Express $\alpha^{83} $ as a product of disjoint cycles

Question

I have $\alpha$ = $(15)(37964)(8)(2)$ and am asked to express it to the power of $83$

This is what I have done so far,

$\alpha ^{83} = (15)^1(37964)^3(8)(2) \: = (51)(46937) $

Am I doing it correctly?

Thanks

Answer 1

The first equality is correct. But you did not compute the resulting cycles correctly.

Note: $(15)^1 = (51)= (15)$. That part is okay.

But $(3 7 9 6 4)^3=(3 7 9 6 4)\cdot(3 7 9 6 4)\cdot(3 7 9 6 4)$.

$3 \to 7 \to 9 \to 6$

$6\to 4 \to 3 \to 7$

$7 \to 9 \to 6 \to 4$

$4 \to 3 \to 7 \to 9$

$9 \to 6 \to 4 \to 3.\;$ So we are done.

$(36749) = (3 7 9 6 4)^3$

$$\alpha ^{83} = (15)^1(37964)^3(8)(2)\: = (15)(36749)(8)(2) = (15)(36749)$$

Answer 2

The first equality sign is correct, but not the second. Compute $(3 7 9 6 4)^3$ as $(3 7 9 6 4)\cdot(3 7 9 6 4)\cdot(3 7 9 6 4)$.