Group of order $60$ [NBHM_2006_PhD screening test_Algebra]

Let $G$ be a group of order $60$, pick out the true statements:
a. $G$ is abelian
b. $G$ has a subgroup of order $30$.
c. $G$ has subgroups of order $2$, $3$, and $5$.
d. $G$ has subgroups of order $6$, $10$, and $15$.


My Attempt:   
a is false because $A_5$ is an non abelian group of order $60$.
For  b,c,d I have no idea.if $G$ was abelian then $c$ is correct by cauchy theorem .
 A: *

*There is a non-abelian simple group of order $60$. Thus, $(a)$ is false. 

*We note that $(b)$ is false for precisely the same reason as $(a)$. There is a non-abelian simple group of order $60$.  

*$(c)$ is true. This is because, by Cauchy's theorem, there is an element of those orders specified. The subgroup those elements generate will respectively be the required subgroup. 

*This is a bit tricky if one does not want to use the fact that the non-abelian simple group of order $60$ is the alternating group on $5$ symbols, $A_5$. It is a straight forward Sylow calculation to show that $A_5$ has no subgroup of order $15$. 
Proof that $A_5$ does not have a subgroup of order $15$ 
Prove that a group of order $15$ is cyclic (Sylow's Theorems). Now prove that no permutation on $5$ symbols can have order $15$. 
A: As $A_5$ is of order $60$ and a-non abelian group, option A is not true .
Also, $A_5$ is a simple group, so it doesn't have subgroup of order $30$, as a subgroup of index $2$ is normal.  So Option B is also not true.
Now by Cauchy's theorem for finite groups, C must be true.
As we know a group of order $15$ is cyclic, if $A_5$ has a subgroup of order $15$ then that subgroup must be cyclic.  Hence $A_5$ has an element of order $15$, which is a contradiction as $A_5$ doesn't have any elements of order $15$. So in this way option D is also not true.
