The group $S_3$ of permutation of $\{1,2,3\}$ acts on the three-dimensional vector space over the finite field $\Bbb F_3$ of three elements, by permuting the vectors in basis $\{e_1,e_2,e_3\}$ by $$\sigma\cdot e_i=e_{\sigma(i)}, \quad i=1,2,3$$ for all $\sigma \in S_3$. The cardinality of the set of the vectors fixed under the above action is:

a) 0
b) 3
c) 9
d) 27

Attempt: If $\sigma$ is identity permutation then $\sigma\cdot e_i=e_{\sigma(i)}=e_i$ so all the basis elements will be fixed. For $\sigma$ non identity we can proceed further but that will be long and time taking, So I am looking for general formula and approach to tackle this type of problem.


You have the set $X=\mathbb{F}_3^3=\{(a,b,c):a,b,c\in\mathbb{F}_3\}$.

You have the group $G=S_3=\{e,(12),(13),(23),(123),(132)\}$.

The set of elements of $X$ fixed by the action is $X^G=\{x\in X:\sigma x= x\text{ for all }\sigma\in G\}$.

Suppose that $(a,b,c)\in X^G$, so that $\sigma\cdot(a,b,c)=(a,b,c)$ for all $\sigma\in G$.

For the identity element, you actually don't get any information: since $e\cdot(a,b,c)=(a,b,c)$, we get that $a=a$, $b=b$, and $c=c$, which tells us nothing. In other words, every element of $X$ is fixed by the identity element of $G$.

However, for $(12)$, we see that since $(12)\cdot(a,b,c)=(b,a,c)$, we must have $(b,a,c)=(a,b,c)$ and hence that $a=b$.

Are there any other restrictions you can conclude about $a$, $b$, and $c$ when $(a,b,c)\in X^G$, or is that all?

Then: how many elements of $X$ meet those restrictions, i.e., how many elements are in $X^G$?

  • $\begingroup$ Thanx for the quick reply. If we take (1,3).(a,b,c), we must have a=c, (2,3).(a,b,c), we must have b=c $\endgroup$ – Chiranjeev_Kumar Dec 25 '17 at 11:23
  • $\begingroup$ Exactly. You can conclude that $(a,b,c)\in X^G\iff a=b=c$. So, what is the cardinality of $X^G$? $\endgroup$ – Zev Chonoles Dec 25 '17 at 11:25
  • $\begingroup$ So cardinality should be 3 $\endgroup$ – Chiranjeev_Kumar Dec 25 '17 at 11:26
  • $\begingroup$ Exactly right! :) $\endgroup$ – Zev Chonoles Dec 25 '17 at 11:26
  • $\begingroup$ Thanx, allot :) your way of helping was awesome. $\endgroup$ – Chiranjeev_Kumar Dec 25 '17 at 11:34

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