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I came across this problem which says:

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 $15.$

To tackle the problem,i have used Sylow's first theorem and concluded that G has subgroups of order $2,3$ and $5.$ But i do not know how to determine whether G is abelian or not. Please help.

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I have got it. Since G is an alternating group (A5) and we know that An is abelian iff n<=3, then G can not be abelian. Am i right? –  learner Nov 21 '12 at 3:51
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3 Answers

up vote 2 down vote accepted

The dihedral group $D_{30}$ has order 60. Dihedral groups $D_n$ for $n>2$ are not abelian.

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What about $\,A_5\leq S_5\,$...? This must answer (a), (b), and (d) (remember:$\,A_5\,$ is a simple group)

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$G$ cannot have a subgroup of order $30$. If so , that should be normal because $[G:H]=2$. But note that there exists a group of order $60$ is simple(eg $A_5$). So option b wrong.

By cauchy's theorem option (c) is true.

Option (d) is not true since $A_5$ has no subgroup of order $15.$

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No, there exists a simple group of order $60$. Not all groups of order $60$ are simple. –  Tobias Kildetoft Jan 23 at 14:12
    
please can you give me a group of order 60 which is not simple. –  Murugan Jan 23 at 14:18
    
All apart from $A_5$ (for example $C_{60}$). –  Tobias Kildetoft Jan 23 at 14:19
    
What is $C_{60}$ is it dihedral group. –  Murugan Jan 23 at 14:20
    
No, the cyclic group (but the dihedral group of order $60$ would of course also be an example). –  Tobias Kildetoft Jan 23 at 14:23
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