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\vspace{-1em}

To see a typical application of
\nameref{halls_thm}, consider $G$ a finite group
and $H$ a subgroup of $G$. We have the left cosets
$L_1, \ldots, L_k$ say $g_1 H, \ldots, g_k H$ and
right cosets $R_1, \ldots, R_k$, say $H g_1',
\ldots, H g_k'$, where $k$ is the index of $H$ in
$G$.

Can we pick the same representatives?
In other words, can we pick some $g_1, \ldots,
g_k$ such that the left cosets are $g_1 H, \ldots,
g_k H$ and the right cosets are $H g_1, \ldots, H
g_k$, up to reordering? Equivalently, can we
reorder the $R_i$ such that $L_i \cap R_i \neq
\emptyset ~\forall i$? In other words, we seek a
matching in the bipartite graph with vertex sets
$X = \{L_1, \ldots, L_k\}$, $Y = \{R_1, \ldots,
R_k\}$, for which $L_i$ is joined to $R_i$ if
their intersection is non-empty.

By \nameref{halls_thm}, it is enough to show
$|\Gamma(A)| \ge |A| ~\forall A \subset X$. In
other words, we need $\bigcup_{i \in A} L_i$ meets
at least $|A|$ right cosets. But $\bigcup_{i \in
A} L_i = |A| |H|$, $\bigcup_{i \in A} L_i$ must
meet at least $|A|$ right cosets as each right
coset has size $|H|$.

\section{Connectivity}

How ``connected'' is a connected graph? A path $P$
is connected, $P - x$ is disconnected for some $x
\in P$. A cycle $C$ is connected, and also $C - x$
is connected $\forall x \in C$, but $C - x - y$ is
disconnected for some $x, y \in C$.

If $G$ is a ``cube'', then $G - x - y$ is
connected $\forall x, y \in G$, but can be
disconnected by removing three vertices.

\begin{flashcard}[connectivity-of-a-graph-defn]
\begin{definition*}[Connectivity]
  \glsadjdefn{k_conn}{$k$-connected}{\gls{graph}}
  \cloze{
  For a graph $G$ with $|G| \ge 2$, the
  \emph{connectivity} of $G$, written $\kappa(G)$,
  is the least size of a set $S \subset V$ such
  that $G - S$ is disconnected or a single point.
  If $\kappa(G) \ge k$, we say $G$ is
  $k$-connected. Equivalently, $G$ is
  $k$-connected if and only if $|G| > k$ and
  for all $S \subset V$ with $|S| < k$, $G - S$ is
  connected.
  }
\end{definition*}
\end{flashcard}

\begin{example*}
  \phantom{}
  \begin{enumerate}[(1)]
    \item No tree is $2$-connected.
    \item $C_n$ is $2$-connected but not
      $3$-connected.
    \item The ``cube'' is $3$-connected.
  \end{enumerate}
\end{example*}

\begin{warning*}
  We can have $\kappa(G - x) > \kappa(G)$. For
  example, consider a cycle joined to a vertex
  $x$. Then $\kappa(G) = 1$, but $G - x = C_n$, so
  $\kappa(G - x) = 2$.
\end{warning*}

\begin{remark*}
  We must have $\kappa(G) \le \delta(G)$, since
  for any $x \in G$, removing $\Gamma(x)$ from $G$
  disconnects it.
\end{remark*}
\vspace{-1em}

We have that if $G$ is $1$-connected, then there
is an $ab$-path in $G$ for all distinct $a, b \in
G$.

It would be nice if $G$ being $k$-connected
implies there exists a family of $k$ independent
paths from $a$ to $b$.

\begin{flashcard}[independent-paths-defn]
\begin{definition*}[independent paths]
  \cloze{We say that two $ab$-paths $P_1$ and $P_2$ are
  \emph{independent} if $P_1 \cap P_2 = \{a,
  b\}$.}
\end{definition*}
\end{flashcard}

\begin{flashcard}[separator-defn]
\begin{definition*}[$ab$-separator]
  \glsnoundefn{separator}{separator}{separators}
  \cloze{For distinct $a, b \in G$, say $S \subset
  V \setminus \{a, b\}$ separates $a$ and $b$, or
  is an $ab$-separator, if $a$ and $b$ are in
  different components of $G - S$.

  Equivalently, every $ab$-path meets $S$.}
\end{definition*}
\end{flashcard}

\begin{flashcard}[mengers-thm]
\begin{theorem}[Menger's Theorem]
  \label{mengers_thm}
  \cloze{
  Let $a$ and $b$ be distinct non-adjacent
  vertices in a graph $G$ such that every
  $ab$-separator has size at least $k$. Then $G$
  contains a family of $k$ independent $ab$-paths.
  }
\end{theorem}
\end{flashcard}

\begin{remark*}
  \phantom{}
  \begin{enumerate}[(1)]
    \item The converse is trivial. It we have $k$
      independent paths from $a$ to $b$, then any
      separator $S$ must meet each path.
    \item An equivalent formulation says that the
      minimum size of an $ab$-separator is the
      maximum size of an independent family of
      $ab$-paths.
    \item We need $a$ and $b$ to be non-adjacent,
      else there would be no separators!
    \item We cannot just ``pick a point on each of
      a maximum-sized independent family of
      $ab$-path'' to prove this.
    \item \nameref{mengers_thm} generalises
      \nameref{halls_thm}. Indeed, given bipartite
      $G$ on $X$, $Y$, form $G'$ by adding $a, b$
      where $\Gamma(a) = X$, $\Gamma(b) = Y$.
      Then, $G$ has a matching if and only if $G'$
      has a family of $|X|$ independent
      $ab$-paths. So, by \nameref{mengers_thm}, it
      is enough to show that each $ab$-separator
      $S$ has size of at least $|X|$. Let $S = A
      \cup B$ be a separator, with $A \subset X$
      and $B \subset Y$. Since $S$ is a separator,
      we must have $\Gamma_G(X \setminus A)
      \subset B$. So $|S| = |A| + |B| \ge |A| +
      \Gamma_G(X \setminus A)| \ge |A| + |X
      \setminus A| = |X|$.
  \end{enumerate}
\end{remark*}