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: $$(1423)(34)(56)(1324)$$ $$=(1423)(34)(56)$$ $$=(1423)(34)$$ $$=(4424)$$
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).