How to calculate "$(1, 2, 3, 4, 5) \cdot (1, 2)$"?

Question

For a $k$-cycle $(1, 2, 3, 4, 5)$ and a transposition $(1, 2)$.Calculation of "$(1, 2, 3, 4, 5) \cdot (1, 2)$". We all know that groups satisfy the associative property, i.e. $(a \cdot b) \cdot c=a \cdot (b \cdot c)$.

I will get $ \left \{ 1, 2, 3, 4, 5 \right \} \longmapsto \left \{ 2, 3, 4, 5, 1 \right \} $, when first calculate the $k$-cycle. Then after the transposition $(1, 2)$ we get $ \left \{ 2, 3, 4, 5, 1 \right \} \longmapsto \left \{ 3, 2, 4, 5, 1 \right \} $.

But according to the associative property, I first calculate the transposition $(1, 2)$ we get $ \left \{ 1, 2, 3, 4, 5 \right \} \longmapsto \left \{ 2,1, 3, 4, 5 \right \} $. Then after the $k$-cycle $(1, 2, 3, 4, 5)$ we get $ \left \{ 2, 1, 3, 4, 5 \right \} \longmapsto \left \{ 1, 3, 4, 5, 2 \right \} $.

(1)Why are the results of these two calculations different? Which step is wrong? (2)Which is the correct calculation?

Answer 1

You seem to be following the convention that $$(1,2,3,4,5)\cdot(1,2)$$ means "do $(1,2,3,4,5)$ first, then do $(1,2)$." That's fine, but you need to state it explicitly.

Your computation is done incorrectly. You correctly say that $(1,2,3,4,5)$ will result in the permutation (using two-line notation) $$\left(\begin{array}{ccccc} 1&2&3&4&5\\ 2&3&4&5&1 \end{array}\right),$$ which you write as "$\{1,2,3,4,5\}\mapsto\{2,3,4,5,1\}$". Note that this is bad notation, because sets (which are denoted with curly brackets) do not have any inherent order....

Then you say that applying the permutation $(1,2)$ exchanges the numbers that are in the first and second position of this, which is why you write

$\{2,3,4,5,1\}\mapsto \{3,2,4,5,1\}$.

That is not what $(1,2)$ does. The permutation does not exchange "whatever is in the first position with whatever is in the second position". This permutations exchanges $1$ and $2$. So what we have is: $$\begin{array}{c} {}\\ (1,2,3,4,5)\\ {}\\ (1,2)\\ {}\\ \end{array} \left(\begin{array}{ccccc} 1&2&3&4&5\\ &&\downarrow&&\\ 2&3&4&5&1\\ &&\downarrow&&\\ 1&3&4&5&2 \end{array}\right)$$ that is, the result is the permutation that you would write as

$\{1,2,3,4,5\} \mapsto \{1,3,4,5,2\}$.

So your first error lies in how to interpret "$(1,2)$". It's not "switch what's in position one with what's in position $2$" (just like $(1,2,3,4,5)$" does not mean "move what's in position 1 to position 2, what's in position 2 to position 3, etc.". The permutation $(1,2)$ means "replace $1$, wherever it is, with $2$; and replace $2$, wherever it is, with $1$". And $(1,2,3,4,5)$ means "replace $1$ with $2$, $2$ with $3$, $3$ with $4$, $4$ with $5$, and $5$ with $1$.

Your second error lies in invoking "associativity". There is no associativity at play here; associativity requires three or more permutation, you only have two. Instead, you are trying to compute $$(1,2)\cdot(1,2,3,4,5)$$ which would be an instance of commutativity,, not associativity. In fact, these two permutations to not commute.

Then you miscompute the result of multiplying what you write as

$\{1,2,3,4,5\} \mapsto \{2,1,3,4,5\}$

which in two-line notation would be $$\left(\begin{array}{ccccc} 1&2&3&4&5\\ 2&1&3&4&5 \end{array}\right),$$ by $(1,2,3,4,5)$.

Now you seem to be interpreting the product by $(1,2,3,4,5)$ as "move what's in position 2 to position 1, what's in position 3 to position 2" etc., which wouldn't even be the correct thing to do under you (already incorrect) interpretation of the permutation. Instead, again understanding that multiplying by this permuation means "replace $1$ by $2$, replace $2$ by $3$", etc, we would get: $$\begin{array}{c} {}\\ (1,2)\\ {}\\ (1,2,3,4,5)\\ {}\\ \end{array} \left(\begin{array}{ccccc} 1&2&3&4&5\\ &&\downarrow&&\\ 2&1&3&4&5\\ &&\downarrow&&\\ 3&2&4&5&1 \end{array}\right)$$ So you would get what you'd write as

$\{1,2,3,4,5\} \mapsto \{3,2,4,5,1\}$

and not, as you claim, the map $$\left(\begin{array}{ccccc} 1&2&3&4&5\\ 1&3&4&5&2 \end{array}\right)$$ which you write as

$\{1,2,3,4,5\}\mapsto \{1,3,4,5,2\}$.

This is an incorrect calculation as well.

Answer 2

You are confusing associativity, which has nothing to do with your question, and commutativity, which indeed does not hold for permutations.

Answer 3

Let $\sigma=(12345)(12)=\sigma_1\circ \sigma_2$, with $\sigma_1=(12345)$ and $\sigma_2=(12)$. The correct calculation is as follows: $$ \sigma(1)=(\sigma_1\circ \sigma_2)(1)=\sigma_1( \sigma_2(1))=\sigma_1(2)=3, $$ and similarly $\sigma(2)=2,\sigma(3)=4,\sigma(4)=5$ and $\sigma(5)=1$. Hence $2$ is fixed by $\sigma$, and we have $\sigma=(2)(1345)=(1345)$, since we delete the fixed ones in the notation of cycles.