For which of the following values of n is there a group of order n with no proper normal subgroups? 1) n=21
2) n=9
3) n=60
4) n=98
As per the link here http://groupprops.subwiki.org/wiki/Subgroup_of_index_equal_to_least_prime_divisor_of_group_order_is_normal there is least prime divisor of group order existing for each of the options. Given n=21 we can say that if there is a sub group of order 3 its normal. But how do we prove that a subgroup order 3 exists? With lagrange's theorem we can say that if there is a subgroup of size 3 it is proper subgroup (3 divides 21) but lagrange's theorem cannot be used to prove existence of the subgroup. Can you help me how to find the right answer?
 A: Well, it is well known that there is a unique simple group of order $60$ upto isomorphism. This group is called the alternating group on $5$ symbols which is denoted by $A_5$.
I am surprised, you pose this as a multiple choice question because the fact I had written above takes a good deal of work. 
Also, your contention that the subgroup of order $3$, (if it exists, you're unsure it does, but it actually does!) is normal. 

The theorem states that Subgroup of index equal to least prime divisor of group order is normal. 

So, the least prime dividing the order of the group is $3$ which must be the index of the subgroup. So, you're looking for the subgroup of order $7$. 
To prove the existence of a subgroup of that order, use Cauchy's Theorem.  

Similarly, a group in which every element has order divisible by $p$ for a prime $p$ is called a $p$-group. A $p$-group has non-trivial center. Use these fact to conclude that a group of order $9$ has a non-trivial normal subgroup. 
Finally for a group of order $98$, use the fact that Sylow $7$-subgroup is order $49$ and hence of index $2$ and hence normal in the whole group. 

Use Google. Good Luck.
(Ping me in case you have some problem.) 
