$\sigma$ a permutation of $A$. Say $\sigma$ moves $a\in A$, if $\sigma(a)\neq a$. How many elements are moved by $\sigma$ of length $n$?

Let $\sigma$ be a permutation of a set A. We shall say "$\sigma$ moves $a\in A$" if $\sigma(a)\neq a$. If $A$ is a finite set, how many elements are moved by a cycle $\sigma \in S_A$ of length $n$?

I am confused about the meaning of the question. A cycle can move $0-n$ element(s) in $A$. Is this correct?

Define a permutation $\sigma$ such that $\sigma(a)\neq a$ for $\forall a\in A$. Does the question mean how many this kind of permutation exist?

• A cycle of length $n$ means a permutation of the form $\sigma = (x_1,x_2,...,x_n)$ (i.e. $\sigma(x_1)=x_2$, $\sigma(x_2)=x_3, ... ,\sigma(x_n)=x_1$ and all other elements are fixed) where $x_i$ are different elements of $A$. Now can you tell me what is the answer? – Levent Oct 9 '16 at 17:57
• Does the text of the exercise mention "a random permutation"? – Did Oct 9 '16 at 18:29
• @Levent So the answer is n? – User90 Oct 9 '16 at 18:46
• @Did No. That's the whole problem. – User90 Oct 9 '16 at 18:48
• Yes, the answer is $n$. – Levent Oct 9 '16 at 18:48

As discussed in the comments, the answer to your question is that a cycle of length $n$ "moves" exactly $n$ elements, i.e. exactly $n$ elements are not left fixed.
In general, we know we can factor a permutation $\sigma$ as a product of disjoint cycles, say of lengths $p_1,p_2,\ldots,p_k$. We can ask the same question: how many elements does $\sigma$ move? The answer is $$p_1+\cdots+p_k$$ It's as simple as that.