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\centerline{MATH 351 -- Modern Algebra I}
\centerline{Selected Solutions for Problem Set 10}
\centerline{December 7, 2018}
\bigskip
\noindent
7.18.  Let $G$ be an infinite group and let $H$ be the subset
of $G$ consisting of the elements with only finitely many conjugates.
Equivalently $H$ is the subset of $a \in G$ such that $|C_a|$ is finite.
We have to show that $H$ is a subgroup of $G$.
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\item{$\bullet$}  First, we consider $e$.  For all $g \in G$, 
$geg^{-1} = e$.  Therefore $C_e = \{e\}$ and $|C_e| = 1$.  Since
this is finite, $e\in H$.  
\item{$\bullet$} Now let $a,b \in H$ and consider $C_{ab}$.  
For all $g \in G$, we have using associativity
$$gabg^{-1} = ga(g^{-1}g)bg = (gag^{-1})(gbg^{-1}).$$
The factor on the left is an element of $C_{a}$ and the factor on
the right is an element of $C_b$.  Hence when we form these
products, we obtain {\it at most} $|C_a|\cdot |C_b|$ different
elements of $G$.  There could be also strictly fewer than 
$|C_a|\cdot |C_b|$ in some cases, but in any case, the number
$|C_{ab}|$ is finite.  This shows $ab \in H$.
\item{$\bullet$} Now let $a \in H$ and consider $|C_{a^{-1}}|$.  The
elements of this class all have the form $ga^{-1}g^{-1}$ for $g \in G$.  
Note that the inverse of this element of $G$ is
$$(ga^{-1}g^{-1})^{-1} = gag^{-1},$$
which is one of the elements of $|C_a|$.  Since we began by 
assuming $a \in H$, there are only finitely many different values
obtained for $gag^{-1}$.  That implies there are only finitely 
many conjugates of $a^{-1}$ too since each element of $G$ has
a unique inverse in $G$.  That shows $a^{-1} \in H$.
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\noindent
Putting together these three points, we get that $H$ is a subgroup of $G$.
\bigskip
\noindent
7.24.  Most people saw correctly that you want to use Sylow II to 
say that any two Sylow 2-subgroups $H_1$ and $H_2$ must be conjugate.  But then 
we must say why (or prove directly that) if $H_2 = gH_1g^{-1}$ for
two subgroups $H_1,H_2$ of $G$ and some element $g \in G$, then 
$H_1$ and $H_2$ must be {\it isomorphic} as groups.  Here's how
to show that.  Consider the conjugation mapping $\alpha$:
$$\eqalign{\alpha : H_1 &\to H_2\cr
                    h &\mapsto ghg^{-1}\cr}$$
We claim that $\alpha$ is an isomorphism.  First, it is a group
homomorphism because if $h,k \in H_1$, then
$$\alpha(hk) = ghkg^{-1} = gh(g^{-1}g)hg^{-1}= (ghg^{-1})(gkg^{-1}) = \alpha(h) \alpha(k).$$
The mapping $\alpha$ is surjective with image $H_2$ by Sylow II.  Moreover, since 
$|H_1| = p^n = |H_2|$ if $|G| = p^n r$ with $\gcd(p,r) = 1$, once
we know $\alpha$ is surjective, then $\alpha$ must be injective as 
well.  Hence $\alpha$ is an isomorphism of groups.

But this shows that it is impossible for one Sylow 2-subgroup $H_1 \cong {\bf Z}_4$
and another Sylow 2-subgroup $H_2 \cong {\bf Z}_2 \times {\bf Z}_2$.  The groups
${\bf Z}_4$ and ${\bf Z}_2 \times {\bf Z}_2$ are not isomorphic, by the 
fundamental theorem of finite abelian groups.  
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\noindent
7.30.  Let $|G| = 56 = 2^3\cdot 7$.  We claim that $G$ must have a normal subgroup
other than $\{e\}$ and $G$, so $G$ is not simple.  Let $p = 7$.  By Sylow III, 
the number of Sylow 7-subgroups is $\equiv 1 \bmod 7$ and that number divides $8$.
Thus both $1$ and $8$ Sylow 7-subgroups are possible.  If there is just one, then
as usual, Sylow II implies that subgroup is normal and we are done.  On the 
other hand, if there are $8$ Sylow 7-subgroups in $G$, each of order $=7$, then since $7$ is prime,
each of the Sylow 7-subgroups is cyclic and there are $6$ elements of order $7$
in each of them.  Distinct Sylow 7-subgroups can only intersect in the identity
element.  Counting up the total number of elements in the union of 
those subgroups we get $1 + 8\cdot 6 = 49$.  This leaves only $7$ other elements
of $G$, and those $7$ elements together with $e$ make up exactly one
subgroup of order $8$ by Sylow I (there exists at least one subgroup with $2^3 = 8$
elements, but there is no room for more than one in this case(!)).  Hence if
there are $8$ Sylow 7-subgroups, there must be only one Sylow 2-subgroup, hence
that Sylow 2-subgroup is normal by Sylow II.  Hence $G$ is not simple in
this case either. 
\bye  
