A_{j+k}$; $1 \leq j \leq n-1$ and $j+k \leq m$ does not have a fixed point.

Does such a sequence of permutation always exist? If yes, is there a good way to generate such permutations? Do we have a name for such a sequence of permutations? Thanks for your time.

-

## 1 Answer

Of course such permutations exist if $n > m$ (just take each $A_i$ to be the same $n$-cycle). On the other hand if $m \le n$, a pigeonhole argument should give some pair of $A_1, A_1 A_2, A_1A_2A_3, \ldots$ mapping element $1$ to the same element, inducing a fixed point in the quotient of this pair. Are you sure this question is well-posed?

-

Yes, you are right. This was easy after all... Thanks. – Hardik Udeshi Jun 13 at 8:46
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 13, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8834207653999329, "perplexity_flag": "head"}