Group of order $60$

Solution 1:

• 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 $15$.

Solution 2:

Here's a computational approach. In fact, there are $13$ isomorphism classes of groups of order $60$. These can be accessed by the GAP AllSmallGroups function.

Here's the input code:

for G in AllSmallGroups(60) do
  Conj:=ConjugacyClassesSubgroups(LatticeSubgroups(G));
  SubgroupSizes:=Set(Conj,C->Size(C[1]));
  Print(StructureDescription(G)," ",IsAbelian(G)," ",SubgroupSizes,"\n");
od;

And here is the output:

C5 x (C3 : C4) false [ 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60 ]
C3 x (C5 : C4) false [ 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60 ]
C15 : C4 false [ 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60 ]
C60 true [ 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60 ]
A5 false [ 1, 2, 3, 4, 5, 6, 10, 12, 60 ]
C3 x (C5 : C4) false [ 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60 ]
C15 : C4 false [ 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60 ]
S3 x D10 false [ 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60 ]
C5 x A4 false [ 1, 2, 3, 4, 5, 10, 12, 15, 20, 60 ]
C6 x D10 false [ 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60 ]
C10 x S3 false [ 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60 ]
D60 false [ 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60 ]
C30 x C2 true [ 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60 ]

Each row identifies one of the $13$ groups of order $60$, writes whether or not it's abelian, and the list of subgroup sizes of that group.

The notation as to which group is which should be largely self-explanatory, except that ":" represents a semi-direct product.

• "$G$ is abelian" is true for $2$ out of $13$ isomorphism classes of groups of order $60$.
• "$G$ has a subgroup of order $30$" is true for $11$ out of $13$. (The two counter-examples are $A_5$ and $C_5 \times A_4$.)
• "$G$ has a subgroup of order $2$, $3$, and $5$" is true for $13$ out of $13$.
• "$G$ has a subgroup of order $6$, $10$, and $15$" is true for $11$ out of $13$. (The two counter-examples are $A_5$, which has no subgroup of order $15$, and $C_5 \times A_4$, which has no subgroup of order $6$.)

Disclaimer: I don't mean to undermine the importance of understanding the theoretical methods for answering these questions.