# Using orbit stabilizer theorem

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!

• 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. Mar 22, 2016 at 0:57
• The size of orbit can not be $8$, because $G$ is subgroup of $S_7$, the permutation group on seven letters. Mar 22, 2016 at 4:13

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.

• 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). Mar 22, 2016 at 5:12
• 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. Mar 22, 2016 at 5:32
• So do you need the Sylow theorem or not? Aug 31, 2020 at 1:13

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).

• I've posted an answer without having realized that you also just had, and yes, yours (and mine ;-) ) is correct.
– user810157
Aug 31, 2020 at 15:30

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.