Examples of a number $d$ that is a divisor of the order of a group $G$, but there is no quotient group $G/N$ whose order is $d$.

Question

My first thought is that if there is an element of $G{/}N$ of any order $n$, then there is an element of order $n$ in $G$. But I am not sure if there is a correlation between this and the proposed question.

Can someone please point me to the right track? Thanks!

Answer 1

Here is a minimal counterexample. 3 divides $|S_3|=6$, but there is no normal subgroup of index 3 in the group $S_3$.

Answer 2

It is well known that the group $A_5$ is simple i.e. it has no nontrivial normal subgroup. So it has no nontrivial quotient group as well.

Now, $|A_5|=60$ so any nontrivial divisor of $60$ (e.g. $5$) works.