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Let the group $G \subset S_7$, $|G|=8$. Show that there exists $i\in{1,...7}$ such that for all $f\in G,f(i)=i$.

I have attempted this problem, but I am not entirely sure if my thinking is correct. Using orbit stabilizer theorem, I write $|G|=|orb_G(i)||stab_G(i)|$, and since $|G|=8, $ then $|stab_G(i)|$ is equal to either $1,2,4,8$ and $|orb_G(i)|$ equal to $8,4,2,1$ respectively. Now, when the order of the orbit is 8 and stabilizer 1, can I conclude that there exists a function that maps $i$ to $i$?

Any help would be appreciated!

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    $\begingroup$ You want the order of the orbit to be 1 not 8. Then the order of the stabilizer is 8 so all 8 elements of g satisfy f(i)=i. $\endgroup$
    – Michael N
    Mar 22, 2016 at 0:57
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    $\begingroup$ The size of orbit can not be $8$, because $G$ is subgroup of $S_7$, the permutation group on seven letters. $\endgroup$
    – p Groups
    Mar 22, 2016 at 4:13

3 Answers 3

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This answer is not only by using orbit stabilizer theorem, but with something other important theorems.

Note that $G$ is a $2$-group, hence it should be contained in some Sylow-$2$ subgroup of $S_7$.

Let $S_6=Stab(7)=$permutation group on first six symbols.

Then a Sylow-$2$ subgroup, say $P$, of $S_6$ is also a Sylow-$2$ subgroup of $S_7$.

Hence $G$ is contained in some conjugate of $P$, say $\sigma P\sigma^{-1}$ for some $\sigma\in S_7$.

Now $P$ stabilizes (i.e. fixes) the letter $7$, hence $\sigma P\sigma^{-1}$ fixes $\sigma(7)$, and consequently $G$ also fixes $\sigma(7)$, which you expected.

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    $\begingroup$ No need to invoke Sylow, the orbit-stabiliser suffices: all orbits have size powers of 2 and to write 7 as a sum of powers of 2, one of the summands must be 1, hence there is an orbit of size 1 (i.e., a fixed point). $\endgroup$
    – verret
    Mar 22, 2016 at 5:12
  • $\begingroup$ Oh yes: a $2$-group acting on set of odd order, so one point should be fixed by the group, which is expected. Thanks for noticing it. I had confused with it while looking orbit under single element and orbit under whole group. $\endgroup$
    – p Groups
    Mar 22, 2016 at 5:32
  • $\begingroup$ So do you need the Sylow theorem or not? $\endgroup$ Aug 31, 2020 at 1:13
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After reading the comments and thinking about it, I decided to try and write a proof of my own. Hopefully it will be of use to someone else.

Proof:

  1. From the Orbit-Stabilizer Theorem - $|O(u)|=|G|/|C(u)|$ where $O$ is the orbit of $u$ and $C$ is the stabilizer of $u$.

  2. By definition of the orbit O, it follows that $\forall u, f(u)=u$ is equivalent to $O(u)={u}$.

  3. $O(u)=\{u\}$ is equivalent to $|O(u)|=1$ since $\forall u, u \in O(u)$, because of the identity permutation in $G$, ie. we can't have $O(u)=\{k\}$ for $k \neq u$

  4. Since $|O(u)|$ is a divisor of $|G|=8$ (by 1.) => $|O(u)|$ can be either 1,2,4,8

  5. But O is a partitioning of $\{1...7\}$. Hence $\sum_{u}|{O(U)}|=|\{1..7\}|=7$.

  6. If $\forall u, |O(u)| \neq 1$, then $|O(u)|$ must be one of 2,4,8 - which are all even.

  7. Hence $\sum_{u}|{O(U)}| \equiv 0(\textrm{mod}\ 2)$. But $7 \equiv 1(\textrm{mod}\ 2)$.

  8. Hence $\exists u, |O(u)|=1$ and we are done.

Note: Basically the proof comes down to the Orbit-Stabilizer Theorem + Orbits are a partition statement(proof omitted, but one can easily verify it's an equivalence relation, hence a partition).

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  • $\begingroup$ I've posted an answer without having realized that you also just had, and yes, yours (and mine ;-) ) is correct. $\endgroup$
    – user810157
    Aug 31, 2020 at 15:30
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Consider the action of $G$ on $\{1,\dots,7\}$ as a group of permutations. Then (Orbit-Stabilizer Theorem), the orbits have size either $1$ or $2$ or $4$ or $8$ (the divisors of $8$). But there is no way to get $7$ (the size of the acted on set) out of $2$ and/or $4$ as summands (let alone $8$). Therefore, $\exists \bar i\in \{1,\dots,7\}$ such that $|O(\bar i)|=1$ and hence (again by the OST) $|\operatorname{Stab}(\bar i)|=8$, which is precisely what you are required to prove.

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