How to proof (step-by-step) that

Question

If $g \in S_n$ and let be $1\leq k \leq n$. Prove that $g(12\cdots k)g^{-1}=(g(1)g(2)\cdots g(k))$.

In $S_n$ the cycle $(12\cdots k)$ means

• $1\longmapsto 2$
• $2\longmapsto 3$
• $\vdots$
• $k-1\longmapsto k$

Answer 1

Hint

There is no specific technic. You must do it brute force calculation.

For example, if $i\notin\text{Supp}(g):=\{j\in \{1,...,n\}\mid g(j)\neq j\},$ then $$g(1\ ...\ k)g^{-1}i=\begin{cases}i&i\notin\{1,...,k\}\\g(i+1)&i\in\{1,...,k-1\}\\ g(1)&i=k\end{cases}.$$

Now if $i\notin\{g(1),...,g(k)\}$ then $i\notin \{1,...,k\}$ and $(g(1)\ ...\ g(k))i=i$. If $i\in\{g(1),...,g(k)\}$ then $i=g(i)$ and thus $$(g(1)\ ...\ g(k))i=\begin{cases}g(i+1)&i\in\{1,...,k-1\}\\ g(1)&i=k\end{cases}.$$

Therefore $$g(1\ ...\ k)g^{-1}i=(g(1)\ ...\ g(k))i,$$ for all $i\notin \text{Supp}(g).$ Now, I let you prove the equality whenever $i\in \text{Supp}(g).$