how to show that $\langle\sigma\rangle\cap\langle\tau\rangle$ is a trivial group? Let $\sigma=(1 2)(3 4 5)$ and  $\tau=(1 2 3 4 5 6)$ in $S_6$, 
then show that 
$\langle\sigma\rangle\cap\langle\tau\rangle$ is a trivial group ?
My try :As i was googling it and find it and it is written that: Any power of $\sigma$ takes $1$ either to $1,$ or to $2.$ The only element of $\langle \tau \rangle$ which takes $1$ to $1$ is $\mathrm{id},$ and the only element which takes $1$ to $2$ is $\tau$ itself. But, on the other hand, any element of $\langle \sigma \rangle$ preserves $\{3,4,5\},$ which $\tau$ certainly doesn't. Thus
$$
\langle \sigma \rangle \cap \langle \tau \rangle =\{\mathrm{id}\}.
$$
I was reading this solution but ididn't understanding  anything..and  what does it mean to say ?
Pliz help me and tell me the proper solution in detail.....
 A: Here is a rephrasing of the proof. If there are sentences or arguments you don't understand, say which ones in a comment, and I'll try to help you if I can.
Take a permutation $\gamma\in \langle \sigma\rangle\cap \langle \tau\rangle$. Because $\gamma\in \langle\sigma\rangle$, we have either $\gamma(1) = 1$ or $\gamma(1) = 2$. 


*

*If $\gamma(1) = 1$, then because $\gamma\in \langle\tau\rangle$, we must have $\gamma = \text{id}$ (there is only one element in $\langle\tau\rangle$ that sends $1$ to $1$, and that's the identity)

*If $\gamma(1) = 2$, then because $\gamma\in \langle \sigma \rangle$, we have that the numbers $\gamma(3), \gamma(4)$ and $\gamma(5)$ are $3, 4$ and $5$ in some order. But this doesn't work with $\gamma \in \langle\tau\rangle$, because the only element in $\langle \tau\rangle$ that sends $1$ to $2$ is $\tau$ itself, which sends $5$ to $6$. Therefore $\gamma(1) = 2$ is impossible.


So the only possible option for $\gamma$ is $\text{id}$. This proves $\langle \sigma\rangle\cap \langle \tau\rangle = \{\text{id}\}$.
A: Both permutations you wrote have order six, but their cycle structure as well as that of all their non-trivial powers are quite different:
Any power of a product of a two-cycle and a three-cycle (disjoint from the two-cycle) is either


*

*the identity permutation,

*a product of a two-cycle and a (disjoint) three-cycle,

*a single two-cycle, or

*a single three-cycle.


On the other hand, any power of a six-cycle is either


*

*the identity permutation,

*a six-cycle,

*a product of three disjoint two-cycles, or

*a product of two disjoint three-cycles.


So, the only element that can be both a power of $(1 2)(3 4 5)$ and of $(1 2 3 4 5 6)$ is indeed the identity.
