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So I am new to cycle notation and needless to say I am finding it a bit confusing. I know that when computing these, I need to work right to left=.

Compute each of the following:

a. $$(12)(1253)$$

1→2, 2→5, 5→3 1→2

So I think this equals (33) because the second term will send both 1 and 2 to 3

$$(12)(1253)=(1253)$$

b. $$(1423)(34)(56)(1324)$$

I am not sure if this is the right method to calculate this when I have more than 2 but I tried this:

1→3,3→2,2→4 5→6 3→4 1→4,4→2,2→3

Maybe I apply the last term to all of them? $$(1423)(34)(56)(1324)$$ $$(4444)(44)(56)$$

And repeat: $$(1423)(34)(56)(1324)$$ $$($4444)(44)(56)$$ $$(4444)(44)$$

And again: $$(1423)(34)(56)(1324)$$ $$(4444)(44)(56)$$ $$(4444)(44)$$ $$(4444)$$

This doesn't look right to me though...

c $$(1254)(13)(25)^2$$

So I assume that $$(1254)(13)(25)^2=(1254)(13)(25)(25)$$

Since I don't think I did b correctly, I am going to try it another way:

$$(1254)(13)(25)(25)$$ $$=(1254)(13)(25)$$ $$=(1254)(13)$$ $$=(3254)$$

This looks like it could maybe me correct? So I am attempting b again: $$(1423)(34)(56)(1324)$$ $$=(1423)(34)(56)$$ $$=(1423)(34)$$ $$=(4424)$$

Math Major
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  • The notation $(33)$ would mean the trivial permutation if it means anything. The notation $(4424)$ is completely meaningless. Basically, inside a single cycle, you never want the same number to appear more than once. Remember, these are permutations ("shuffles "), so there is no such thing as sending both $1$ and $2$ to $3$. Every number is sent somewhere, and no two numbers are sent the same place. – Arthur Mar 05 '15 at 23:16
  • So are you saying that a should be simply (3) and b (42)? – Math Major Mar 05 '15 at 23:30
  • In the first part of your question, you said "because the second term will send both 1 and 2 to 3". Note that each of those cycles indicates a bijection between the set of numbers (objects) you are working on, and their multiplication is nothing but composition of bijections. So, it is impossible for a bijection (in this case cycle) to map two different things to the same thing! – Kaveh Mar 05 '15 at 23:49

2 Answers2

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The permutation $\pi = (12)(1253)$ can be written as disjoint cycles the following way. Let $\sigma = (12)$ and $\tau = (1253)$. Thus, our permutation $\pi$ is the composition $\sigma \circ \tau$, which we'll write as $\sigma\tau$.

To see what $\pi(1)$ is, we need to know that

$$\pi(1) = \sigma(\tau(1)) = \sigma(2) = 1.$$

Thus, we know $\pi$ starts off as $(1)$ in cycle notation. Since $\pi$ fixes $1$, we'll move on to $2$, and compute

$$\pi(2) = \sigma\tau(2) = \sigma(\tau(2)) = \sigma(5) = 5.$$ Thus, we know that $\pi$ looks like $(1)(25 \ldots )$ so far. Next, we'll see where $\pi$ sends $5$.

$$\pi(5) = \sigma\tau(5) = \sigma(\tau(5)) = \sigma(3) = 3.$$ So now we know $\pi$ looks like $(1)(253 \ldots)$. We'll compute

$$\pi(3) = \sigma\tau(3) = \sigma(\tau(3)) = \sigma(1) = 2,$$ and we're back to where we began, in our second cycle, and can close it up: $(1)(253)$.

Of course, $\pi$ fixes $4$, since both $\sigma$ and $\tau$ fix $4$. Thus we can write

$$\pi = (1)(253)(4) = (253),$$ since we drop cycles containing only a single number (unless it's the identity of $S_n$, in which case we usually write $(1)$ if we need to write it in cycle notation).

pjs36
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    This was really helpful! Following your method I got: a. (253) b. (12)(56) c. (13254) does this look correct? – Math Major Mar 06 '15 at 00:45
  • @MathMajor I'm glad it was helpful! I always learned to trace paths "backwards" through the various cycles, and my formal answer amounts to this. Anyway, those look great for b) and c), provided they really do mean to square only $(25)$, and not the entire product. If it is the entire product squared, I got $(12)(34)$, which is of course subject to errors. Nicely done! – pjs36 Mar 06 '15 at 02:04
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Use the expansion $(ab)(bcd)=(abcd)$, so $(12)(1253)=(12)(12)(253)=(253)$.

Bill Kleinhans
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