# Lesson 24 Solutions

Exercise 1) Write down the permutation given by the product $(1, 2, 3)\cdot (2, 3, 4)$.

Solution: Recall what the notation $(1, 2, 3)$ means.  It means “move the element in compartment 1 to compartment 2, that in 2 to 3, and that in 3 to 1”.  Similarly for $(2, 3, 4)$.  Thus, let us see what happens if we first apply $(2, 3, 4)$ and then apply $(1, 2, 3)$.  We can do this one element at a time.  Let’s see where the element in compartment 2 goes.  First, it goes from 2 to 3 (via the permuation $(2, 3, 4)$, and then it goes from 3 to 1 (via the permutation $(1, 2, 3)$).  Thus, whatever our final permutation looks like, it’ll have something like $(..., 2, 1, ...)$ in it, because we know that whatever is in 2 gets sent to 1.  Let us now follow 1 through.  Whatever is in compartment 1 doesn’t get touched by the first permutation, and then in the second permutation it gets sent to 2.  Thus we know that whatever is in compartment 1 goes to compartment 2, and so 1 and 2 actually just get swapped (because we’ve already shown that whatever is in 2 goes to 1).  So part of our permutation will include $(1, 2)$.

Now let us foll0w 3 through.  First it gets sent to 4 (by $(2, 3, 4)$), and then compartment 4 is not touched by the second permutation.  Thus, our total permutation looks something like $(1, 2)(..., 3, 4...)$.  Let us now follow 4 through.  By the first permutation, whatever is in 4 gets sent to 2, and then the second permutation sends whatever is in 2 to 3, so the total effect on compartment 4 is to send its contents to compartment 3.  Thus 3 and 4 just get swapped, since 4 goes to 3 and 3 goes to 4.  Thus we have that $(1, 2, 3)\cdot (2, 3, 4)=(1,2)(3, 4)$.  Note that it doesn’t matter in which order we write the two swaps on the right hand side of this equation, because they don’t interact with each other.

Exercise 2)  What is the inverse of the permutation $(2, 3, 4)$?  What about $(1, 2, 3, 4)$?  What about $(1, 2, 4)\cdot (2, 3, 1)$?

Solution: Let’s first find the inverse of $(2, 3, 4)$.  Well, as mentioned in the lesson, we simply reverse the order of the contents of the parentheses, so that the inverse is $(4, 3, 2)$.  We can check this by combining them: $(2, 3, 4)\cdot (4, 3, 2)$.  The first sends 4 to 3, and the second sends 3 to 4, so it fixes 4.  Similarly, it fixes 3 and 2, so that it is indeed the identity permutation.  Similarly, $(4, 3, 2)\cdot (2, 3, 4)$ is the identity, so that we’ve successfully found the identity.  Similarly, the inverse of $(1, 2, 3, 4)$ is just $(4, 3, 2, 1)$, for the exact same reasons.

Now, we can find the inverse of $(1, 2, 4) \cdot (2, 3, 1)$, in two different ways.  Perhaps the more enlightening way would be to first re-express this product as a single permutation, much like we did in exercise 1.  In particular, we have $(1, 2, 4)\cdot (2, 3, 1)=(2, 3)(1, 4)$, where we used the same logic as in exercise 1 (namely, just following each element through the two permutations).  Then it’s automatic that the inverse of this element is just $(2, 3)(4, 1)$, since the order doesn’t matter (because the swaps don’t interact with each other in this case) and since the inverse of a swap is just the same swap.  The other way that we could find this is to follow the procedure that we derived in the lesson.  Namely, we reverse the order of the permutations themselves, and then reverse the order of their internal elements, so that the inverse of $(1, 2, 4)\cdot (2, 3, 1)$ is $(1, 3, 2)\cdot (4, 2, 1)$.  Now, if we re-express this last product as a single permutation, we’ll find the same result that we found above, namely that $(1, 3, 2)\cdot (4, 2, 1)=(2, 3)(4, 1)$.

Exercise 3) Think about how this construction can be generalized to a set with arbitrarily many (but only finitely many) elements.  (Okay, this isn’t really an exercise, but it’ll help to warm up for the next lessons).

Solution:  Well, this isn’t really an exercise, so I can’t really provide a solution, but hopefully these few sentences will help guide the way.  To generalize what we’ve done in the case of permuting 4 elements to the case of permuting an arbitrary (finite) number of elements, we simply consider a wider class of permutations.  Namely, if our set has 5 elements, we can consider permutations like $(1, 2, 5, 3, 4)$ or $(1, 5)$ or $(3, 4, 5)$ or $(5, 2, 1, 3)$.  For 67 elements, well, there are a lot more options, but the ideas are the exact same.  Note that the group of permutations of 2 objects is pretty boring.  Namely, all we have is the identity permutation and the swap permutation $(1, 2)$.  We’ll discuss more general permutations in the next lesson.

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