# Problem understanding abstract algebra permutation problem

So we have two permutations, composed of $k$ and $s$ disjoint cycles with increasing lengths (number of elements in each cycle). We want to prove:

They have the SAME amount of cycles. The distance of each cycle is the same.

As long as $\alpha$ and $\beta$ are conjugate.

This is my interpretation of the problem. However to get started, I need to understand it better, especially the conjugate part and how it relates to my problem.

Any help appreciated, thanks!

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The key observation is that for a single cycle $(123\dots k)$ and arbitrary $\varphi\in S_n$, we have $$\varphi\circ(123\dots k)\circ\varphi^{-1}\ =\ \left(\varphi(1)\,\varphi(2)\,\dots\varphi(k)\right)\,.$$ Now, this also holds with a product of disjoint cycles. This shows that the conjugate of $\alpha$ by $\varphi$ has the same cyclic structure.
For the other direction, suppose that $\alpha$ and $\beta$ has the same cyclic structure. Choose an element $a_{i1}$ from each given cycle $\alpha_i$, and let $a_{i,n+1}:=\alpha_i(a_{in})$,
that is, $\ \alpha_i=(a_{i1}a_{i2}\dots a_{id_i})$. Similarly with the $\beta$'s: $\ \beta_i=(b_{i1}b_{i2}\dots b_{id_i})$, and let $\varphi$ move each element $a_{in}$ to $b_{in}$ (and act arbitrarily otherwise). Then we have $\varphi\circ\alpha\circ\varphi^{-1}=\beta$.
In your example, you mean for (123...k) to be $\alpha$, yes? Because you say that this shows the conjugate of $\alpha$ and $\varphi$ have the same cyclic structure. –  user3200098 Feb 19 '14 at 23:19
Well, say $\alpha_1$. –  Berci Feb 19 '14 at 23:27
Why $\alpha$1? You say "$\alpha$ by $\varphi$ has the same cyclic structure." –  user3200098 Feb 19 '14 at 23:35
For disjoint cycles, we still have $\varphi\circ(a_{11}\,a_{12}\,\dots)\,(a_{21}\,a_{22}\dots)\,\dots\circ\varphi^{‌​-1}=(\varphi a_{11}\,\varphi a_{12}\dots)\,(\varphi a_{21}\,\varphi a_{22}\dots)\dots$. –  Berci Feb 19 '14 at 23:59