Abstract Nonsense

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

Point of post: In this post we discuss several different formulations of normal subgroups and discuss how one can define a group structure on the set of cosets of such a normal subgroup.

Motivation

One sees the idea of a “quotient structure” pop up frequently in mathematics. There are quotient spaces in both topology and linear algebra for example. One mods out by the set-theoretic kernel of semi-metric to turn it into a metric etc. Of course, for groups the same holds. It turns out though that like algebra there are some pretty serious theorems to do with quotient groups (the isomorphism theorems) which we shall see in subsequent posts. The problem is that unlike both linear algebra (where all the underlying group structure of the vector spaces was assumed abelian) and topology the way we “want to” define a quotient group doesn’t always work; there is a necessary and sufficient condition on a subgroup such that the quotient group structure “makes sense”. This condition is normality which, in a nutshell, says that the subgroup is invariant under conjugation (or that it’s invariant under all inner automorphisms).

Inner Automorphisms

Let $G$ be a group, and fix $g_0\in G$. We define the inner automorphism induced by $g_0$, denoted $i_{g_0}$, to be the map

$i_{g_0}:G\to G:g\mapsto g_0^{-1}g g_0$

It’s not difficult to see that $i_{g_0}$ is deserving of it’s name in that it really is an automorphism. Moreover, consider that for any fixed $g_1,g_2\in G$ and variable $g\in G$ we have that

$i_{g_1}\left(i_{g_2}\left(g\right)\right)=i_{g_1}\left(g_2^{-1}gg_2\right)=g_2^{-1}g_1^{-1}gg_1g_2=\left(g_1g_2\right)^{-1}\left(g_1g_2\right)=i_{g_1g_2}(g)$

so that the map

$\Phi:G\to \text{Aut}(G):g\mapsto i_g$

is a homomorphism. We call $\Phi\left(G\right)\leqslant \text{Aut}(G)$ the inner automorphism group of $G$ and denote it $\text{Inn}(G)$.

Normal Subgroups

Let $G$ be a group and $N\leqslant G$. We say that $N$ is a normal subgroup of $G$, and denote this $N\trianglelefteq G$, if for every $g\in G$ the following holds:

$i_g\left(N\right)=N$

There are several equivalent definitions of normality. Namely:

Theorem: Let $G$ be a group and $N\leqslant G$. Then, the following are equivalent

\displaystyle \begin{aligned}&\mathbf{(1)}\quad N\trianglelefteq G\\ &\mathbf{(2)}\quad gN=Ng\textit{ for all }g\in G\\ &\mathbf{(3)}\quad \textit{The set of left cosets of }N\textit{ coincides with the set of right cosets}\\ &\mathbf{(4)}\quad i_g\left(N\right)\subseteq N\textit{ for all }g\in G\\ &\mathbf{(5)}\quad \textit{There exists a group }G'\textit{ and a homomorphism }\theta:G\to G'\textit{ such that }N=\ker\theta\end{aligned}

Proof:

$\mathbf{(1)}\implies \mathbf{(2)}$: Let $g\in G$ be arbitrary. We know that

$i_g(N)=gNg^{-1}=N$

so, let $h\in gN$ then $h=gn$ for some $n\in N$ and so $g^{-1}h\in N=gNg^{-1}$. Thus, there exists some $n'\in N$ such that $g^{-1}h=g^{-1}n'g$ and so $h=n'g$ so that $h\in Ng$. The reverse inclusion is done identically.

$\mathbf{(2)}\Longleftrightarrow\mathbf{(3)}$: The forward implication is trivial. To do the reverse implication note that since $gN$ is a right coset we have that $gN=Nh$ for some $h\in G$. But, since $g\in gN$ it follows that $g\in Nh$. But, recall that the right cosets of a subgroup partition $G$ and since $g\in Ng$ it follows that $Nh=Ng$

$\mathbf{(2)}\implies\mathbf{(4)}$: Let $h\in g^{-1}Ng$, then $h=g^{-1}ng$ for some $n\in G$. In particular we have that $gh=ng$. Thus, $gh\in Ng=gN$ and thus there exists some $n'\in N$ such that $gh=gn'$ and thus $h=n'\in N$.

$\mathbf{(4)}\implies\mathbf{(1)}$: Note that since $i_g\left(N\right)\subseteq N$ for all $g\in G$ we need only prove the reverse inclusion for all $g\in G$. It clearly suffices to do this for a fixed but arbitrary $g_0\in G$. To do this let $n\in N$. By assumption we have that $i_{g_0^{-1}}\left(N\right)\subseteq N$ and so there exists $n'\in N$ such that $g_0 n g_0^{-1}=n'$ or equivalently $n=g_0^{-1}n'g_0$ and thus $n\in i_{g_0}\left(N\right)$ as desired.

$\mathbf{(1)}\Longleftrightarrow \mathbf{(5)}$: To prove the reverse implication we note that if $\theta:G\to G'$ is a homomorphism with $N=\ker\theta$ then for every $g\in G$ we have that

$\theta\left(g^{-1}Ng\right)=\theta\left(g^{-1}\right)\theta\left(N\right)\theta(g)=\theta\left(N\right)=e_{G'}$

so that $g^{-1}Ng\subseteq N$. The conclusion follows by appealing to our last implication.

We stall the proof of the converse to a later point in this post. $\blacksquare$

We next wish to know how normal subgroups interact with homomorphisms. In particular we note that:

Theorem: Let $G$ and $G'$ be groups and $N\trianglelefteq G$ and $N'\trianglelefteq G'$. Then, if $\theta\in\text{Hom}\left(G,G'\right)$; $\theta^{-1}\left(N'\right)\trianglelefteq G$. If $\theta$ is surjective then $\theta\left(N\right)\trianglelefteq G'$.

Proof: To prove the first part we let $g\in G$ and note that

\begin{aligned}\theta\left(g^{-1}\theta^{-1}\left(N'\right)g\right) &= \theta\left(g^{-1}\right)\theta\left(\theta^{-1}\left(N'\right)\right)\\ &= \theta(g)^{-1}\theta\left(\theta^{-1}\left(N'\right)\right)\theta(g)\\ & \subseteq \theta(g)^{-1}N'\theta(g)\\ &= N\end{aligned}

and thus $i_g\left(\theta^{-1}\left(N'\right)\right)\subseteq \theta^{-1}\left(N'\right)$ for every $g\in G$. It follows from our earlier characterization of normality that $\theta^{-1}\left(N'\right)\trianglelefteq G$.

Suppose now that $\theta$ is surjective. We note then that for any $g'\in G'$ we have that $g'=\theta(g)$ for some $g\in G$. Thus,

$\theta(g)^{-1}\theta\left(N\right)\theta(g)=\theta\left(g^{-1}N\theta(g)\right)=\theta(N)$

from where normality follows. $\blacksquare$

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

December 31, 2010 -

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