$$ \sigma = \begin{pmatrix} 1& 2& 3& 4& 5& 6& 7& 8& 9& 10& 11& 12& 13& 14 \\ 7& 5& 2& 6& 4& 3& 10& 13& 11& 9& 1& 8& 14& 12 \\ \end{pmatrix} $$ Let $\sigma \in S_{14}$ and $\langle \sigma \rangle =:H$, where $H$ is a subgroup of $S_{14}$. Find all the $\tau \in H$ such that $\tau(1)=1$.
Hello everyone, I found many exercises on the Symmetric Groups $S_n$ like this one i wrote above.
I first write $\sigma$ as product of his cycles: $\sigma=(1\hspace{1mm} 7\hspace{1mm} 10\hspace{1mm} 9\hspace{1mm} 11)(2\hspace{1mm} 5\hspace{1mm} 4\hspace{1mm} 6\hspace{1mm} 3)(8\hspace{1mm} 13\hspace{1mm} 14\hspace{1mm} 12) $. Since the cardinality of $H$ is equal to the order of $\sigma$, then his cardinality is equal to the $lcm(5,5,4)=20$, where $"5,5,4"$ are the lengths of the cycles of $\sigma$.
I have many doubts about this below:
So, if $H=\{\sigma^i | i \in \mathbb{Z} \}$ I can just write the powers of $\sigma$ (and this is very painful) and I will have a certain power, suppose $\sigma^5$, where $\sigma(1)=1$ and then in all the other powers between $5$ and $20$ (that is the order of $\sigma$) I will have $\sigma(1)=1$.
I would really appreciate some help.
P.s: sorry for my english
You want $\tau=\sigma^r$ to fix $1$; clearly, from the expression for $\sigma$ as a product of disjoint cycles only $\sigma^{5s}$ will do this. That is $\tau=e,\sigma^5, \sigma^{10},\sigma^{15}$, or $\tau=e, (8\ 13\ 14\ 12), (8\ 14)(13\ 12), (8\ 12\ 14\ 13)$.
I don't think it's very painful.