How is $g(12)(34)g^{-1} = (g(1) g(2))(g(3)g(4))$ Firstly I think (please confirm) the notation $g(x)$ (for $x \in \{1,2,3,4\}$) caters to the position $y \in \{1,2,3,4\}$ such that $x \to y$.

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*Is it true that $g((12)) = (g(1) g(2))$? I can not wrap around my head if this is even defined. If so, an explanation would be really helpful.

*How is $g(12)(34)g^{-1} = (g(1) g(2))(g(3)g(4))$ where $g \in S_4$?

I am fine if hints or heuristics are provided so I can think how it works. A definition for $(1)$ would enable me to think $(2)$.
 A: The definitions you may be missing: On the set $A=\{1,2,3,4\}$:

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*$g$ is any permutation (i.e. any bijection $A\to A$). $g^{-1}$ is its inverse.

*$(1 2)$ is the permutation that swaps $1$ with $2$ (i.e. maps $1$ to $2$, $2$ to $1$, $3$ to itself and $4$ to itself). Similarly, $(3 4)$ is the permutation that swaps $3$ with $4$. Similarly again, $(g(1) g(2))$ is a permutation that swaps $g(1)$ with $g(2)$ - whatever those two elements of $A$ are. I am leaving to you to guess what is $(g(3) g(4))$.

*Multiplying permutations means composing them into new permutations by applying the composed permutations one after another, right-to-left. This means: if $g, h$ are permutations, then $gh$ is a permutation such that $gh(x)=g(h(x))$.

Does this help? Please try to take it from there, and the rest of the solution is hidden in the spoiler below.

 Let's see how $g(1 2)(3 4)g^{-1}$ works. What does it map the element $g(1)\in A$ into? Remember we apply the maps right-to-left, so, first, $g^{-1}$ acts on $g(1)$, mapping it into $1$. Then, $(3 4)$ maps $1$ into itself. Then, $(1 2)$ maps $1$ into $2$. Finally, $g$ maps $2$ into $g(2)$. Can you finish this now by proving that $g(2)$ also maps to $g(1)$, that $g(3)$ maps to $g(4)$ and $g(4)$ maps to $g(3)$? In other words, the whole permutation swaps $g(1)$ with $g(2)$ and $g(3)$ with $g(4)$, which is the same thing that $(g(1) g(2))(g(3) g(4))$ does.

A: In general, when you conjugate a $k$-cycle (say $(12\dots k)$) by a permutation $\sigma$, you get $(\sigma(1)\sigma(2)\dots\sigma(k))$.
Secondly, conjugation by a permutation is a homomorphism.
