Permutations: If $3\le m\le n$, Find $\sigma\tau^{-1}$ for the cycles $\sigma=(1\,2\,\ldots,m-1)$ and $\tau=(1\,2\,\ldots,m-1\,m)$

Question

My attempt at the question:

$\tau^{-1} =$ $$m\mapsto m-1\mapsto m-2\mapsto \cdots 2\mapsto 1$$

So starting from

$$1\mapsto 2\mapsto 3\mapsto 4\mapsto \cdots \mapsto m-1\mapsto m$$

Apply $\tau^{-1}$

$$m\mapsto 1\mapsto 2\mapsto 3\mapsto \cdots\mapsto m-2\mapsto m-1$$

Apply $\sigma$

$$m\mapsto 2\mapsto 3\mapsto 4\mapsto \cdots\mapsto m-1\mapsto 1$$

So $\sigma\tau^{-1}= (1,m)(2)(3)\cdots(m-1)$

I'm very new to abstract algebra and am not sure how to incorporate the given fact that $3 \le m\le n$

Answer 1

You have done it right.

The algebra here, fortunately, is not abstract. As $\tau$ moves all numbers forward by 1 step (except the last which is sent to 1) $\tau^{-1}$ moves backward by 1 (except . . .).

Now $\sigma$ again moves forward. So $\sigma\circ \tau^{-1}$ should largely restore everything to its place, except the numbers at the edge. Pay attention to those edges. So this is not really abstract as it can be described with words, rather than symbols. Here the number $n$ plays no role.

Answer 2

Notice $\tau$ just shifts every element to the right once, moving $m$ back over to the first entry. $\sigma$ does the same as $\tau$ except it pretends the $m$th entry isn't there: it rotates all elements to the right once except the $m$th entry, pushing $m-1$ into the first entry.

Applying $\tau^{-1}$ is supposed to get you back where you started if you had applied $\tau$, so that's rotating everything left once (putting the first place value in the $m$th place). We then have $$\tau^{-1}(1, 2, \ldots, m) = (2, 3, 4, \ldots, m-1, m, 1)$$ and then have $$\sigma \left(\tau^{-1}(1, 2, \ldots, m)\right) = (m, 2, 3, 4, \ldots, m-2, m-1, 1)$$ which just ultimately swaps the first place with the $m$th place, that is, $$\sigma \tau^{-1}= (1 m).$$

Answer 3

The cycle $\sigma=(1,\,2,\,\ldots,\, m-1)$ can be written as $$ 1 \mapsto 2\mapsto 3 \mapsto \cdots \mapsto (m-1) \mapsto 1 \quad\text{and } \quad m\mapsto m,\ m+1\mapsto m+1,\ \ldots,\ n\mapsto n. $$ And $\tau = (1,\,2,\,3,\,\ldots,\,m-1,\, m)$ is $$ 1 \mapsto 2\mapsto 3\mapsto \cdots \mapsto (m-1)\mapsto m \mapsto 1 $$ So $\tau^{-1}$ is $$ 1 \mapsto m \mapsto (m-1) \mapsto (m-2) \mapsto\cdots \mapsto3\mapsto 2\mapsto 1. $$ Applying $\tau^{-1}$ takes $1$ to $m$ and then applying $\sigma$ takes $m$ to $m$.

Applying $\tau^{-1}$ takes $m$ to $m-1$ and then applying $\sigma$ takes $m-1$ to $1$.

So $1$ and $m$ are interchanged.

Applying $\tau^{-1}$ takes $m-1$ to $m-2$ and then applying $\sigma$ takes $m-2$ to $m-1$.

So $m-1$ is fixed by $\sigma\tau^{-1}$.

Applying $\tau^{-1}$ takes $m-2$ to $m-3$ and then applying $\sigma$ takes $m-3$ to $m-2$.

So $m-2$ is fixed by $\sigma\tau^{-1}$.

Just keep going like that until you understand what $\sigma\tau^{-1}$ does.