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I'm doing this problem:

Let $G$ be a group of order 21, which acts on thhe set $S$.

a. Show that if $|S| =8$, then $S^G \neq \varnothing$.

b. For what other integers $n$ between 1 and 100 can you prove that if $|S| =n$, then $S^G \neq \varnothing$?

c. For the remaining integers between 1 and 100, show that there is a set S with |S| = n and $S^G = \varnothing$.

$S^G $ is the fixed subset of S fixed by $G$, which contains all the stubborn elements in $S$ that doesn't change when any $g \in G$ acts on them.

So far this is what I have: Since 3 and 7 are prime divisors of 21, we know by Cauchy's Theorem that G has an element of order 3 and an element of order 7. I probably need to connect this to the class equation. Can anyone give me a hint?

Thanks

Quy Xuan Cao
  • 95

1 Answers1

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Hints:

  • The orbit stabilizer theorem tells you what sizes the orbits of $G$ can have. List them all!
  • A point is fixed under $G$ if and only if it belongs to an orbit of size $1$. Can you write $8$ as a sum of orbit sizes without using orbits of size $1$?
  • To answer the general version you need to either experiment (recommended) or study Frobenius' Coin Problem. A motivator for the study of numerical semigroups.

Completing the arguments (now that the OP has solved their problem).

  • The size of an orbit is a factor of $|G|$. In this case the options are thus $1,3,7$ and $21$. The OP observed that $G$ has subgroups of all the "complementary" orders ($21,7,3,1$) so all those may occur with our group $G$.
  • All the ways to write $8$ as a sum of numbers from the set $\{1,3,7,21\}$ involve $1$, namely $8=1+1+1+1+1+1+1+1$, $8=3+1+1+1+1+1$,$8=3+3+1+1$ and $8=7+1$. So orbits of size one are forced upon us. Therefore in a set of eight elements $G$ necessarily has a fixed point.
  • In parts b and c we must look for ways to write $n$ as sums of threes and sevens only. $21$ doesn't reall add anything. A little bit of experimenting shows that the small sums are $0,3$, $6=3+3$, $7$, $9=3+3+3,10=7+3,12=3+3+3+3,13=7+3+3$, $14=7+7$. At this point we observe that we have three consecutive integers in our list ($12,13,14$). By adding a necessary number of threes to them we can reach any natural number. We can conclude that the only natural numbers missing from the list are $1,2,4,5,8$ and $11$. It is not difficult to show that if $a$ and $b$ are two coprime natural numbers, then we can write all but $G(a,b):=(a-1)(b-1)/2$ natural numbers in the form $ai+bj,i,j\in\Bbb{N}$. In the present case we found all the $G(3,7)=6$ of them. The largest missing number is $ab-a-b$. See Robjohn's answer for the argument.
Jyrki Lahtonen
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