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.....

  • $\begingroup$ The solution you googled is correct and written down properly. I'm not sure what you expect that we can add to that. Which parts don't you get? $\endgroup$ – Mathematician 42 Nov 16 '17 at 11:57

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 $\langle \sigma\rangle\cap \langle \tau\rangle = \{\text{id}\}$.

  • $\begingroup$ @ Arthur im not getting ur 2 statement ,,that it is written that which send 5 to 1,,as from my view 6 is sending to 1,, $\endgroup$ – user476275 Nov 16 '17 at 12:11
  • $\begingroup$ @Michael You're right, for some reason I thought $\tau = (12345)$ while writing that paragraph. I fixed it. $\endgroup$ – Arthur Nov 16 '17 at 12:15
  • $\begingroup$ Thanks for giving the answer@ Arthur but i have one doubt $\langle \tau\rangle$ that send 1 to 2 and 6 to 1,,,my question is that .....only element which sends 1 to 2 is $\tau$ itself,,,that mean i can say that $\gamma(1) = 2$ ? but u say it is impossible,,,,,im not still getting $\endgroup$ – user476275 Nov 16 '17 at 12:30
  • $\begingroup$ @Michael The second point is an argument by contradiction to reach the conclusion that $\gamma(1) = 2$ is impossible. We start by assuming that $\gamma(1) = 2$, and we arrive at the contradiction that $\gamma(5)$ must be either $3, 4$ or $5$ (since it's an element of $\langle \sigma \rangle$), but at the same time it must also be $6$ (because $\gamma$ must be equal to $\tau$). $\endgroup$ – Arthur Nov 16 '17 at 12:35
  • $\begingroup$ @ Arthur ..thanks a lots i got it,,,,,thank u $\endgroup$ – user476275 Nov 16 '17 at 12:47

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.


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