# Abstract Algebra: Equivalence Classes Under ~(conjugate) Relation

Sorry. Don't know the latex.

Let $g=(1 \ 2 \ 3 \ 4 \ 5 \ 6 \ 7 \ 8)$ in $S_8$

Find a specific permutation $f$ in $S_8$ so that $fgf^{-1} =(1 \ 5 \ 6 \ 3 \ 8 \ 7 \ 2 \ 4)$ (note: $fgf^{-1}$ is short for composition)

Use the conjugation rule $fgf^{-1} = (f(1)\; f(2)\; f(3)\; f(4)\; f(5)\; f(6)\; f(7)\; f(8))$.

EDIT: Upon request, I give more details (which in principle are already given by Berci in the comments): If f and g are any permutations, you can compute $fgf^{-1}$ by simply writing down $g$ as a product of disjoint cycles, and then applying $f$ entry-wise to this representation of $g$. So in your case, $fgf^{-1} = (f(1)\; f(2)\; f(3)\; f(4)\; f(5)\; f(6)\; f(7)\; f(8))$. Since your exercise states $fgf^{-1} = (1 \ 5 \ 6 \ 3 \ 8 \ 7 \ 2 \ 4)$, we get $$(f(1)\; f(2)\; f(3)\; f(4)\; f(5)\; f(6)\; f(7)\; f(8)) = (1 \ 5 \ 6 \ 3 \ 8 \ 7 \ 2 \ 4).$$ Now we can simply read off that $f(1) = 1$, $f(2) = 5$, $f(3) = 6$, $f(4) = 3$, $f(5) = 8$, $f(6) = 7$, $f(7) = 2$, $f(8) = 4$. That is (compare Berci's post) $$f = \begin{pmatrix}1 & 2 & 3 & 4 & 5 & 6 & 7 & 8\\1 & 5 & 6 & 3 & 8 & 7 & 2 & 4\end{pmatrix}$$ or as a product of disjoint cycles $$f = (2 \ 5\ 8\ 4\ 3\ 6\ 7).$$

• Great hint/useful to know! +1${}{}{}{}{}{}$ – Namaste Feb 26 '13 at 20:54
• Thanks for the reply. I'm not sure how to apply this. I'm really lost here. Maybe you could give me an example. – mathguy Feb 26 '13 at 22:28
• The example is already given. This is a general fact, that conjugation by $f$ within $S_n$ means that we apply $f$ to each point in the cycle factorisation. – Berci Feb 27 '13 at 10:55
• Ok, so, assuming composition of permutation is written from right to left (as for functions), let's follow the cycle structure of $fgf^{-1}$, starting from (e.g.) $f(1)$: Where is it mapped? $f(1)\mapsto fgf^{-1}(f(1))=fg(1)=f(2)$ since $g=(1\ 2\ 3\ 4\ 5\ 6\ 7\ 8)$ mapping $1\mapsto 2$.. The same way, $fgf^{-1}(f(2)=f(3)$, and so on... – Berci Feb 27 '13 at 10:58
• This is great! Thanks so much! – mathguy Feb 28 '13 at 3:11

Let $f$ map $1\mapsto 1,\ 2\mapsto 5,\ 3\mapsto 6,\ 4\mapsto 3,\ ...,\ 8\mapsto 4$.

• Thanks for the reply. I don't know how you found this. Could you be more specific or explain the mechanics of how you obtained the answer? – mathguy Feb 26 '13 at 22:30
• Also check out the other answer. We keep the order (though it could be rotated) and map the $8$ values $(1,2,3,4,5,6,7,8)$ given in the cycle of $g$ to the desired other $8$ values $(1,5,6,3,8,7,2,4)$ in the given order. Please check that this $f$ will satisfy the requirement. – Berci Feb 27 '13 at 10:53