\documentclass{article}
\newtheorem{problem}{Problem}
\title{CHM: A.10}
\author{Po-Shen Loh}
\date{5 May 2001}
\begin{document}
\maketitle

\begin{problem}[A.10, Schweitzer 1967/2]
Let $K$ be a subset of a group $G$ that is not a union of left cosets
of a proper subgroup.  Prove that if $G$ is a torsion group or if $K$
is a finite set, then the subset

\begin{displaymath}
\cap_{k \in K} k^{-1}K
\end{displaymath}

consists of the identity alone.  [L. R\'{e}dei]

\end{problem}

\noindent \textbf{Solution:}

We first deal with the case $|K| < \infty$.  Let $K = \{a_1, a_2,
\ldots, a_n\}$.  Suppose that there is some element $x \neq 1$ in the
intersection; then for each $i \in \{1, 2, \ldots, n\}$, there exists
some $j$ for which $a_i^{-1}a_j = x$.  Let us renumber the elements of
$k$ so that this is true:

\begin{eqnarray*}
x &=& a_1^{-1}a_2 \\
x &=& a_2^{-1}a_3 \\
x &=& a_3^{-1}a_4 \\
&\ldots& \\
x &=& a_r^{-1}a_1
\end{eqnarray*}

Since we have a finite subset, we are guaranteed to return to $a_1$
eventually.  Multiply these $r$ equations; we find that $x^r = 1$.
Since we took $r$ iterations to get back to $a_1$, we also know that
no smaller power of $x$ is the identity.  Hence, $|x| = r$.

Let us repeat the process starting with $a_{r+1}$ instead of $a_1$.
Since $|x| = r$, we must have a cycle of $r$ elements.  In this way,
we separate the elements of $K$ into $n/r$ disjoint subsets.  Yet each
subset is a coset of $\langle x \rangle$, so we have a union of left
cosets, contradicting our given information.  Therefore, the
intersection is just the identity.

In the other case, we find a similar result; suppose again that $x$ is
in the intersection.  Now we know that $|x| = r < \infty$ for some $r$
because our group is a torsion group.  Thus for each element in $K$,
we will be able to iterate it and get an $r-$cycle, which is a left
coset in $G$; we again find that we have a union of left cosets, a
contradiction.

And we are done.



\end{document}

