# Thread: composition of permutation groups

1. ## composition of permutation groups

hi, could anyone explain to me how you combine those? i just don't get it..
for example:
(1 2 3) o (1 2 3) = (1 2 3)
(2 1 3)__(3 2 1)___(3 1 2)
or
(p q r) o (p q r) = (p q r)
(q r p)__(r p q)___(p q r)
(ignore the underscores)
thanks!

2. You know how function composition works.
If $f(x) = x^2 \,\& \,c(x) = \cos (x)$ then $f \circ c(x) = \cos ^2 (x)\,\& \,c \circ f(x) = \cos \left( {x^2 } \right)$.
The “inter” function operates first.
So if $p = \left( {\begin{array}{*{20}c} 1 & 2 & 3 & 4 \\ 3 & 2 & 4 & 1 \\
\end{array}} \right)\,\& \,q = \left( {\begin{array}{*{20}c} 1 & 2 & 3 & 4 \\
2 & 4 & 1 & 3 \\ \end{array}} \right)$
then in $p \circ q$ the q permutation acts first and then the p acts.
Thus we have $p \circ q = \left( {\begin{array}{*{20}c} 1 & 2 & 3 & 4 \\
2 & 1 & 3 & 4 \\ \end{array}} \right)\,\& \,q \circ p = \left( {\begin{array}{*{20}c} 1 & 2 & 3 & 4 \\ 1 & 4 & 3 & 2 \\\end{array}} \right)\,$

3. so when you combine P and Q you have
1->2 followed by 4->1 gives us 1->2
2->4 followed by 3->4 gives us 2->1
3->1 followed by 2->2 gives us 3->3
4->3 followed by 1->3 gives us 4->4
is that right? but I just dont get it how you end up with those numbers. do you multiply, add, subtract or do something else?

4. Originally Posted by fzr
so when you combine P and Q you have
1->2 followed by 4->1 gives us 1->2
2->4 followed by 3->4 gives us 2->1
3->1 followed by 2->2 gives us 3->3
4->3 followed by 1->3 gives us 4->4
is that right? but I just dont get it how you end up with those numbers. do you multiply, add, subtract or do something else?
NO, not that at all.
When we do $p \circ q$ we get
1->2 followed by 2->2 gives us 1->2
2->4 followed by 4->1 gives us 2->1
3->1 followed by 1->3 gives us 3->3
4->3 followed by 3->4 gives us 4->4.

5. oh now i get it... thanks m8