Formula for conjugating a permutation

Question

I wish to conclude that $g(12)(34)g^{-1}$ is the permutation $(g(1)g(2))(g(3)g(4))$ for $g \in S_4$. I found a proof of this fact for a general permutation here, but it is quite long and I would like a quick justification. I have tried decomposing the permutation into transpositions but couldn't see how it was helpful.

Answer 1

Note if $\sigma$ sends $i$ to $j$ then $g\sigma g^{-1}$ sends $g(i)$ to $g(j)$, because

$$(g\sigma g^{-1})\big(g(i)\big)=g(\sigma(g^{-1}(g(i)))=g(\sigma(i))=g(j). $$

Therefore, applying the permutation

$$\tau=g(12)(34)(5)(6)\cdots(n)g^{-1}$$

to the values

$$g(1),~g(2)~,g(3)~,g(4),~g(5),~g(6),~\cdots,~g(n)$$

yields the values

$$\color{Green}{g(2),~g(1)},~\color{Blue}{g(4)~,g(3)},~g(5),~g(6),~\cdots,~g(n) $$

(in that order, obviously).

That is, $\tau$ swaps $g(2)\leftrightarrow g(1)$ and $g(3)\leftrightarrow g(4)$ and leaves the rest $g(5),g(6),\cdots,g(n)$ alone.