I need to find the normal subgroup lattice of the group $U_{6n} = \langle a,b | a^{2n} = b^ 3= 1, a^{-1}ba = b^{-1}\rangle$. To the best of my knowledge this group was introduced at first by GORDON JAMES and MARTIN LIEBECK in their book "Representations and Characters of Groups". I have recently been working on normal subgroup lattices of finite groups. I can obtain the normal subgroup lattice structure of all groups in the mentioned book other than $U_{6n}$.

What is the normal subgroup lattice of $U_{6n}$ or where can I find it?


I am very thankful for the answer and the comment given by Derek Holt_ but his formula for computing the number of normal subgroups seems to be incorrect. To show this, I consider a simple GAP program (written by my student Ms. Fatemeh Moftakhar) to compute the number of normal subgroups and 2n + d(n), where d(n) denotes the number of divisors of n. My program and its output is as follows:

local f,a,b,r,v;
return v;


for i in [1,2..20] do

for i in [1,2..20] do
t:=2*i + Size(DivisorsInt(i));

Output of the Program:

[ 3, 5, 6, 7, 6, 10, 6, 9, 9, 10, 6, 14, 6, 10, 12, 11, 6, 15, 6, 14 ],
[ 3, 6, 8, 11, 12, 16, 16, 20, 21, 24, 24, 30, 28, 32, 34, 37, 36, 42, 40, 46 ].



Firstly you get the set $S$ of all subgroups containing $b$, which have the form $\langle a^i,b \rangle$ for various $i$, and correspond to the subgroups of the cyclic group $\langle a \rangle$ of order $2n$.

Now if $H \lhd U_{6n}$ and $H \not\le C_G(b) = \langle a^2,b \rangle$, then $b \in [b,H]$, so $b \in H$.

So the set $T$ consisting of the remaining normal subgroups $H$ of $U_{6n}$ (i.e. those that do not contain $b$) are subgroups of the abelian group $\langle a^2 \rangle \times \langle b \rangle$. But, if $a^ib \in H$ then so is its conjugate $a^ib^{-1}$, so $b \in H$, contradiction. Hence $T$ just consists of the subgroups of the cyclic group $\langle a^2 \rangle$ of order $n$. Note that each subgroup in $T$ is contained in some of the sugbroups of $S$.

  • $\begingroup$ If the OP wants the lattice structure of normal subgroup I think this does not answer the question completely. For example I'm not sure if it shows how many normal subgroups exist. $\endgroup$ – Minimus Heximus Oct 19 '14 at 18:21
  • $\begingroup$ But I am sure that it does answer the question completely. The number of normal subgroups if the sum of the numbers of divisors of $n$ and $2n$. $\endgroup$ – Derek Holt Oct 19 '14 at 21:26
  • $\begingroup$ Dear Derek, Thanks for your comments. I edited my question containing some counterexamples for your formula. Could you please see edited question? $\endgroup$ – Ali Reza Ashrafi Oct 20 '14 at 15:10
  • $\begingroup$ @Ali: You misinterpreted Derek's formula, which is $d(2n) + d(n)$, not $2n + d(n)$... $\endgroup$ – Tom De Medts Oct 20 '14 at 16:03
  • $\begingroup$ Everything is OK! Thanks to Tom for his comment and Thanks to Derek for his nice argument. $\endgroup$ – Ali Reza Ashrafi Oct 20 '14 at 16:35

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