Can a non-trivial conjugacy class in $A_n$ contains $In one of the proofs of simplicity of $A_n$ ($n\geq 5$), a fact used is the following:

There is no (non-identity) conjugacy class in $A_n$ containing $<n$ elements.

I initially was using simplicity of $A_n$; but that was the aim behind use of this fact. Any hint for proving this fact? 
Also, is there always a conjugacy class in $A_n$ with exactly $n$ elements?
 A: The theorem is false eg. for $n=3$, so we'll consider only $n \geq 5$.
I'll make heavy use of the following fact about conjugacy classes in $A_n$:

Conjugacy classes are either of the form "everything of a given cycle type", or "half the things of a given cycle type"; this latter occurs iff the cycle type is all odd and all distinct.

If the former, we're immediately done, because there are at least $n$ things of any given cycle type. Indeed, let $A_n$ act on the set $\{1, 2, \dots, n \}$. Then by just varying the image of the element $1$, we get $n$ different elements.
If the latter: Let $\sigma \in A_n$ have the given cycle type $(m_1, m_2, \dots, m_k)$, where $m_1 > m_2 > \dots > m_k$. The cycle type is not just $(1)$ because the class is not the identity. Moreover, since $n \geq 5$, we have $m_1 \geq 3$. Wlog the $m_1$-cycle in $\sigma$ moves the element $1$.
We have two cases. If there are at least two of the $m_i$ which are bigger than $2$, then we have at least $n$ elements given by varying the image of $1$.
Indeed, say $a, b, c$ are in a cycle which does not contain $1$ in $\sigma$: that is, $$\sigma = (1 x y z \dots)(a b c \dots)\dots$$
with the first cycle longer than the second.
If $\sigma : 1 \mapsto 2$ (which wlog it does), we can conjugate $\sigma$ by $(2m)(ab)$ to obtain another member of $\sigma$'s conjugacy class which sends $1 \mapsto m$ for each $m \not = a, b$. In order to get $1 \mapsto a$, we can use instead $(2 a)(b c)$, and in order to get $1 \mapsto b$, we can instead use $(2b)(ac)$. Therefore we have found $n$ distinct elements of $\sigma$'s conjugacy class.
The only remaining case is that at most one of the $m_i$ is bigger than 2. That is, the cycle type of $\sigma$ is of the form $(n)$ or $(n-1, 1)$. 


*

*In the former case, the cycle type is $(n)$. There are $\frac{n!}{2n}$ elements of the conjugacy class, so we're done: $\frac{n!}{2n} = \frac{(n-1)!}{2} \geq n$ for $n \geq 5$.

*In the latter case, the cycle type is $(n-1, 1)$. There are $(n-1)!/2$ of these also, so again we're done.


We have found $n$ distinct elements of the conjugacy class of $\sigma$.
