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[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 .

  • 1
    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](http://meta.math.stackexchange.com/q/3153/264)), and there's nothing anyone can do to make people only vote for "good" reasons so [don't worry about it](http://meta.math.stackexchange.com/a/1722/264).2012-07-22

2 Answers 2

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  • 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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    There's a much easier way to see $A_5$ has no subgroup of order 15. If it had, it would have index 4, and $A_5$ would act transitively on its cosets, providing a non-trivial homomorphism from $A_5$ to $S_4$, contradicting that $A_5$ is simple.2018-04-21
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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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    Updated. Thanks for pointing that out.2013-04-23