If $b\in <S_{6},\bullet>$ , $b= \left[ \begin{array}{cc} 1&2&3&4&5&6\\ 2&3&4&5&6&1 \end{array} \right] $
and the subgroup
$H=<b>= $$ \left\{ \begin{array}{c} b= \left[ \begin{array}{cc} 1&2&3&4&5&6\\ 2&3&4&5&6&1 \end{array} \right] ,b^{2}= \left[ \begin{array}{cc} 1&2&3&4&5&6\\ 3&4&5&6&1&2 \end{array} \right]\\b^{3}= \left[ \begin{array}{cc} 1&2&3&4&5&6\\ 4&5&6&1&2&3 \end{array} \right] ,b^{4}= \left[ \begin{array}{cc} 1&2&3&4&5&6\\ 5&6&1&2&3&4 \end{array} \right]\\ b^{5}= \left[ \begin{array}{cc} 1&2&3&4&5&6\\ 6&1&2&3&4&5 \end{array} \right],b^{6}= \left[ \begin{array}{cc} 1&2&3&4&5&6\\ 1&2&3&4&5&6 \end{array} \right]=e\\\end{array}\ \right\}$
I need to find all non trivial subgroup of $H$.
Using Lagrange's theorem and the fact that any subgroup of $H$ (mark them $K_i$) is non trivial if: $$\{e\}\subset K_{i}$$ and $$K_{i}\neq\{e\},K_{i}\neq\ H$$ that means $o(K_{i})\neq 1,o(K_{i})\neq 6$
so it must be that $o(K_{i})=2$ or $o(K_{i})=3$
if $o(K_{i})=2$,then I found one subgroup:$K_{1}=\{b^{6}=e,b^{3}\}$
if $o(K_{i})=3$,then I found one subgroup:$K_{2}=\{b^{6}=e,b^{2},b^{4}\}$
How can I find more?
Is there any fast way to find them all?
Thanks.
$o(H)=6$,$H$ is cylic,so $\phi(6)=2$.
$H$ has 2 generators. $<b>$ is a generator,so: $o(b)=6$ and $$o(b^{t})=\frac{o(b)}{gcd(o(b),t)}=\frac{6}{gcd(6,t)}$$ to find the second generator:$o(b^{t})=6$ this happens iff $gcd(6,t)=1$
so,$t=1$ and $t=5$ means that $<b^{5}>=H$
There is only 2 non trivial subgroups.