What is the order of the cycle $(1,4,5,7)$?

Question

I tried to find some power that equals the identity element of permutation, but I couldn't do it. Like $(1,4,5,7)^2=(1,5)(4,7)$.

Answer 1

$(1,4,5,7)\neq ()$

$(1,4,5,7)^2=(1,5)(4,7)\neq ()$

$(1,4,5,7)^3=(1,7,5,4)\neq ()$

$(1,4,5,7)^4=()$, where $()$ denotes the identity cycle.

So we conclude: The order of the cycle $(1,4,5,7)$ is $4$.

Answer 2

Just think about this logically. The cycle is completely disjoint and is well-formed in that every element maps to another unique element, so no "cycle" exists within the cycle.

In order for $1$ to go back to itself (and thus constitues an identity cycle), then applying the cycle rules again and again, it has to go to $4$...then $4 \rightarrow 5 \rightarrow 7 \rightarrow 1$, so $$1 \rightarrow 4 \rightarrow 5 \rightarrow 7$$ The same argument can be used for each element in the cycle, any element not in the cycle is fixed so it does not affect the order of the cycle. So for each element, the permutation $(\alpha_1\alpha_2\alpha_3\alpha_4)$ has to be preformed $4$ times for each element to map back to itself. Indeed, in general, if $\alpha_i \neq \alpha_j, i \neq j$ then $$|(\alpha_1\alpha_2\alpha_3\alpha_4.......\alpha_n)| = n$$

Answer 3

Conjugating by $(42)(53)(74)$ gives $(1234)$; conjugates have the same order, so we're just looking for the order of $(1234)$. That's the same as its order in $S_4$, since $S_4$ contains every power of $(1234)$. (Every power of $(1234)$ must move $1$, $2$, $3$, and/or $4$.)

Now think of $S_4$ acting (faithfully) on the vertices of a square. What is $(1234)$ doing to the vertices?