If $h$ is the $k$-cycle and g element of $S_n$, prove that $ghg^{-1}=(g(a_1)g(a_2)...g(a_k))$

Question

$g$ element of $S_n$, prove that $g(a_1 a_2 ... a_k)g^{-1}=(g(a_1)g(a_2)...g(a_k))$

I'm more concerned with how I can derive the prove of this question

Please clarify your question: what are $h$ and $a_i$? From the general setting, I assume $a_i$ are transpositions and $h=a_1a_2...a_k$ — AspiringMathematician, Apr 04 '17 at 03:08
Possible duplicate of Why are two permutations conjugate iff they have the same cycle structure? — Wang Kah Lun, Apr 04 '17 at 03:13

score 2 · Answer 1 · answered Apr 04 '17 at 03:10

2

Hint:

$$g(a_1a_2\dots a_k)g^{-1}(g(a_i))=g(a_1a_2\dots a_k)(a_i)=g(a_{i+1})$$

answered Apr 04 '17 at 03:10

Wang Kah Lun

1 Answers1