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QUESTION


Say, giving your reasons, whether the set $G$ forms a group under
the given operation in each of the following cases:

\begin{description}

\item[(i)]
$G$ is the set of subsets of a set $X$, with the operation of
union of sets.

\item[(ii)]
$G$ is the set of subsets of a set $X$, with the operation of
intersection of sets.

\item[(iii)]
$G$ is the set of subsets of a set $X$, with the operation of
symmetric difference, i.e., $A\triangle B=A\cup B-A\cap B$. (You
nay assume that this operation is associative.)

\item[(iv)]
$G=\mathbf{R}$ with the operation $x*y=min(x,y)$.

\item[(v)]
$G=\mathbf{R}$ with the operation of subtraction.

\item[(vi)]
The set of functions from a set $X$ to $\mathbf{R}$ with the
operation $(f+g)$ defined by $(f+g)(x)=f(x)+g(x)$ for all $x\in
X$.

\item[(vii)]
The set of odd numbers under addition.

\end{description}




ANSWER


\begin{description}

\item[(i)]

\begin{eqnarray*}
(X'\cup Y)\cup Z&amp;=&amp;X\cup(Y\cup Z)\\
X'\cup\emptyset&amp;=&amp;\emptyset\cup X'=X'\\ X'\cup
Y=\emptyset&amp;\rightarrow&amp;\{x|x\in X'\textrm{ or }x\in
Y\}=\emptyset\\ &amp;\rightarrow&amp;X=Y=\emptyset
\end{eqnarray*}

so unless $X'=\emptyset,\ G$ is not a group.

\item[(ii)]

\begin{eqnarray*}
(X'\cap Y)\cap Z&amp;=&amp;\{x\in X'\cap Y|x\in Z\}\\ &amp;=&amp;\{x|x\in X',x\in
Y\textrm{ and }x\in Z\}\\ &amp;=&amp;\{x\in X'\cap(Y\cap Z)\}
\end{eqnarray*}

$X'\cap X=X'$ so there is an identity but $X'\cap Y=X=X'=Y=X$ so
there are no inverses so $G$ is not a group.

\item[(iii)]
Operation is associative.

$X'\triangle Y=X'\Leftrightarrow X'\cup Y\backslash X'\cap
Y=X\Leftrightarrow \{y\in Y\backslash X'\}=\{y\in X'\cap Y\}$ so
$Y$ must be a subset of $X'$. Hence identity$=\emptyset$.

$X\triangle X=\emptyset$ so $X'$ has an inverse so $G$ is a group.

\item[(iv)]
$min(x,y)=x\Rightarrow y\geq x$ since $\mathbf{R}$ has no maximal
element there is no identity. So $G$ is not a group.

\item[(v)]
$(x-y)=x\Rightarrow y=0$ and 0 is the identity.

$(x-y)-z\neq(x-(y-z))$ unless $z=0$ so not associative
$\Rightarrow$ not a group.

\item[(vi)]

\begin{eqnarray*}
((f+g)+k)(x)&amp;=&amp;(f+g)(x)+k(x)\\ &amp;=&amp;f(x)+g(x)+k(x)\\ &amp;=&amp;(f+(g+k))(x)
\end{eqnarray*}

so the operation is associative

$(f+0)(x)=f(x)$ if $0(x)=0\forall x$ so + has an identity.

Define $(-f)(x)=-f(x)\forall x$ so $f(x)+(-f)(x)=0\forall x$ so
inverses exist $\forall f$ so $G$ is a group.

\item[(vii)]
$3+5=8\not\in\{$ odd numbers $\}$ so this is therefore not a
group.

\end{description}



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