Let $n\geq 5$ be odd, What is a presentation of $A_n$ with generators $a_n=(123),b_n=(1,2,\ldots,n)$?

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    $\begingroup$ You need to distinguish between $n$ being odd and $n$ being even cases. In particular, $b_n=(1,2,\dots,n)$ for odd $n$ is even, and $c_n=(2,3,\dots,n)$ for even $n$ is even (how odd). Otherwise, this appears to be a difficult question: what is your motivation for asking? $\endgroup$ – Vladimir Sotirov Mar 21 '12 at 2:13
  • $\begingroup$ Sorry i was thinking this for odd $n$. $\endgroup$ – Balin Mar 22 '12 at 2:23

I suggest you look at http://www.math.auckland.ac.nz/~obrien/research/an-sn-present.pdf to get some idea of the current state of knowledge about this question. Theorem 1.3 states that $A_n$ has a 2-generator presentation with $O(\log n)$ relations and length $O((\log n)^2)$.

Typing $$\rm presentation\ alternating\ group$$ into Google got me this and other references.


There are few known presentations for $A_n$. You can express their generators in $a_n=(123)$, $b_n=(1,2,\ldots,n)$ and obtain some presentations in these generators. For example, take the Carmichael's presentation for $A_n$ (for any $n$) from p.172 of:

Carmichael R.D. Introduction to the theory of groups of finite order. Boston. 1937.

Generators: $V_i=(1,2,i+2)$, $i=1,...,n-2$.

Relators: $V_1^3=V_2^3=\dots=V_{n-2}^3=1$, $(V_iV_j)^2=1$ for $1\le i<j\le n-2$.

Now express $3$-cycles $(1,2,i)$ inductively in your generators $a_n$, $b_n$ as follows: $(1,2,3)=a_n$, and $$ (1,2,i+1)=a_nb_n^{-1}(1,2,i)b_na_n^2 $$ (we multiply permutations left-to-right, as if they are acting on the right). This way you get some presentation in your generators. It may be not the shortest one, but it will work.


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