Why is the permutation $(a,c,d,e)(a,b)=(a,d)(b,c,e)$

Question

I'm working through a proof in my notes. We already know that the transposition $(a,b)\in G$ and $(a,b,c,d,e)\in G$, where $G$ is a group of permutations of the elements $a,b,c,d,e$, so it's a subgroup of $S_5$.

By closure, we have $(a,b,c,d,e)(a,b)=(a,c,d,e)\in G$, which makes perfect sense to me, since $a$ is mapped to $b$ which is then mapped to $c$ from the left transposition, so overall $a$ is mapped to $c$, and so on for the other elements.

Again by closure, we have $(a,c,d,e)(a,b)=(a,d)(b,c,e)\in G$. I think this may be a typo, but if it is true, can someone please explain why, because there's no way $a$ can be mapped directly to $d$. I'm pretty sure $(a,c,d,e)(a,b)=(a,b,c,d,e)$. But the proof in my textbook relies on this single element of $G$ to have order $6$, but $(a,b,c,d,e)$ has order $5$.

Answer 1

You are correct, $(a,c,d,e)(a,b)=(a,b,c,d,e)$. In fact, this is easy to see since $(a,b)$ is its own inverse, and therefore by what you found previously, $$(a,c,d,e)(a,b)=(a,b,c,d,e)(a,b)(a,b)=(a,b,c,d,e).$$

Answer 2

The proof relies on finding elements in $G$ of orders $4,5$ and $6$, therefor $|G|=lcm(4,5,6)=60$, so either $G=A_5$ or $G=S_5$, but $(a,b)\in G$ and $(a,b)\notin A_5$. Therefore $G=S_5$.

Although it isn't true that $(a,c,d,e)(a,b)=(a,d)(b,c,e)$, we can still show that $(a,d)(b,c,e)\in G$ since $(a,b,c,d,e)(a,c,d,e)=(a,d)(b,c,e)$, hence $(a,d)(b,c,e)\in G$.