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\centerline{Mathematics 243, section 1 -- Algebraic Structures}
\centerline{Selected Solutions -- Problem Set 8}
\centerline{December 2, 2003}
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\noindent
{\it General Comment:}  
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It seems from the work I saw on this 
problem set that some very basic material
(such as what ``set builder notation'' means, what
it means for a mapping to be 1-1 or onto, what
an inverse function is, what composition of mappings means, and 
how to write these things down in a correct fashion) has 
become hazy again in some people's minds.  If you are in that 
category, {\it begin your review for Exam 3 by looking back 
at Chapter 1 and learning that material once and for all now. 
It is part of the basic language of mathematics that you must
have reliably at your fingertips now and for the rest of 
your careers as mathematics majors.  Even though I did not
mention this explicitly on the Review Sheet for Exam 3, it is 
a certainty that there will be questions dealing with these ideas!}
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\noindent
4.1/14.  The problem asks for all of the {\it distinct cyclic subgroups} of 
the alternating group $A_4$ (all even permutations of $\{1,2,3,4\}$).
Many people just wrote down the cycle decompositions of the 
12 elements of $A_4$.  Recall that the cyclic subgroup generated
by an element $a$ in a group $G$ is the set
$$\langle a\rangle = \{a^k | k \in {\bf Z}\}$$
This means that you must take all of the elements $a$ in 
$A_4$, determine their powers, and find the elements in $\langle a \rangle$.
There are exactly {\it eight} distinct cyclic subgroups:
$$\eqalign{
\langle () \rangle &= \{()\}\cr
\langle (12)(34) \rangle &= \{ (), (12)(34) \}\cr
\langle (13)(24) \rangle &= \{ (), (13)(24) \}\cr
\langle (14)(23) \rangle &= \{ (), (14)(23) \}\cr
\langle (123) \rangle &= \{ (), (123), (132) \} = \langle (132)\rangle \cr
\langle (124) \rangle &= \{ (), (124), (142) \} = \langle (142)\rangle \cr
\langle (134) \rangle &= \{ (), (134), (143) \} = \langle (143)\rangle \cr
\langle (234) \rangle &= \{ (), (234), (243) \} = \langle (243)\rangle \cr}$$
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\noindent
4.1/16.  The full group operation table for the ``octic'' group 
can be found inside the back cover of the textbook (in the ``endpapers'').
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\noindent
4.2/1,4.  Most people followed the format of the proof of Cayley's
Theorem in the text for these, and expressed the permutations
as the one-to-one and onto mappings $f_a : G \to G$ 
(or $m_a$ using the notation
I used in class).  That's fine, but you should also be aware that
you can also get subgroups of $S_n$ (with elements written in cycle 
decomposition form)
isomorphic to each of these groups.  This way of writing the 
groups comes by labeling the elements of the group using 
the integers $\{1,2,3,\ldots,n\}$ in some fixed fashion, and
seeing what the rows of the group operation table look like
as permutations of $\{1,2,3,\ldots,n\}$.  For instance, for the 
Klein 4-group, letting 
$$\eqalign{e \leftrightarrow 1\cr
a \leftrightarrow 2\cr
b \leftrightarrow 3\cr
ab \leftrightarrow 4\cr}$$
the group table becomes:
$$\matrix{ &1&2&3&4\cr
          1&1&2&3&4\cr
          2&2&1&4&3\cr
          3&3&4&1&2\cr
          4&4&3&2&1\cr }$$
So we also get an isomorphism
$$\phi : G \to \{(), (12)(34), (13)(24), (14)(23)\} \subset S_4$$
by letting $\phi(e) = (), \phi(a) = (12)(34)$, etc.
The subgroup of $S_6$ corresponding to the group in problem 4 in this way 
is
$$\{(), (12)(34)(56), (13)(26)(45), (164)(235), (15)(24)(36), (146)(253)\}$$
written in the same order as the permutation matrices (i.e.
$\phi(I_3) = (), \phi(P_1) = (12)(34)(56)$, etc. defines the isomorphism).
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\noindent
4.1/5.  a)  Let $h_a : G \to G$ be the mapping defined by 
$h_a(x) = xa$. The problem says to show that $h_a$ is a permutation
of $G$. This means that we must show that $h_a$ is a
1-1 and onto mapping.  To show that $h_a$ is 1-1, consider
the equation $h_a(x) = h_a(y)$ for $x,y\in G$.  Then
$xa = ya$, so multiplying on the right by $a^{-1}\in G$ we
get 
$$(xa)a^{-1} = (ya)a^{-1} \Rightarrow x(aa^{-1}) = y(aa^{-1}) \Rightarrow
x = y.$$
Hence $h_a$ is 1-1.  To show $h_a$ is onto, let $y\in G$ be
arbitrary and consider $x = ya^{-1}$.  Then 
$$h_a(x) = h_a(ya^{-1}) = (ya^{-1})a = y(aa^{-1}) = y$$
Since $y$ was arbitrary, we see that every element of $G$ is in the 
image of $h_a$, which is what we wanted to show.  

b)  Consider $H = \{h_a | a \in G\}$.  This means the set of all the 
mappings $h_a$ where $a$ runs over all the elements of $G$ (``set 
builder notation!'').  We must show that this set is a group 
under the operation of composition of mappings.  That is, we need
to show:
\item{i.} $H$ is closed under composition
\item{ii.} composition is associative as a binary operation on $H$
\item{iii.}  there is an identity element for composition in 
$H$
\item{iv.} every element of $H$ has an inverse in $H$ (under composition).
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\item{i.}  We must show $h_a\circ h_b$ is one of the elements of $H$,
that is that $h_a\circ h_b = h_c$ for some $c \in G$.  If $x$ is
any element of $G$,
$$(h_a\circ h_b)(x) = h_a(h_b(x)) = h_a(xb) = (xb)a = x(ba) = h_{ba}(x)$$
This means that $h_a\circ h_b = h_{ba}$ (i.e. the $c$ above is 
the product $c = ba \in G$).  Hence $H$ is closed under composition.
\item{ii.}  Composition is always associative:
$$((h_a\circ h_b)\circ h_c)(x) = h_a(h_b(h_c(x))) = (h_a\circ(h_b\circ h_c))(x)$$
(Alternatively, you could work this out and use associativity of multiplication in $G$
to see that $x((ab)c) = x(a(bc))$ for all $x\in G$.)
\item{iii.}  Since $G$ is a group, it has an identity element $e$.
The mapping $h_e(x) = xe = x$ for all $x\in G$.  Hence $h_e$
is the identity mapping from $G$ to itself, which is 
the identity element for composition:  $h_a \circ h_e = h_a = h_e \circ h_a$
for all $a$.
\item{iv.}  The inverse mapping of $h_a$ is the mapping $h_{a^{-1}}$
since for all $x \in G$,
$$(h_a\circ h_{a^{-1}})(x) = x(a^{-1}a) = xe = h_e(x)$$
Similarly, $h_{a^{-1}}\circ h_a = h_e$.

c)  The mapping $\phi : G \to H$ defined by $\phi(a) = h_a$
is {\it not necessarily an isomorphism of groups} because
$$\phi(ab) = h_{ab} = h_b \circ h_a = \phi(b) \circ \phi(a)$$
by what we said in b i. above.  For a group isomorphism we would 
need $\phi(ab) = \phi(a)\circ \phi(b)$.
\bye


