Express $(152)(45)$ as a product of powers of $(12)$ and $(12345)$ I'm looking at an old group theory question that I botched, and I'm wondering how I should proceed. The question asks to write $(152)(45)$ as a product of powers of $(12)$ and $(12345)$.
The 2-cycle $(45)$ is easily enough written as $(12345)^3(12)(12345)^{-3}$. It's also easy to break down the 3-cycle $(152)$ into $(12)(15)$, which partly satisfies the requirements. I just need to figure out how to express $(15)$ as a product of exponents of $(12)$ and $(12345)$. This is where I'm a bit stuck.
I thought about expressing $(15)$ as $(12345)^j(12)^k(12345)^{-m}$ for positive integers $j$, $k$ and $m$, but I couldn't see if/how that would help. Then I tried $(45)(34)(23)(12)$ (knowing that I could decompose each of those transpositions as I did above) but it doesn't produce a 2-cycle. Is there another way to make $(15)$ that I'm missing?
Thanks.
 A: $(15)=(15432)(12)(12345)$ if I'm not mistaken.
Edit: To clarify, this corresponds to $m=j=4$ I guess.
Edit #2: It might be worth mentioning that $\sigma(12)\sigma^{-1}=(\sigma(1)\sigma(2))$ for any permutation $\sigma$ (and more generally, this holds for any cycle type).
A: Have you ever played the Game of 15 (Fifteen Puzzle)? You randomly move the tiles and then have to solve the puzzle: move them back to the initial position.
The reason I am asking you is that one way (not always optimal) to solve the puzzle is row by row. It is quite easy to fill the first two rows with the right tiles using allowed permutations, and then move the last two rows in a cycle and permute them until they eventually take the right positions.
Similarly you can solve your problem. Your permutation is $12345\rightarrow 51324$ and you have two cycles to represent it.
Let us do it step by step.
We have initial permutation: $12345$.
First, put $1$ after $5$: no action ($1$ is already after $5$ because we have the large cycle $(12345)$ which we are to apply at the very last step).
Then, put $3$ after $1$ by swapping $3$ and $2$: $(12345)(12)$ $\rightarrow32451$.
Now, put $2$ after $3$ and $4$ after $2$: already there.
Now cycle to the final position: $(12345)^3$ $\rightarrow 51324$.
To sum up: we have obtained a solution: $$(152)(45)=(12345)(12)(12345)^3$$ which looks quite optimal to me. Just compare it to what you currently have: $(152)(45)=(12)(12345)^4(12)(12345)^4(12)(12345)^2$.
