Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $E$ be a set and $f:E\to E$ be a function such that $f\circ f=Id$.

Let $A=\{x\in E, f(x)\neq x\}$.

Suppose that $A$ is finite.

Prove that the cardinality of $A$ is even.

My idea is to rewrite $A$ as a disjoint union of sets with even cardinality, but I've been unsuccessful so far.

I noticed that $f$ acts as a permutation over the elements of $A$.

What should I do next ?

share|cite|improve this question
Hint: Think about what happens when $f(x)\ne x$. – paw88789 Aug 31 '14 at 9:37
Hint: $\forall x \in A$ holds $f(x) \in A$, hence $f$ permutes all elements of $A$ without fixing any point of $A$. – Crostul Aug 31 '14 at 9:43
@Crostul what then ? Think about $\{1,2,3\}$ – LeGrandDODOM Aug 31 '14 at 9:44
up vote 4 down vote accepted

For each $a\in A$ let $S_a=\{a,f(a)\}$.

1) Show that $|S_a|=2$ for all $a\in A$.

2) Show that $S_a\cap S_b\ne \emptyset$ implies $S_a=S_b$, for all $a,b\in A$.

3) Count the number of elements in $A=\bigcup_{a\in A}S_a$.

In a bit more sophisticated language, there is an action of $\mathbb Z_2$ on $A$ given by applying $f$. The orbits are the $S_a$. By definition, all stablizers are trivial, and thus all orbits have size $|\mathbb Z_2|=2$. The entire set $A$ is the disjoint union of orbits, and thus even.

share|cite|improve this answer

$f|_A:A \longrightarrow A$ permutes all elements of $A$ and has no fixed points and $f^2 = id_A$.

I want to prove that if $S\subseteq A$ and $f$ permutes the elements of $S$, then the cardinality of $S$ is even. Taking $S=A$ we get the thesis.

I work by contradiction. Suppose there exists $S \subseteq A$ such that $f$ permutes the points of $S$ and $S$ has odd cardinality. Then you can consider such a set with minimal (odd) cardinality. Clearly, since the cardinality of $S$ is odd, we have $S \neq \emptyset$.

But now, pick $x \in S$. Then $S'=S \setminus \{ x, f(x)\}$ has odd cardinality and $f$ permutes all elements of $S'$, contradicting the minimality of $S$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.