// Numbas version: finer_feedback_settings {"name": "Conjugate of permutations", "extensions": ["permutations"], "custom_part_types": [{"source": {"pk": 68, "author": {"name": "Christian Lawson-Perfect", "pk": 7}, "edit_page": "/part_type/68/edit"}, "name": "Write a permutation in cycle notation", "short_name": "write-a-permutation-in-cycle-notation", "description": "

The student must write a permutation in cycle notation. You can optionally require the cycles to be disjoint and/or all of length 2.

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You could find $\\tau \\sigma \\tau^{-1}$ by first finding $\\tau^{-1}$ and then performing the composition by hand, but there's a quicker way: $\\tau \\sigma \\tau^{-1}$ has the same cycle structure as $\\sigma$, but each $n$ is replaced by $\\tau(n)$.

\n
Proof
\n

Assume, without loss of generality, that $\\sigma$ is a single cycle $(b_1, \\dots, b_n)$, and let $\\pi = \\tau \\sigma \\tau^{-1}$. Look at what happens to $\\tau(b_i)$ when acted on by $\\sigma$:

\n

\\[ \\tau(b_i) \\overset{\\tau^{-1}}\\mapsto b_i \\overset{\\sigma}\\mapsto b_{i+1} \\overset{\\tau}\\mapsto \\tau(b_{i+1}) \\]

\n

So, for any $1 \\lt i \\leq n$, $\\tau(b_i) \\overset{\\pi}\\mapsto \\tau(b_{i+1})$.

\n

For any $x \\notin \\{b_1,\\dots,b_n\\}$, $\\sigma(x) = x$, so $\\pi(x) = x$.

\n

Hence, $\\pi$ is the cycle $(\\tau(b_1), \\dots, \\tau(b_n))$.

\n

a)

\n

$\\var{conjugate1}$

\n

b)

\n

$\\var{conjugate2}$

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Where $\\sigma = \\var{sigma1}$ and $\\tau = \\var{tau1}$, write down the permutation $\\tau \\sigma \\tau^{-1}$.

\n

$\\tau \\sigma \\tau^{-1}$ = [[0]]

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Where $\\sigma = \\var{sigma2}$ and $\\tau = \\var{tau2}$, write down the permutation $\\tau \\sigma \\tau^{-1}$.

\n

$\\tau \\sigma \\tau^{-1}$ = [[0]]\n

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