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

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.

• I think you're on the right track. To me, the natural candidate for a conjugate of $\lambda$ would be $\lambda^{2}$, but that's a guess on my part. (+1) Oct 15, 2020 at 0:31
• Using the map $\lambda\mapsto\lambda^2$, you can iterate to find cycles: $\lambda^1\mapsto\lambda^2\mapsto\lambda^4\mapsto\lambda^1$ and $\lambda^3\mapsto\lambda^6\mapsto\lambda^5\mapsto\lambda^3$. This gives the cycle decomposition $(124)(365)$. Oct 15, 2020 at 3:49

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 $$\langle\lambda\rangle$$.