2

Let us consider $S_8$ the symmetric group of all permutation of the symbols $\{1, 2, ..., 8\}$ of order $8!$. And given that for some $\tau\in S_8$ of order 2, there exists $\sigma\in S_8$ such that $\sigma\tau\sigma=\tau$ holds.

What are the possible choice of $|\sigma|$?

Note that here $|\theta|$ denotes the order of the permutation $\theta\in S_n$.

My try: If $\sigma$ and $\tau$ be disjoint permutation meaning by $\tau$ and $\sigma$ no common symbols from $\{1, 2, ..., 8\}$ then they commute. So that we shall get $\sigma\tau\sigma=\tau$ i.e. $\tau\sigma^2=\tau$ i.e. $\sigma^2=\varepsilon$ which shows that $\sigma$ must be a permutation of order 1 or 2. We are done in this case.

But what if $\sigma$ and $\tau$ share some common entry/ symbol?

Then ? What to do ?

KON3
  • 4,111
  • 1
    $\sigma\tau\sigma=\tau$ and $|\tau|=2$ and also $$\sigma\tau\sigma\tau=\tau\tau$$ this means that $|\sigma\tau|=2$. –  Sep 09 '15 at 07:47
  • @EugenCovaci I am not clear properly. How come $|\sigma|$ becomes 1, 2, 4 only, nothing else? $8!=40320$ is a huge number. Would you mind to explain me your words a more please ? :-) – KON3 Sep 09 '15 at 09:55
  • @Anjan3 Actually I was trying to build a 8 elements subgroup containing $\sigma$ –  Sep 09 '15 at 11:37

2 Answers2

1

Since $\tau=\tau^{-1}$, therefore we can rewrite $\sigma\tau\sigma=\tau$ as $\tau \sigma \tau^{-1}=\sigma^{-1}$. Thus we want those $\sigma$'s which upon conjugation by $\tau$ give $\sigma^{-1}$. Note that in the disjoint cycle representation of $\sigma$ and $\sigma^{-1}$ not only the cycle structure remains the same but the symbols within each cycle will also be the same (except in the ordering).

Let $\tau=(ij)$.

  1. If $i,j$ are not in any of the cycles of the disjoint cycle representation of $\sigma$ then $\tau\sigma\tau^{-1}=\sigma$. In which case we will have $\sigma=\sigma^{-1}$. Thus $|\sigma|=1 \text{ or }2$.
  2. If only one of $i$ or $j$ is in the disjoint cycle representation of $\sigma$ then the resulting permutation $\tau\sigma\tau^{-1} \neq \sigma^{-1}$ because either $i$ will be replaced by $j$ or vice versa and the conjugate of $\sigma$ will no longer have the same symbols in the cycle structure.

  3. So both $i,j$ should be in the the disjoint cycle representation of $\sigma$.

  4. But if $i$ and $j$ were in different cycles of the the disjoint cycle representation of $\sigma$ then too upon conjugation the symbols in the cycle will change and it can no longer be equal to $\sigma^{-1}$.

  5. So both $i,j$ have to be in the same cycle of the disjoint cycle representation of $\sigma$. If that cycle has length $\geq 4$, then by the exchange of $i$ with $j$, it no longer will result in the cycle structure that can possibly give $\alpha^{-1}$. Therefore that cycle can only be of length $2$ or $3$. The case of length $2$ can be easily dealt with. So let us consider the cycle of length $3$. If it is of the form $( i k j)$, then after the action of $\tau$ via conjugation it will become $(jki)$ which is the inverse of $(ijk)$. Now all the other cycles of the disjoint cycle representation of $\sigma$ cannot be of length more than $2$ (because if they were then upon conjugation by $\tau$ they will remain unchanged and hence cannot give us $\sigma^{-1}$. Thus the order of such a $\sigma$ can be only either $3$ or $6$.

Thus $|\sigma|=1,2,3,6$.

Anurag A
  • 41,067
  • Thats nice. But what if $\tau=(ab)(cd)$ form cause in this case also the order is 2. – KON3 Sep 09 '15 at 09:51
  • One more request to you. Kindly specify what is $\alpha$. I hope some edit is required, am i right ? – KON3 Sep 09 '15 at 09:52
  • If $\tau=(ab), \sigma=(axby)$ what will happen in $\sigma\tau\sigma=\tau$ ? Which mean $\sigma$ has the possibility to be a 4-cycle that you declined :-( – KON3 Sep 09 '15 at 15:35
  • @Anjan3 you are right. I have to think more before I can fill that gap. – Anurag A Sep 11 '15 at 04:59
1

Given $\tau^2=1$. Then $\sigma\tau\sigma=\tau$ is equivalent to $\sigma\tau\sigma\tau=1$ i.e. $(\sigma\tau)^2=1$.

In dihedral groups of order $2n$, we have a situation where $r$ and $sr$ are both reflections, and then $s$ is a rotation, whose possible orders are $n$ or its divisors.

Thus, we can think of which dihedral groups can be embedded in $S_8$. The answer is: the dihedral groups of order $16,14,...,6$ (and of course $4,2$) can be embedded in $S_8$.

Then for taking $\sigma$ to be rotation of maximum order in the diedral groups (considered as subgroup of $S_8$) we can see that the orders of these $\sigma$ are $\frac{16}{2}=8, \frac{14}{2}=7,\cdots, 1$.

Thus, $1,2,...,8$ are among possible orders of $\sigma$.

I think, these are the only possible orders: since, a finite group generated by two elements of order $2$ is dihedral. (see these links 1, and 2).

Let me know if some part is not clear.

Community
  • 1
Groups
  • 10,238