permutation cycle group of a large power

Question

I was going to add this to my previous question here, but I didn't think that was allowed.

The question I have here:

(1254)^1000

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How would I find something of this large? Would I be able to do (1254)^4 and find it that way since 1000 is divisible by 4?

Answer 1

Indeed, you can use $(1254)^4$. In particular, we have $$ (1254)^{1000} = ((1254)^{4})^{250} = ? $$

Answer 2

The order of the element in a group, o(a), is the smallest positive integer such that $a^{n} = e_{G}$, the identity. The order of a cycle happens to be the number of elements it has. So o(1254) = 4. So let's work it out:

(1254)(1254) = (15)(24)

(15)(24)(1254) = (1452)

(1452)(1254) = (1)(2)(5)(4) = (1)

So here's a hint- think about rules of exponents. If you're familiar with modular arithmetic and Fermat's Little Theorem, all of that applies here.