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

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

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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.

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

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

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Is this clear now, bobdylan? –  amWhy Feb 11 '13 at 15:59
ohh no not at all, I would have sworn i clicked it but alas, my age is catching up to me, all 25 years of it lol –  bobdylan Feb 12 '13 at 17:36
@bobdylan not a problem! I just wanted to make sure I hadn't offended you in some way. You're young! And busy! (and mind full of group theory, etc!) –  amWhy Feb 12 '13 at 17:38

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

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