7
$\begingroup$

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

$\endgroup$
  • 1
    $\begingroup$ 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
3
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.