\chapter{Product Groups}

\section{Direct Product Groups}



There are several group constructions that are called 'product groups', however the simplest is perhaps the \emph{direct prodcut group}; a group formed by taking the cartesian product of two groups (hence creating ordered pairs) with the new operation applying both group's operations component wise.

\begin{definition}[Direct product of 2 groups]
Let $(G,\circ_G)$ and $(H,\circ_H)$ be groups, the \emph{direct product of $G$ and $H$} is a group $(G \times H, \circ)$ where $G\times H$ and $\circ : G \times H \to G \times H$ are defined as such.
\begin{itemize}
	\item $G \times H$ is the cartesian product of the group sets
	\item $ (g_1,h_1) \circ (g_2,h_2)=(g_1\circ_G g_2 , h_1 \circ_H h_2) $
\end{itemize}
\end{definition}

This construction is indeed always a group; it can be checked rather quickly. It's really the simplest way to slap groups together; there isn't even a condition for $H,G$ to be related to eachother in any way!

\begin{proposition}
\begin{itemize}
\item $(1_G ,1_H)$ is the identity element
\item $(g,h)^{-1}=(g^{-1} ,h^{-1})$
\item $G \cong G \times \{1_H\}$ and $H \cong \{1_G\} \times H)$ 
\end{itemize}
\end{proposition}





%\subsection{A 'Chinese remainder theorem' for groups}

%One interesting problem is to 'factor' groups; find an isomorphic representation of a group via products of more basic groups.

%This isn't 


%Readers familiar with elementary number theory may recall the Chinese remainder theorem; it can be restated in terms of group theory thanks to product groups.



\begin{proposition}[Chinese remainder theorem (group theory)]
Let $(n_i)_{i=1}^{k} \in \mathbb{Z}$ be coprime integers and $N = \prod^{k}_{i=1} n_i$. Then $ \mathbb{Z} / N \mathbb{Z} \cong \large\times^{k}_{i=1} \mathbb{Z} / n_i \mathbb{Z}$.
\end{proposition}



\section{Product of group subsets}

Factoring a group into a direct product representation is a colorful process; .

Since factoring groups into direct product representations is quite an unruly process, let's consider just the case of factoring groups into subgroups. There is a theorem that states exactly when this can be done, however it requires delving into the idea of \emph{products of subgroups}.

For completeness, we first introduce the less powerful \emph{product of group subsets}; a precursor to product of subgroups.

\begin{definition}[Product of group subsets]
A \emph{product of group subsets} is the set closed by composing elements of two subsets together.
\[ S,T \subseteq G \implies ST= \{ st : s \in S \land t \in T \} \]
\end{definition}

There are a few important points to note about the definition we've just created. Firstly, $S$ and $T$ are subsets of $G$ and not necessarily subgroups of $G$. Secondly, we can interpret $ST$ as the composition closure of elements of $S$ and $T$ from the left and right respectively.

When checking to see if inverses exist, the identity exists, and composition is well defined, we can see that this doesn't necessarily yield a group. 


\subsection{Product of subgroups}

Since $S$ and $T$ aren't subgroups, we're restricted from invoking many properties of groups; all we really know is that $s \circ t$ is well defined, but that's about it. This construction is a group merely when it is closed under inversion, contains the identity, and composition is well defined. To make our studies less insipid, let's look at the case when forming \emph{product of subgroups}.


\begin{definition}[Product of subgroups]
A \emph{product of subgroups} is the set closed by composing elements of two subgroups together.
\[ H,K \leq G \implies HK= \{ hk : h \in H \land k \in K \} \]
\end{definition}


Unfortunately products of subgroups aren't necessarily subgroups either, but the following proposition tells us exactly when they are groups.

\begin{proposition}
Let $G$ be a group with subgroups $ H,K \leq G$ Then $HK \leq$ iff $HK=KH$
\[ HK \leq G \iff HK=KH \]
Furthermore, if $HK \leq G$ then $\langle H \cup K\rangle = HK$.
\end{proposition}

If $HK$ is a group, then for any $hk\in HK$ we also have $(hk)^{-1}=k^-{-1}h^{-1} \in HK$, however $k^{-1}h^{-1} \in KH$. Since the inverse map $f(hk)=k^{-1}h^{-1}$ is a bijection from $HK$ to $KH$, we see that a necessary condition is that $HK=KH$.

To prove that $HK=KH$ is also a sufficient condition is left as an exercise for the reader; it's not too difficult (plus I can't be fucked explaining).



\begin{proposition}[Dedekind's modular law]
Let $G$ be a group with subgroups $ H,K,L \leq G$ and $K \subseteq L$.
\[ (HK)\cap L = (H\cap L )K\]
\end{proposition}


\subsection{Second isomorphism theorem}


Second isomorphism theorem is a pleothora of results pertaining to products of subgroups where one of the subgroups is normal.

Consider $HN$, by corollary of our previous proposition, $HN \leq G$. 

