Find a normalizer of order 3 in $S_7$ of cyclic subgroup in $S_7$

Question

Let $\lambda = (1 \ 2 \ 3 \ 4 \ 5 \ 6 \ 7)$ be a cycle in $S_7$. Find an element of order 3 in $S_7$ that normalizes the cyclic subgroup generated by $\lambda (\langle\lambda\rangle)$.

I have written the elements in $\langle \lambda\rangle= \{1, (1 \ 2 \ 3 \ 4 \ 5 \ 6 \ 7), (1 \ 3 \ 5 \ 7 \ 2 \ 4 \ 6),(1 \ 4 \ 7 \ 3 \ 6 \ 2 \ 5),(1 \ 5 \ 2 \ 6 \ 3 \ 7 \ 4),(1 \ 6 \ 4 \ 2 \ 7 \ 5 \ 3),(1 \ 7 \ 6 \ 5 \ 4 \ 3 \ 2) \}$

Let's $\sigma$ be the element, then we have $\sigma\lambda\sigma^{-1} = \lambda^{i}$, with $i$ being an integer.

Since the element has order 3, it can either be a single or double 3-cycle, I have tried with a few singles and found that none of them works on $(1 \ 2 \ 3 \ 4 \ 5 \ 6 \ 7)$ because it permutes at most 3 elements in the cycle, which doesn't fall into the set $\langle\lambda\rangle $.

So the answer must be an double 3-cycle, or am I doing anything wrong?

I have not learned the methods for finding things like this, so I would really appreciate some guidance. Thank you.

Answer 1

Let $\sigma=(235)(476)$ (Refer here for the method to find $\sigma)$. By direct calculation, it can be shown that $\sigma\lambda\sigma^{-1}=\lambda^2$.
Hence $$\sigma\lambda^i\sigma^{-1}=(\sigma\lambda\sigma^{-1})^i=\lambda^{2i}\in \langle\lambda\rangle.$$ So we have $\sigma\langle\lambda\rangle\sigma^{-1}\subseteq\langle\lambda\rangle$. Since both sets have the same size and are finite, we have $\sigma\langle\lambda\rangle\sigma^{-1}=\langle\lambda\rangle$.
Thus $\sigma$ is an element of order $3$ that normalizes $\langle\lambda\rangle$.