Every finite group $H$ is a quotient of $A_n$ for some $n \geq 1$. Every finite group $H$ is a quotient of $A_n$ for some $n \geq 1$.
Can anyone help me to prove or disprove this? I do not understand the term "a quotient of $A_n$"? Does this mean  that there will be a normal subgroup $N$ such that  $A_n/N$ is isomorphic to  $H$?
 A: It is a well known result that $A_n$ is simple if $n \geq 5$. That is, there are no non trivial normal subgroups: if $H \triangleleft A_n$ then $H = 0$ or $H = A_n$. Hence, if $n \geq 5$, any quotient of the form $A_n/N$ with $N \triangleleft A_n$ is either zero or isomorphic to $A_n$ itself, and in particular it has order either $0$ or $\frac{n!}{2}$. It is clear from here, thus, that not every finite group satisfies this property: if for example $G$ is a group of order $100$,  
$$
\frac{5!}{2} < |G| < \frac{6!}{2}
$$
so $G$ cannot be isomorphic to a quotient of $A_n$ for some $n \geq 1$ (note that here we discard $n\leq 4$ because in that case $G$ has more elements than $A_n $).
More generally, if your group has order different from $0$ or $\frac{n!}{2}$ for $n \in \mathbb{N}$, it will not be isomorphic to any quotient of $A_n$ for $n \neq 5$. It could be, however, that your group is isomorphic to a quotient of $A_4$. The only proper non-trivial normal subgroups of $A_4$ are of order $4$, so we have the following conclusion

Let $G$ be a group with $|G| \not \in \{0,3\} \cup \ \{\frac{n!}{2}\}_{n \in \mathbb{N}}$. Then, $G$ cannot be isomorphic to a quotient of some $A_n$, that is, 

$$
G \not\simeq A_n/N 
$$

for any $n \in \mathbb{N}$, $N \triangleleft A_n$.

