Enigma and Rejewski's theorems (Permutation-Cycles) Rejewski stated the following

Theorem on the Products of Transpositions
If two permutations of the same degree consist solely of disjoint transpositions, then their product will consist of disjoint cycles of the same length in even numbers.

As example we consider the following
\begin{align*}
   X &= (a_1a_2)(a_2a_3)(a_4a_5) \ldots (a_{2k-3}a_{2k-2}) (a_{2k-2}a_{2k-1}) \\
   Y &= (a_2a_3)(a_4a_5)(a_5a_6) \ldots (a_{2k-2}a_{2k-1}) (a_{2k-1}a_{2k}) 
  \end{align*}
Then
\begin{align*}
   XY = (a_1 a_3 a_5 \ldots a_{2k-3}a_{2k-1})(a_{2k}a_{2k-2} \ldots a_6 a_4 a_2)
  \end{align*}
Additionally we assume the reverse to be true.
Now it is stated

Letters entering into one and the same transposition of permutation $X$ or $Y$, enter always into two different cycles of permutation $XY$.

and

If two letters found in two different cycles of the same length of the permutation $XY$ belong to the same transposition, then the letters adjacent to them (one to the right, the other to the left) also belong to the same transposition.

I have a hard time to understand the latter statements, especially in the light of the given theorem.
I'd be happy to see some in depth answer, since I did not have had any course in abstract algebra yet.
 A: Your example $X$ and $Y$ don't look correct to me.  They should be products of disjoint 2-cycles, so $X$ would be $(a_1, a_2)(a_3, a_4) \cdots$  Perhaps this is a copying error, or a typo in the source.
With my corrected notation, the first sentence you are asking about means that for all $i$, $a_{2i-1}$ and $a_{2i}$, which belong to the same 2-cycle in $X$, belong to different cycles in $XY$.
The second sentence means that if $XY$ contains two cycles $(r_1, \ldots, r_k)$ and $(s_1, \ldots, s_k)$ and if $(r_1, s_1)$ is a 2-cycle in $X$, then also $(r_2, s_k)$ is a 2-cycle in $X$.
Here is a typical example.  Let $X = (1, 2)(3, 4)(5, 6)(7, 8)(9, 10)$ and $Y=(2, 7)(1, 4)(3, 5)(6,10)(8, 9)$. I suspect Rejewski is multiplying in the traditional left-then-right order, in which case $XY$ is $(1, 7, 9, 6, 3)(2, 4, 5, 10, 8)$.  Note that 1 and 2 belong to different cycles in $XY$, as do 3 and 4, 5 and 6, 7 and 8, and 9 and 10, illustrating the first result.  Now looking at $XY$, 1 and 2 belong to the same cycle in $X$. Moving to the right in the first cycle and to the left in the second (so wrapping back to the end of the second cycle), 7 and 8 belong to the same cycle in $X$. Taking another step, 9 and 10 belong to the same cycle in $X$, and so on.
All this is going to be hard to understand without having taken a course covering the basics of permutation groups.  I'd recommend that first.
