Group of order $60$

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 .

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Mind that Cauchy's Theorem (If $p$ divides the order of a group $G$, then there is an element in $G$ of order $p$) applies to any finite group. Statement b is false, again $A_5$ gives the counterexample (why?). Can you work out statement d for the group $A_5$? – Andrea Mori Jul 21 '12 at 10:57
Yes another badly worded question! For a,b, and d, the answer is that it depends on which group of order 60 $G$ is. They are all true for the cyclic group of order 60. (But I didn't downvote it!) – Derek Holt Jul 21 '12 at 12:23
I've deleted some off-topic comments. Patience, please either politely ask for the reason for a downvote, or don't say anything. Also, as a reminder to everyone, downvoting without leaving a comment is acceptable here (see e.g. the answers on this meta post), and there's nothing anyone can do to make people only vote for "good" reasons so don't worry about it. – Zev Chonoles Jul 22 '12 at 14:49

• 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 $15$.

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thanx for the answer, yes I need help proving last part – Un Chien Andalou Jul 21 '12 at 10:56
@Patience I have left an hint. It must be easy to take it from there. I can however elaborate further but I am sure you can take it yourself. – user21436 Jul 21 '12 at 11:06
@Tojamaru : Another way (if you already know one way) of seeing $A_5$ has no subgroup of order $15$ is to see that anyy group of order $15$ is cyclic.. this follows similar steps but then it is a little bit generalized case... – Praphulla Koushik Jan 9 '14 at 7:02

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.

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Thank you, @Douglas. Let me add for the reader of this post who will try to use Subgroups from the SONATA package: as SONATA's manual confirms, Subgroups is intended for small examples. Please read How do I get the subgroups of my group from the GAP F.A.Q. for further hints. – Alexander Konovalov Apr 22 '13 at 20:54
Updated. Thanks for pointing that out. – Douglas S. Stones Apr 23 '13 at 11:27