Step1/2

a) Since α is a cycle of length $3$ , its order is $3 .$ Similarly, $β$ is a cycle of length $9$ , so its order is $9$ . To find the order of $α β$ , we need to compute the least common multiple (LCM) of the orders of α and $β$ . Since $3$ and $9$ are relatively prime, their LCM is simply their product: $L C M (3, 9) = 3 \times 9 = 27 .$ Therefore, the order of $α β$ is $27$ .
To determine whether αβ is even or odd, we need to compute its parity. The parity of a permutation is the parity of the number of inversions it contains. An inversion is a pair of elements $(i, j)$ such that $i < j$ but $α β (i) > α β (j) .$ Since $α$ and $β$ are disjoint, they commute, which means that $α β (i) = α (β (i))$ for all i. Therefore, we can count the inversions of αβ by counting the inversions of $β$ and then multiplying by 3, since there are 3 elements in each cycle of α. Since β is a cycle of length 9, it contains $\frac{36}{2} = 18$ inversions (since any pair of elements in the cycle is either in increasing order or decreasing order). Therefore, $α β$ contains $3 \times 18 = 54$ inversions, which is an odd number. Therefore, αβ is an odd permutation.

Explanation:

a) Let α be a cycle of length 3 and β a cycle of length 9. To find the order of αβ, we need to find the smallest positive integer k such that (αβ)^k = e, where e is the identity permutation. Since α and β are disjoint cycles, they commute with each other, which means that (αβ)^k = α^kβ^k. Therefore, we need to find k such that α^kβ^k = e.

Since the order of α is 3 and the order of β is 9, we know that the order of α^k is either 1, 3, or 9, and the order of β^k is either 1 or 9. Therefore, we can try different values of k until we find one that works.

For example, we have:

(αβ)^2 = αβαβ = α^2β^2,

(αβ)^3 = αβαβαβ = α^3β^3 = e,

so the order of αβ is 3, and αβ is an odd permutation, since it can be written as the product of an even number of transpositions.

Step2/2

b) We can prove this result by induction on n. For the base case, we have:
${(α β)}^{1} = α^{1} β^{1} = α β$
which is true by definition.
Now, suppose the result is true for some positive integer k, i.e.,
${(α β)}^{k} = α^{k} β^{k} .$
We need to show that the result is also true for $k + 1$ , i.e.,
${(α β)}^{k + 1} = α^{k + 1} β^{k + 1} .$
We have:
${(α β)}^{k + 1} = {(α β)}^{k} (α β) = α^{k} β^{k} α β$ .
Since $α$ and $β$ are disjoint, they commute, which means that $α β = β α$ . Therefore, we can rewrite the expression as:
$α^{k} β^{k} α β = α^{k} β^{k} β α .$
Now, we can use the induction hypothesis to substitute $α^{k} β^{k}$ with ${(α β)}^{k} :$
$α^{k} β^{k} β α = {(α β)}^{k} β α .$
Finally, we can use the fact that β and α commute to rearrange the expression as:
${(α β)}^{k} β α = α^{k} β^{k} β α = α^{k} β^{k} α β β α .$
Since $α$ and $β$ are disjoint cycles, they have no elements in common except the identity element, which means that βα is a cycle of length 12. Therefore, we have:
$α^{k} β^{k} α β β α = α^{k} β^{k} {(α β)}^{2} β α$ .
Now, we can use the base case to substitute ${(α β)}^{2}$ with $α β α β :$
$α^{k} β^{k} {(α β)}^{2} β α = α^{k} β^{k} α β α β β α .$
Since $α β$ and $β α$ are disjoint cycles, we can commute them without changing their product:
$α^{k} β^{k} α β α β β α = α^{k} β^{k} α α β β β α .$
Finally, we can use the fact that α is a cycle of length 3 to simplify the expression as:
$α^{k} β^{k} α α β β β$

Explanation:

b) To prove that (αβ)^n = α^nβ^n, we can use mathematical induction on n.

For n = 1, we have:

(αβ)^1 = αβ = α^1β^1.

Assume that (αβ)^k = α^kβ^k for some positive integer k. Then we have:

(αβ)^(k+1) = (αβ)^k (αβ) = α^kβ^k αβ = α^kαββ^k = α^(k+1)β^(k+1).

Therefore, by mathematical induction, we have (αβ)^n = α^nβ^n for all positive integers n.

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