Reducing Normal Subgroup Calcs: Finite Groups Only?

[landor]

My abstract algebra book is talking about reducing the calculations involved in determining whether a subgroup is normal. It says:

If N is a subgroup of a group G, then N is normal iff for all g in G, gN(g^-1) [the conjugate of N by g] = N.

If one has a set of generators for N, it suffices to check that all conjugates of these generators lie in N to prove that N is a normal subgroup (because the conjugate of a product is the product of the conjugates and the conjugate of the inverse is the inverse of the conjugate).

Similarly, if generators for G are known, then it suffices to check that the conjugates of N by these generators all equal N [these generators for G normalize N].

In particular, if generators for both N and G are known, then this reduces the calculations to a small number of conjugations to check. If N is a FINITE group then it suffices to check that the conjugates of a set of generators for N by a set of generators for G are again elements of N.

My question is: why does it stipulate that N be finite? Wouldn't this work if N has infinite order as well?

Thanks!

 

[Martin Rattigan]

Not quite.

Let ${C_2=\{e,x\}}$ be the cyclic group of order 2. Let ${H=\{\lambda|\lambda:\mathbb{Z}\rightarrow C_2\wedge\lambda^{-1}(x)\text{ is finite}\}}$ with pointwise multiplication, i.e. for ${\lambda,\mu\in H,\lambda\mu:n\in\mathbb{Z}\mapsto \lambda(n)\mu(n)}$. (That is H is a direct product of copies of $C_2$ with $\mathbb{Z}$ as index set). For ${\lambda\in H}$, let ${\lambda_k:\mathbb{Z}\rightarrow C_2}$ be ${\lambda_k:n\mapsto \lambda(n-k)}$ and for ${k\in\mathbb{Z}}$ let [itex]{\phi_k\rightarrow H}[/itex] be ${\phi_k:\lambda\mapsto \lambda_k}$.

Then $H$ is a group and each $\phi_k$ is an isomorphism of $H$. Morover $\phi_j\circ\phi_k=\phi_{j+k}$, so ${\Phi=\{\phi_k|k\in\mathbb{Z}\}}$ is a group isomorphic to $\mathbb{Z}$ under addition, with the isomorphism $\theta:k\in\mathbb{Z}\mapsto \phi_k$.

We can then form the semidirect product ${G=\mathbb{Z}\times_\theta H}$ as $\{(k,\lambda):k\in\mathbb{Z}\wedge\lambda\in H\}$ with the product ${(k,\lambda)(k',\lambda')=(k+k',\lambda_{k'}\lambda')}$.

The set ${\{(1,\epsilon)\}\cup\{(0,\lambda^{(r)})|r\in\mathbb{Z}\}}$ generates $G$, where ${\epsilon:n\in\mathbb{Z}\mapsto e}$ and ${\lambda^{(r)}:r\mapsto x,\lambda^{(r)}:s\neq r\mapsto e}$.

If we consider the subgroup $N$ of $G$ generated by ${\{(0,\lambda^{(r)}):r\geq 0\}}$ this is not normal in G, because, e.g. ${(-1,\epsilon)^{-1}(0,\lambda^{(0)})(-1,\epsilon)=(0,\lambda^{(-1)})}$ and ${(0,\lambda^{(-1)})\notin N}$.

On the other hand ${(1,\epsilon)^{-1}(0,\lambda^{(r)})(1,\epsilon)=(0,\lambda^{(r+1)})}$ and ${(0,\lambda^{(s)})^{-1}(0,\lambda^{(r)})(0,\lambda^{(s)})=(0,\lambda^{(r)})}$, so conjugation of each member of a set of generators of $N$ by each member of a set of generators of $G$ remains in $N$.

If you check both $g^{-1}ng$ and $gng^{-1}$ for each of the generators $g$ of $G$ and $n$ of $N$ this should work in all cases.

 

[landor]

Thanks!

Your final sentence might be explaining this, but why are you using $$g^{-1}ng$$ as your criterion for normality instead of $$gng^{-1}$$?

 

[Martin Rattigan]

Thanks!

Your final sentence might be explaining this, but why are you using $$g^{-1}ng$$ as your criterion for normality instead of $$gng^{-1}$$?

Sorry. I did notice I'd switched sides, but I didn't want to go back and revise it.

I was brought up at a time when many algebraists liked to write their mappings on the right, so that composite mappings are applied in the order you read them instead of right to left. In this case to get a homomorphism $\phi$ from a group to its inner automorphisms the inner automorphism $g^\phi$ should be $x\mapsto g^{-1}xg$, when $x^{gh^\phi}=x^{g^\phi h^\phi}$. In this case also the normal subgroup should be the right hand factor in the semidirect product.

If you read it in a mirror you should have no problem. Obviously the two conventions are equivalent.

The point of the final sentence is that if the group of automorphisms of $N$ may be infinite (which would not be the case were $N$ finite), then you need to also check that the inverses of the inner automorphisms corresponding to the generators $G$ map $N$ into itself to prove $N$ is normal.

 

[blue_raver22]

I would respond by saying that the stipulation that N be finite is necessary in order for the statement to hold true in all cases. The reasoning behind this is that for finite groups, the set of generators is a finite set, and therefore the number of conjugates that need to be checked is also finite. However, for infinite groups, the set of generators may also be infinite, and therefore the number of conjugates that need to be checked may also be infinite. This makes it impossible to reduce the calculations to a small number of conjugations, as stated in the book. Therefore, in order for the statement to hold true, it is necessary for N to be a finite group.