Find the number of elements of $A_4$ composed of the product of two non-disjoint transpositions.

Question

Find the number of elements of $A_4$ composed of the product of two non-disjoint transpositions.

There are following $12$ elements in $A_4=\{e, (123), (132), (124), (142), (134), (143), (234), (243), (12)(34), (13)(24), (14)(23)\},$ of which the required ones are $8.$

Instead I get $(4C2\cdot 3C2)/2= (6\cdot 3)/2= 9.$

Why the difference of $9-8=1?$

Answer 1

The right formula is $$\frac{4C2\cdot 4}{3}= 8$$ You pick the first transposition $(ab)$, for that you have $4C2=6$ choices, then You have restrictions on the second transposition,its must intersect the first transposition $(ab)$, so once you have fixed the first transposition,you have four possibilities , two transpositions containing $a$ and two others conatining $b$, finally, any non two disjoint gives u a three cycle(conversely any such cycle can be written as a product of two non disjoint transpositions) and a three cycles can be written in three ways.

Answer 2

You have listed all 12 elements of $A_{12}$ and counted which elements are 3-cycles. These are certainly composed of two non-disjoint ("joint"?) transpositions. It's a matter of semantics whether the identity element (which is equal to $(12)(12)$, for instance) should count. Let's say here that it isn't counted.

What has your second attempt counted? I actually don't know. The $4C2$ term picks out one transposition, and then presumably the $3C2$ picks out another transposition that's non-disjoint. But that's not entirely right. Indeed, what are the $3$ from which you $C2$? In fact, assuming we don't want the identity element included in our count, the two elements we pick for the second transposition should consist of one element from the support of the first transposition (the support is set of element actually moved), and one element from outside the support. Thus we get $2^2=4$, rather than $3C2=3$ options.

So now we have $24$ ways to pick an ordered pair of joint transpositions. However, several of these result in the same final permutation. It remains to see how many times each resulting permutation is counted, and we have our answer (since we already know from brute force counting that the end result is $8$, there better be 3 of each permutation here).

Choose some such pair $(ab)(ac)$. Which other pairs yield the same permutation? It is clear that the same three elements $a, b$ and $c$ must have been chosen. It is also clear that the other pair with $a$ appearing twice (which is to say $(ac)(ab)$) yields a different permutation. So the only possibility are the four pairs where $b$ or $c$ were chosen twice. And we see that indeed, the other pairs that yield the same permutation are $(bc)(ba)$ and $(ca)(cb)$.

So the final answer we get is $$ \frac{4C2\cdot 2^2}3=8 $$ entirely in line with the brute force counting.