# Abstract Nonsense

## 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 that$ng^{-1}\in (ng^{-1})N$ 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 -

## 9 Comments »

1. […] from our last post that if and are groups and then . We claim first that in essence if then for some . In […]

Pingback by Review of Group Theory: The First Isomorphism Theorem « Abstract Nonsense | January 2, 2011 | Reply

2. […] Note that since we have that induces the canonical projection . Note though then […]

Pingback by Review of Group Theory: Interesting Consequence of the First Isomorphism Theorem « Abstract Nonsense | January 2, 2011 | Reply

3. […] We recall that since and that . Note though that and thus it follows that […]

Pingback by Review of Group Theory: The Second Isomorphism Theorem « Abstract Nonsense | January 2, 2011 | Reply

4. […] We recall that since and that . Note though that and thus it follows that […]

Pingback by Review of Group Theory: The Second Isomorphism Theorem « Abstract Nonsense | January 2, 2011 | Reply

5. […] Suppose first that and let be the canonical projection. We note then that for any one has that (since ) and so . Then, since is surjective the […]

Pingback by Review of Group Theory: The Commutator Subgroup and the Abelianization of a Group « Abstract Nonsense | February 27, 2011 | Reply

6. […] first that is solvable and . Let be the sequence of subgroups such that is abelian. Define . By first principles we know that . Moreover, we claim that is abelian for . Indeed, by the second isomorphism theorem […]

Pingback by Review of Group Theory: Solvable Groups « Abstract Nonsense | March 11, 2011 | Reply

7. […] , since is odd and we know   we know that is odd and so for some . Thus, that said if is the canonical projection onto then and thus by definition . Since was arbitrary the conclusion […]

Pingback by University of Maryland College Park Qualifying Exams (Group Theory and Representation Theory) (August-2004) « Abstract Nonsense | May 6, 2011 | Reply

8. […] what we’d like to do is make a construction similar to that of a quotient group but for rings. Namely, for some subset of a ring we’d like to define a ring structure on […]

Pingback by Quotient Rings (Pt. I) « Abstract Nonsense | June 25, 2011 | Reply

9. […] math, in [6]). Consider then the quotient group with the quotient topology inherited by the canonical projection –note that is connected, being the quotient of a connected space. We’d like to endow […]

Pingback by Riemann Surfaces (Pt. II) « Abstract Nonsense | October 2, 2012 | Reply