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Let τ = $(a_1 a_2...a_k)$ be a cycle of length k.

(a) Prove that if σ is any permutation, then στσ−1 = (σ$(a_1),σ(a_2$),...,σ$(a_k$)) is a cycle of length k.

(b) Let µ be a cycle of length k. Prove that there is a permutation σ such that στσ−1 = µ

a) i have part a done

By multiplying by σ on the right, we can see that (a) is true if and only if $στ = (σ(a_1), σ(a_2), . . . , σ(a_k))σ.$

proof

If x 6∈ ${a_1, a_2, . . . , a_k}, $then στ (x) = σ(x) because τ (x) = x.

On the other hand $(σ(a_1), σ(a_2), . . . , σ(a_k)$)σ(x) = σ(x), because the cycle $(σ(a_1), σ(a_2), . . . , σ(a_k$)) only acts on elements of the form $σ(a_i)$ and fixes everything else.

Since x 6∈ {a1, . . . , ak}, then the cycle fixes σ(x). So when x 6∈ {a1, a2, . . . , ak}, στ (x) = σ(x) = (σ(a1), σ(a2), . . . , σ(ak))σ(x).

If x ∈ ${a_1, a_2, . . . , a_k}, then x = a_i $for some i ∈ {1, 2, . . . , k}. If i 6= k, then τ$ (a_i) = a_{i+1} so$ στ (x) = στ $(a_i) = σ(a_{i+1}) and (σ(a_1), σ(a_2), . . . , σ(a_k))σ(a_i) = σ(a_{i+1}).$

If i = k, then τ $(a_k) = a_1,$ so στ $(a_k) = σ(a_1) and (σ(a_1), σ(a_2), . . . , σ(a_k))σ(a_k) = σ(a_1)$.

Therefore στ (x) = $(σ(a_1), σ(a_2), . . . , σ(a_k))σ(x) $for all x, hence the two functions are the same

now i know you can use part a to find b but i am not sure how..

b)Suppose µ = $(b_1, b_2, . . . , b_k)$.

Now let σ be the permutation that satisfies $σ(a_i) = b_i and σ(x) = x$ otherwise.

Then by (a), στσ−1 = $(σ(a_1), σ(a_2), . . . , σ(a_k)) = (b_1, b_2, . . . , b_k)$ = µ.

is this right?

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