4
$\begingroup$

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?


Edit:

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:

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,

$\endgroup$
5
$\begingroup$

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 $S$.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – user47958 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.