How is parity determined? Is it simply even when the # of transpositions is even and odd when the # of transpositions is odd?
I'm able to compute transpositions when the permutations are disjoint but otherwise run into problems. Any insight on this?
On
Rather than working with decompositions into transpositions, it is faster to count the number of inversions of that permutation. This means the numbers of pairs of numbers $(i,j)$ from that permutation such that $i>j$. If this number is odd, then the permutation is odd; and even for even. An example, to clarify: for $(17254)$ the inversions are: $(72), (75), (74), (54)$ - 4 of them. This one is even. For $(1423)$ we have $(42), (43)$ - again even. Finally, for $(154632)$ we get $(54), (53), (52), (43), (42), (63), (62), (32)$ - 8 inversions, even again. So their product is also even.
"Contemporary Abstract Algebra", by Dr. Joseph Gallian, describes an even permutation as: A permutation that can be expressed as a product of an even number of 2-cycles is called an even permutation.
Note: In the examples below, I will compose cycles from right-to-left.
The definition implies that in your first example: $(156)(234)$ is even. $$(156)(234)=(16)(15)(24)(23)$$ We can easily verify that. Start with $1$ and see how that is transformed. In the far right cycle, there is no $1$, so it maps to itself. Likewise in the next cycle (second to end). In the third cycle, $1 \mapsto 5$. Move to the fourth cycle. Now, there is no $5$, so it maps to itself. Thus, the start of the permutation looks like: $(15...)$. So now we consider how $5$ maps. Following the same procedure, $5 \mapsto 6$, and then $6 \mapsto 1$. Since we have returned to our starting number, we "close" the cycle. That is, the cycle now is: $(156)$. Starting with the next number not in the previous cycle, which is $2$, we get $2 \mapsto 3$, $3 \mapsto 4$, and $4 \mapsto 2$. So, the final cycle is: $(156)(234)$, which is the same as the starting one. Thus, $(156)(234) = (16)(15)(24)(23)$. Because there is an even number of 2-cycles (four), the permutation is even.
For the second example it's a little more difficult, but not impossible. We just start by following the same approach as above. Start with $1$, it maps to $5$ in the first cycle (far right). Then the $5$ maps to itself in the second. And finally, $5$ maps to $4$ in the last cycle. So, $1 \mapsto 4$. Continuing this, we obtain: $$(17254)(1423)(154632) = (14672)(3)(5)$$ By converting to disjoint cycles, it is now much easier transform to 2-cycles. $$(14672)(3)(5) = (12)(17)(16)(14)(13)(31)(15)(51)$$ [Note: A 1-cycle (identity for an element) can be expressed as the product of two 2-cycles by selecting one element (must be different) and creating the two 2-cycles. $(x) = (xy)(yx)$]
There are eight 2-cycles, so the above permutation is even.
atherton: Your product of 2-cycles for the second example is not correct.
Working right-to-left: $$(17254)(1423)(154632) = (14672)(3)(5)$$ $$(12)(13)(16)(14)(15)(13)(12)(14)(14)(15)(12)(17) = (1724635)$$ $$(14672)(3)(5) \ne (1724635)$$
Working left-to-right: $$(17254)(1423)(154632) = (1724635)$$ $$(12)(13)(16)(14)(15)(13)(12)(14)(14)(15)(12)(17) = (15364271)$$ $$(1724635) \ne (15364271)$$
Notice that working right-to-left on the product of 2-cycles is the same as working left-to-right of the original product. That - to my knowledge - is just a weird coincidence. Also, it is not valid to switch the order you compose mid-problem.