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I am looking for finite groups $G$ such that the number of subgroups is equal to $|G|$. Examples are:

  • the trivial group

  • $\mathbb{Z}/2\mathbb{Z}$

  • $S_3$

Does anyone know some more examples or can provide some insight? Are there any bounds on the number of subgroups (maybe using Sylow's theorems or whatever)?

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$\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/4\mathbb{Z}$ – Chris Eagle Oct 14 '12 at 18:44
1  
The trivial group qualifies too, of course. – Chris Eagle Oct 14 '12 at 18:45
    
$D_6$ is another one. – Babak S. Oct 14 '12 at 19:14
    
But $D_6\cong S_3$. – anon Oct 14 '12 at 19:33
    
@IHaveAStupidQuestion: no, it has the trivial subgroup, the whole group, $\langle (0, 1)\rangle$, $\langle (1,1)\rangle$, $\langle (0,2)\rangle$, $\langle (1, 0)\rangle$, $\langle (1,2)\rangle$, and $\langle (1,0),(0,2)\rangle$. – Chris Eagle Oct 14 '12 at 20:03
up vote 9 down vote accepted

Here is list of all examples $\leq 100$, found with GAP (EDIT: I added some more, but I didn't check for groups of order $128$). The notation $:$ means semidirect product.

$$\begin{array}{|c|c|} \text{StructureDescription} & \text{Order} \\ \hline \text{Trivial group} & 1 \\ \hline C_2 & 2 \\ \hline S_3 & 6 \\ \hline C_4 \times C_2 & 8 \\ \hline D_{28} & 28 \\ \hline C_6 \times S_3 & 36 \\ \hline (C_{10} \times C_2) : C_2 & 40 \\ \hline C_2 \times (C_5 : C_4) & 40 \\ \hline (C_3 \times Q_8) : C_2 & 48 \\ \hline ((C_3 \times C_3) : C_3) : C_2 & 54 \\ \hline C_6 \times A_4 & 72 \\ \hline C_2 \times ((C_4 \times C_4) : C_3) & 96 \\ \hline (C_5 \times C_5) : C_4 & 100 \\ \hline D_{104} & 104 \\ \hline S_3 \times D_{22} & 132 \\ \hline C_3 \times D_{48} & 144 \\ \hline (C_{40} \times C_2) : C_2 & 160 \\ \hline (C_5 \times (C_8 : C_2)) : C_2 & 160 \\ \hline ((C_2 \times (C_5 : C_4)) : C_2) : C_2 & 160 \\ \hline (C_4 \times (C_5 : C_4)) : C_2 & 160 \\ \hline (C_{40} \times C_2) : C_2 & 160 \\ \hline (C_8 \times D_{10}) : C_2 & 160 \\ \hline (C_2 \times (C_5 : Q_8)) : C_2 & 160 \\ \hline (C_2 \times (C_{11} : C_4)) : C_2 & 176 \\ \hline (C_{15} \times C_3) : C_4 & 180 \\ \hline \end{array}$$

Random related fact: the number of subgroups in the dihedral group $D_n$ of order $n$ is $\sigma(n/2) + d(n/2)$, where $\sigma$ is the sum of divisors function and $d$ is the divisor count function. Thus the dihedral group $D_n$ of order $n$ is an example for the problem when $$n = 2,\ 6,\ 28,\ 104,\ 260,\ 368,\ 1312,\ 17296,\ 24016,\ 69376,\ \ldots$$

I don't know if this sequence is infinite. For more terms, it is $2 \cdot$ $A083874$ from OEIS. Seems that really large examples exist, for example $9223653647124987904$ is in the sequence.

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3  
The appearance of $D_6$ and $D_{28}$ made me think we'd see $D_m$ for all perfect even numbers $m$, so I was surprised 496 wasn't in the list. – Gerry Myerson Oct 14 '12 at 21:57

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