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?




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**Edit:**


I am very thankful for the [answer](https://mathoverflow.net/a/184794/47958) and the comment given by [Derek Holt](https://mathoverflow.net/users/35840/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:



```
g:=function(n)
local f,a,b,r,v;
f:=FreeGroup("a","b");
a:=f.1;
b:=f.2;
r:=[a^(2*n),b^3,b*(a^(-1))*b*a];
v:=f/r;
return v;
end;

AD:=[];BD:=[];

for i in [1,2..20] do
t:=NormalSubgroups(g(i));
b:=Size(t);
Add(AD,b);
od;

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

```

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 ].

```

Best, 



Normal subgroup lattice of the group $U_{6n}$
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 $S$.


