# Order of a permutation which produces a 5-cycle

Question: Let $$\alpha = \left ( 1\,3\,5\,7\,9 \right )\left ( 2\,4\,6 \right )\left ( 8\,10\right )$$. If $$\alpha^{m}$$ is a 5-cycle, what can you say about m?

Attempt:

$$\left | \alpha \right |=lcm\left ( 5,3,2 \right )=30$$

$$\Rightarrow \alpha ^{30}=\varepsilon$$

Unfortunately, I am unable to further beyond this point.

Any hint is appreciated. Thanks in advance.

• Should there be $\;m\;$ instead of "," in the third cycle in $\;\alpha\;$ ? Also, in what group $\;S_n\;$ are you working? That could be relevant. Mar 5, 2017 at 10:27
• A more difficult problem is to figure out what you meant to write. I think you meant "if $\alpha^m$ is a $5$-cycle then what can you say about $m$". Mar 5, 2017 at 10:28
• @DerekHolt That is a good point, as the order of $\;\alpha\;$ itself cannot be less than $\;15\;$ no matter what the last cycle in it is. Mar 5, 2017 at 10:30
• There was a missing "^{m}". I have since added. Mar 5, 2017 at 10:31
• As a hint, try $m=6$ and then $m=12$. Mar 5, 2017 at 10:32

Since $\alpha$ is expressed as the product (which actually means a composition) of disjoint cycles, that is, cycles that have no elements in common, and we know that disjoint cycles are conmutative, we can assure that $\alpha^m=(13579)^m (246)^m (8,10)^m$. Since we need this product to be a 5-cycle, we must consider the following:

• The product $(246)^m (8,10)^m$ is a 2-cycle if $m$ is a multiple of 3 but not a multiple of 2; it is a 3-cycle if $m$ is a multiple of 2 but not a multiple of 3; it is the product of a 3-cycle and a 2-cycle if $m$ is neither a multiple of 2 nor of 3; and it is the identity if $m$ is a multiple of 6. Therefore, it can never be a 5-cycle, and since it is disjoint with $(13579)$, the only way to obtain a 5-cycle as the result for $\alpha^m$ is for these two factors to disappear, that is, for them to equal the identity. We then conclude that $m$ must be a multiple of 6.

• If $m$ is also a multiple of 5, then $(13579)^m$ equals the identity, and since the identity is not a 5-cycle we must exclude this possibility.

Then, the answer is that $\alpha^m$ will be a 5-cycle if, and only if, $m$ is a multiple of 6 but not a multiple of 5. This can also be expressed by saying that $m$ must be a multiple of 6 but not a multiple of 30.

• I like this more "hands-on" solution! Though I think you can simplify it a bit. The key is that powers of $p$-cycles are $p$-cycles (or the identity) for any prime $p$. Since the cycles are disjoint, you just need the $2$- and $3$-cycles to go to identity and the $5$-cycle not to go to identity. That translates to $m$ being divisible by $2$ and $3$ but not by $5$. Mar 6, 2017 at 9:45

As you wrote, $\;ord(\alpha)=30\;$ and thus it generates a group of order $\;30\;$ in $\;S_{10}\;$ , so if

$$\;\langle\alpha^m\rangle=5\implies \alpha^{5m}=1\implies 30\,\mid\,5m\implies\;\ldots \;$$

Observe you have more than one option...

• I think you mean $30\mid5m$. Mar 5, 2017 at 11:46
• @GerryMyerson Oh, of course...hehe. Thanks,. editing. Mar 5, 2017 at 11:56