I've got H = <(2 1 3 4 5 6)(7 8 10 9)> a subgroup of the symmetric group S_10 , and I have to calculate all the elements of H which have got order 6
If I'm not wrong, H can be written like this: H={b^i a^i | i {0,1,2,3} and j {0,1,2,3,4,5}}
I've defined: a=(1 3 4 5 6 2) and b=(7 8 10 9)
I've calculated all the elements:
b^0 a^0
b^0 a^1
b^0 a^2
.
.
.
b^3 a^4
b^3 a^5
And what I've obtained is that the elements of order 6 are:
b^0 a^1 = (1 3 4 5 6 2)
b^0 a^5 = (1 2 6 5 4 3)
b^2 a^1 = (7 10)(8 9)(1 3 4 5 6 2)
b^2 a^2 = (7 10)(8 9)(1 4 6)(2 3 5)
b^2 a^3 = (7 10)(8 9)(1 5)(2 4)(3 6)
b^2 a^4 = (7 10)(8 9)(164)(253)
b^2 a^5 = (7 10)(8 9)(1 2 6 5 4 3)
But I'm not sure this is OK... Anyway, I would like to know a shorter way to find the elements of order 6 without calculating all the elements of H (if it's possible).
Because the cycle are disjoint they commute, so the order of the element $\sigma = \sigma_{1}\cdot \sigma_{2} = lcm(o(\sigma_{1}),o(\sigma_{2}))$, where $\sigma_{1} = (2\hspace{0.1cm} 1\hspace{0.1cm} 3 \hspace{0.1cm}4\hspace{0.1cm} 5 \hspace{0.1cm}6), \sigma_{2} = (7 \hspace{0.1cm} 8 \hspace{0.1cm} 10 \hspace{0.1cm}9)$. Since $o(\sigma_{2}) = 6, o(\sigma_{1}) = 4$ we have that $o(H) = o(\langle \sigma \rangle) = o(\sigma) = 12$, hence $H \cong \mathbb{Z}_{12}$, cyclic.
It follows that for each divisor $n$ of $12$ there are exatcly $\phi(n)$ elements of order $n$, where $\phi$ denotes the Euler function.
Edit : Since you have to count the elements when $n = 6$, if you know that there are $\phi(6) = 2$ such elements of order $6$ you know even which, since there are $2$ obvious choices for it, which are $\sigma^{2} = (2\hspace{0.1cm}3\hspace{0.1cm}5)(1\hspace{0.1cm}4\hspace{0.1cm}6)(7\hspace{0.1cm}10)(8\hspace{0.1cm}9)$ and its inverse.