$N \lhd NH$ since $N \subseteq NH$ and $N \lhd G$
.
Consider the 'reduced' quotient map $\pi : H \to G / N $ one sees that $\mathrm{ker}(\pi)= H \cap N$ and hence $H \cap N \lhd H$ (since the kernel is a normal subgroup of the domain group).

The fact that $HN /N \cong H/ (H \cap N)$ follows by applying the first isomorphism theorem to our $\pi$, noting that $\mathrm{Im}(\pi)= HN / N$

\begin{proposition}[Second isomorphism theorem]
Let $H \leq G$, $N \lhd G$, then the following hold.
\[HN \leq G\]
\[N \lhd HN\]
\[H \cap N \lhd H\]
\[HN /N \cong H/ (H \cap N)\]
\end{proposition}




\subsection{Third isomorphism theorem}

When considering product of normal subgroups, even more can be proven

\begin{proposition}[Third isomorphism theorem]
Let $N,M \lhd G$, then the following hold.
\[M / N \lhd G / N\]
\[(G / N) / (M /N) \cong G / M\]
\end{proposition}










\subsection{Product of subgroups and direct product}

An interesting question to ask is when our product of subgroups is isomorphic to the direct product of those subgroups. The following proposition reveals all.

\begin{proposition}
Let $G$ be a group and $N,M \lhd G$ normal subgroups with $N\cap M =\{1_G\}$. Then we have $N \times M \cong NM $
\[ N,M \lhd G \land N \cap M = \{1_G\} \implies NM \leq G \land M\times N \cong MN \]
\end{proposition}

Why does this hold? By the second isomorphism theorem we definitely have $NM \leq G$ (since it would be enough to have only one normal subgroup). We then consider the function $f(n,m)=nm$, for which we want to prove to be an isomorphism. While proving that $f(n_1 n_2 , m_1 m_2 ) = f(n_1,m_1) f(n_2,m_2)$, we end up with $f(n_1 n_2 , m_1 m_2 ) = n_1 n_2 m_1 m_2$; we must commute those middle terms to prove the result (i.e we need $n_2 m_1 = m_1 n_2$, or equivalently $n_2 m_1 n_{2}^{-1} m_{1}^{-1}=1_{G}$).

Since $N$ and $M$ are both normal, we see that $n_2 m_1 n_{2}^{-1} m_{1}^{-1}$ is in both $N$ and $M$ since we can consider $n_2 m_1 n_{2}^{-1} = m_{*}$ and $m_1 n_{2}^{-1} m_{1}^{-1} = n_{*}$. But because $N \cap M =\{1_G\}$, this must be the identity (or equivalently, the elements commute). $f(1_N,1_M)=1_N 1_M$ is trivial, so $f$ is a homomorphism.

To check for bijectivity, we note that surjectivity is trivial and start checking injectivity. The trivial intersection ensures that elements of $NM$ have a unique representation of the form $nm$, since if $n_1 m_1 = n_2 m_2$, we then have $n_{2}^{-1} n_1 = m_{2} m_{1}^{-1}$, so $n_{2}^{-1} n_1 \in N \cap M$ and therefore the trivial intersection implies that $n_{2}^{-1} n_1 =1_G$ and $n_2 = n_1$. A similar argument shows that $m_2 = m_1$.


If our product of normal subgroups equals the group, then it turns out that we've found a direct product representation for the group from its own normal subgroups!

\begin{corollary}
\[ N,M \lhd G \land N \cap M = \{1_G\} \land NM=G \implies N\times M \cong G \]
\end{corollary}



It is worth noting that this is only a sufficient condition and is not the only condition by which a group can be factored as a direct product; finding direct product group representation can be quite a difficult problem. Recall that we also have CRT from earler, which works with completely different conditions and on an narrower class of groups (cyclic ones with composite order).

That said, our little theorem here is a good connection between the notion of a product of subgroups and a direct product.



\section{Inner semidirect product group}


\begin{definition}[Inner semidirect product]
Let $N \lhd G$, $H \leq G$, and $N \cap H = \{1_G\}$, then the inner semidirect product $N \rtimes H$ is defined as such.
\[N \rtimes H = NH\]
\end{definition}


The innder semidirect product is a generalization of the pro
\begin{proposition}
\[ H \leq G \land N \lhd G \land HN \cong G \iff H \cap N =\{1_G\} \land HN=G \]
\end{proposition}

This proposition gives a sufficient condition for the product of group subsets to form a group; this is known as the inner semidirect product group and is made by 'factoring' the group into a subgroup and normal subgroup as discussed.


\begin{definition}[Inner semidirect product group]
\[H \leq G \]
\[N \lhd G \]
\[N \rtimes H \iff NH=G \land N \cap H = \{1_{G}\}\]
\end{definition}

Note that $N \rtimes H$ is trivially a group since by definition it equals $G$ (which is a group by default).

There are equivalent ways of defining a semidirect prouct group by restating our definition directly in terms of group elements or by homomorphisms or isomorphisms.






\section{Outer semidirect product group}








