Abstract Nonsense

Crushing one theorem at a time

Review of Group Theory: Normal Subgroups and Quotient Groups (Pt. II)


Point of post: This post is a continuation of this one.

There are also some structure properties of normal subgroups.

Theorem: Let G be a group and \left\{N_\alpha\right\}_{\alpha\in\mathcal{A}} be such that N_{\alpha}\trianglelefteq G for every \alpha\in\mathcal{A}. Then, \displaystyle \bigcap_{\alpha\in\mathcal{A}}N_\alpha\trianglelefteq G.

Proof: To do this we merely note that for any g\in G

\displaystyle i_g\left(\bigcap_{\alpha\in\mathcal{A}}N_\alpha\right)\subseteq\bigcap_{\alpha\in\mathcal{A}}i_g\left(N_\alpha\right)=\bigcap_{\alpha\in\mathcal{A}}N_\alpha

from where the conclusion follows. \blacksquare

 

Another clear one is

Theorem: Let G be a group with N\trianglelefteq G and H\leqslant G. Then, N\cap H\trianglelefteq H.

Proof: We merely note that for any h\in H

\displaystyle i_h\left(N\cap H\right)\subseteq i_h\left(N \right)\cap i_h\left(H\right)=N\cap H

and thus from our early characterization of normality it follows that N\cap H\trianglelefteq H. \blacksquare

Remark: The converse of this clearly does not hold.

 

Lastly, we prove a commonly used theorem (which is a specific case of a more general theorem I hope I will eventually get to discuss):

Theorem: Let G be a group and H\leqslant G with \left(G:H\right)=2. Then, H\trianglelefteq.

Proof: It suffices to show (by our characterizations of normality) that the set of right cosets of H and set for left cosets of H coincide. Note though that the set of left cosets is of the form \left\{H,aH\right\} for some a\in G and the set of right cosets are of the form \left\{H,Hb\right\} for some b\in G. But since both of these are partitions it follows that aH=G-H=Hb from where the conclusion follows. \blacksquare

 

 

Quotient Groups

We now wish to define a group structure on G/N for normal N. In particular:

Theorem: Let G be a group and N\leqslant G. Then, the operation \left(g N\right)\left(hN\right)=(gh)N on G/N is well-defined (in the sense that if gN=g'N and hN=h'N then (gh)N=(g'h')N)  if and only if N\trianglelefteq. Moreover, if N\trianglelefteq G then G/N is a group under this operation with \left(gN\right)^{-1}=g^{-1}N and e_{G/N}=eN.

Proof: Assume first that the described relation is well-defined and let g\in G and n\in N be arbitrary. Note then that by assumption

(eg^{-1})N=(ng^{-1})N

note though thatng^{-1}\in (ng^{-1})N and thus there exists some n'\in N such that g^{-1}n'=ng^{-1}.Multiplying on the right gives g^{-1}ng=n' and so g^{-1}ng\in N. Since n and g were arbitrary it follows that i_g\left(N\right)\subseteq N for every g\in G. By our previous theorem we may conclude that N\trianglelefteq G.

Conversely, suppose that N\trianglelefteq G and suppose that g,g',h,h'\in G are such that

gN=g'N\text{ and }hN=h'N

we prove that

(gh)N=(g'h')N

To do this we first note that clearly there exists n_1,n_2\in N such that g'=gn_1 and h'=hn_2. Note though that

\begin{aligned}g'h' &= (gn_1)(g'n_2)\\ &= g(hh^{-1})n_1hg_2\\ &= gh\left(h^{-1}n_1h\right)n_2\\ &= \left(gh\right)\left(n_3n_2\right)\end{aligned}

where n_3=h^{-1}nh\in N. Thus, it follows that g'h'\in (gh)N and since the left cosets of N partition G it follows that (gh)N=(g'h')N as desired.

The verification of the group axioms as are the stated properties about inverses and identity elements. \blacksquare

We call the above group structured defined on G/N to be the quotient group of G mod N (or just G mod N for short). We also never denote the identity element of G/N by eN but rather by just N.

 

For finite groups there is a nice formula for the order of a quotient group. Namely:

Theorem: Let G be finite and N\trianglelefteq G. Then,

\displaystyle \left|G/N\right|=\left(G:N\right)=\frac{|G|}{|N|}

Proof: This is evident from Lagrange’s Theorem.

While not particularly profound the above theorem has an interesting corollary:

Corollary: Let G be a finite group and N\trianglelefteq G. Then, for any g\in G, g^{\left(G:N\right)}\in N.

Proof: By Lagrange’s theorem we have that

g^{\left(G:N\right)}N=\left(gN\right)^{\left(G:N\right)}=N

But, we know that this will be true if and only if g^{\left(G:N\right)}\in N. \blacksquare

 

It’s clear from definition that we have a nice surjective homomorphism from G onto G/N. Namely we constructed the operation on G/N in such a way that the following theorem is true:

 

Theorem: Let G be a group and N\trianglelefteq G. Then, the map

\pi:G\to G/N:g\mapsto gN

is a surjective homomorphism. Moreover, \ker\pi=N.

Proof: By definition we have that

\pi(ab)=(ab)N=\left(aN\right)\left(bN\right)=\pi(a)\pi(b)

and \pi is surjective since every element of G/N is of the form gN for some g\in\ G. To prove that N=\ker(\pi) we clearly note that N\subseteq\ker(\pi) and if g\in\ker(\pi) then N=\pi(g)=gN and this is true if and only if g\in N. The conclusion follows. \blacksquare

 

Remark: This mapping is called the canonical projection of G onto G/N.

 

Corollary: The converse of \mathbf{(5)} in our characterization of normality is now proven since every normal subgroup may be realized as the quotient of its canonical projection.

 

Corollary: If G is an abelian group and N\trianglelefteq G then G/N is abelian. If G is cyclic then G/N is cyclic. In general if G has a property invariant under epimorphisms (surjective homomorphisms) then G/N has the same property.

 

References:

1.  Lang, Serge. Undergraduate Algebra. 3rd. ed. Springer, 2010. Print.

2. Dummit, David Steven., and Richard M. Foote. Abstract Algebra. Hoboken, NJ: Wiley, 200



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January 1, 2011 - Posted by | Algebra, Group Theory, Uncategorized | , , ,

9 Comments »

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