Sign up ×

Just wondering how to find the order of each element in this group:

$A_4 = \{e,(123),(132),(124),(142),(134),(143),(234),(243),(12)(34),(13)(24),(14)(23)\}$

I tried writing each elements not in disjoint cycle but it didn't look right to me. I got 3 for all the cycles with 3, and 4 for the last cycles

share|cite|improve this question
You can take the element $x\in A_{4}$, and then look at $x, x^2, x^3, …$ until you reach $x^{n}=1$ (the identity element). The smallest such $n$ is the order of $x$. –  Prism Mar 31 '14 at 2:13
The order of $(12)(34)$ is $2$. And same for $(13)(24)$ and $(14)(23)$. Your answer for $3$-cycles is right. All the $3$-cycles have order $3$. –  Prism Mar 31 '14 at 2:13

4 Answers 4

In general, the order of any $k$-cycle is $k$.

However, if you have a composition of disjoint cycles, say a $k$-cycle with an $l$-cycle, then the order of the composition will be $\mathrm{lcm}(k, l)$.

(Prove this!)

share|cite|improve this answer

You are almost right. Remember that disjoint cycles commute, and that a $2$-cycle has order $2$. So the cycles of the form $(i,j)(k,l)$ actually satisfy $((i,j)(k,l))^2=\text{Id}$, where $\text{Id}$ is the identity permutation. Thus they have order $2$, not order $4$.

share|cite|improve this answer

Notice for any 3-cycle $(abc)$, $(abc)^2=(abc)(abc)=(acb)$, and $(abc)^3=(abc)^2(abc)=(acb)(abc)=e$, the identity. Thus the order of any 3-cycle is 3.

Noting that disjoint cycles commute, it is easy to see that $((ab)(cd))^2=(ab)(ab)(cd)(cd)=e*e=e$, so the order of any product of two disjoint transpositions is 2.

share|cite|improve this answer
Thx that was a lot of help –  Mary Mar 31 '14 at 5:39
No problem, Mary. I would recommend pondering @mathematics2x2life's response, too. He or she presents a much more interesting result than I presented. –  user138907 Apr 2 '14 at 2:28

The order of any $k$-cycle is $k$, try seeing why by trying a few examples for yourself. Now the disjoint product of $k$-cycles commutes; therefore, if we have the product of $n$, $k$-cycles, $x_1x_2,x_3,\cdots$, we would have $$ (x_1x_2\cdots x_n)^k=x_1^kx_2^k\cdots x_n^k $$ so that the order of a product of disjoint $k$-cycles (not necessarily all of the same length) must be the least common multiple of their orders (their length). Since we write the elements of $A_n$ in disjoint cycle notation, this makes the computation of the orders a quick task. Just be sure to understand why the above things are true (try proving them formally for yourself!).

share|cite|improve this answer

Your Answer


