3
$\begingroup$

I have seen several references to "order" of an element in the Symmetric Group. Specifically, that the order of a cycle is the least common multiple of the lengths of the cycles in its decomposition.

But the Symmetric Group is not cyclic, and I'm only familiar with the concept of "order" for cyclic groups. So what does it mean in this context?

$\endgroup$
9
$\begingroup$

The order of an element $g$ in a finite group $G$ is the smallest integer $n \in \mathbb{N}^*$ such that $g^n = e$ (the neutral element of the group). This is well-defined for every finite group $G$, so in particular for the Symmetric group as well.

The Lagrange theorem implies indeed, that for any finite group $G$ with $p$ elements and any $g \in G$, then $g^p = e$, which shows that every $g$ has finite order, less or equal to $p$.

$\endgroup$
  • $\begingroup$ " smaller than p" Should be "less than or equal to p". $\endgroup$ – Acccumulation Oct 24 '18 at 15:15
  • $\begingroup$ That's true (a french deformation, in french we usually say smaller to mean less or equal to :) ) $\endgroup$ – seamp Oct 24 '18 at 15:31
6
$\begingroup$

Since you know about cyclic groups.

Think of order of an element in the symmetric group $S_n$ as the size of the cyclic group generated by it.

Example 1: Let us take $S_3$. Take $g=(123)$. Consider the cyclic group generated by $g$, that is $\langle g\rangle $. $\langle g\rangle =\{(123),(132),(1)\}$. So order of $g$ is 3 since cardinality of $\langle g\rangle $ is 3.

Example 2: Consider $S_4$ and take $g=(12)(34)$ Then $\langle g\rangle =\{(12)(34),(1)\}$. Since cardinality of $\langle g\rangle $ is 2, order of $g$ is $2$.

Example 3(Take it as an exercise): Consider $S_5$ and take $g=(123)(45)$. Find $\langle g\rangle $ by explicitly writing each element and then calculate the order from $g$ from there

$\endgroup$

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.