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I am asked the find the possible orders of subgroups of a group of order 60.

By Lagrange's theorem: $\left | H \right | | \left | G \right |$ Any positive integer n that is a divisor of $\left | G \right |=60$ is a possible order of a subgroup of a group of order 60.

The possible orders are 1,2,3,4,5,6,10,12,15,20,30,60.

Question: Find a group of order 60 that has subgroup of all possible orders.

Is there a quick way to determine this?

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    $\begingroup$ How about the cyclic group of order 60? $\endgroup$ – carmichael561 Mar 30 '16 at 2:33
  • $\begingroup$ @carmichael561 What is a quick way to determine that it has subgroup of all possible orders? $\endgroup$ – Mathematicing Mar 30 '16 at 2:36
  • $\begingroup$ If $x$ is a generator of $C_{60}$, then $x^d$ has order $\frac{60}{d}$ for each divisor $d$ of 60, hence so does the subgroup it generates. $\endgroup$ – carmichael561 Mar 30 '16 at 2:39
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A finite supersolvable group has a subgroup for every possible order. $11$ of the $13$ groups of order $60$ are supersolvable, so you can use any of them.

The remaining groups, $A_5$ and $C_5\times A_4$, do not have the required property.

Note, that there are finite groups having a subgroup for every possible order which are not supersolvable, so the converse of the initial claim is false.

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    $\begingroup$ Now this is one as correct as uninstructive answer ... $\endgroup$ – Hagen von Eitzen Mar 30 '16 at 11:23

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