1

Given that $x= (1 2 3)(4 5 6 7 8)$, what is the order of $x$? (i.e the smallest integer $k$ such that

Hassan Muhammad
  • 4,282
jj103
  • 155
  • There is a theorem, mentioned below but you can tackle this intuitively...note that if $n$ is not a multiplw of $3$ then $x^n$ cannot be the identity (since the $(1 2 3)$ bit will be non-trivial). Similarly if $n$ is not a multiple of $5$ then $x^n$ is not the identity (since the $(4 5 6 7 8)$ bit will be non-trivial). So $n$ must be a multiple of $15$ for $x^n$ to be the identity. It is simple to see that $x^{15}$ is the identity so the order is $15$. – fretty Sep 22 '14 at 12:12
  • Alternatively calculate $x,x^2, ..., x^{14}$ and find these aren't the identity but that $x^{15}$ is. – fretty Sep 22 '14 at 12:14

3 Answers3

4

When a permutation is written as a product of disjoint cycles, its order is the least common multiple of those cycles' lengths (easy proof by induction).

In your case the order is $\;l.c.m (3,5)=15\;$

Timbuc
  • 34,191
3

The order of a cycle is its length.

The order of a product of disjoint cycles is the least common multiple of the cycles' lengths.

For example, $$\operatorname{order}(1, 2, 3) = 3,\quad \operatorname{order}(4,5, 6, 7, 8) = 5.$$

$$\operatorname{order}(123)(45678) = \operatorname{lcm}(3, 5) = 15.$$

amWhy
  • 209,954
2

We have this general result:

If in a group $G$ we have two elements $x$ and $y$ that commute i.e. $xy=yx$ and with coprime orders $m$ and $n$ respectively : $\gcd(m,n)=1$ then $mn$ is the order of $xy$.

Proof: We solve the equation:

$$(xy)^k=e\iff x^ky^k=e\tag1$$

$$(1)\implies x^{kn}=e\implies m|kn\tag2$$ and $$(1)\implies y^{km}=e\implies n|km\tag3$$ and since $m$ and $n$ are coprime so with $(2)$ and $(3)$ we get $$n|k\;\text{and}\; m|k\implies mn|k$$ so the order of $xy$ is $mn$.

Can you apply this result to your case?