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\begin{center} {\Large\bf Honors Algebra I, Assignment 7} \\ Solutions
\end{center}
\ben
\item Let $G$ denote the set of linear functions $f(x)=mx+b$ on
the real line with $m\neq 0$. Denote such a function by $(m,b)$.
Define a product $f^*g$ as the function $h(x)=g(f(x))$. (Check
assignment 1 and the solutions for earlier work on this group.)
Recall $C(f),N(f)$ denote the conjugate class and
the normalizer of $f$.
\\ {\tt Solution Note:} We'll write $g(x)=cx+d$ below and look at when $fg=gf$.
We have $g^{-1}(x)=(x-d)/c$.
\ben
\item Describe $C(f)$ and $N(f)$ when $m\neq 1$ and $b\neq 0$.
\So For $g\in N(f)$ we want $gf=fg$ so
\[ m(cx+d)+b=c(mx+b)+d \]
so that $md+b=cb+d$ or $(m-1)d=b(c-1)$. Note $m-1\neq 0$ and $b\neq 0$. So
for each $c\neq 1$ we have $d=b(c-1)/(m-1)$. This can also be written
\[ \frac{-b}{m-1} = \frac{-d}{c-1} \]
which has a nice interpretation. Set $\beta=\frac{-b}{m-1}$ so that $f(\beta)=\beta$.
Then $g\in N(f)$ if and only if $g(\beta)=\beta$. For the centralizer in general
\[g^{-1}fg(x) = g(f(g^{-1}(x)))= c\left[m\left(\frac{x-d}{c}\right)+b\right]+d =
m(x-d)+bc+d \]
This gives the centralizer as all functions $mx+\kappa$.
\item Describe $C(f)$ and $N(f)$ when $m\neq 1$ and $b=0$.
\So Here $f(x)=mx$. For $g\in N(f)$ we want $gf=fg$ so
\[ m(cx+d)=c(mx)+d \]
so that $md=d$. As $m\neq 1$ this holds only when $d=0$. So
$N(f)$ is all the maps $g(x)=cx$. For the centralizer in general
\[g^{-1}fg(x) = g(f(g^{-1}(x)))= cm\left(\frac{x-d}{c}\right)+d = mx + d(1-m) \]
As $m\neq 1$ the value $d(1-m)$ can be any real so the centralizer consists
of the functions $g(x)=mx+\kappa$.
\item Describe $C(f)$ and $N(f)$ when $m=1$ and $b\neq 0$.
\So These are the translates $f(x)=x+b$. So for $g\in N(f)$ we want
$cx+d+b=c(x+b)+d$ (that is, that they are the same function)
so we need $b=bc$ so $c=1$. So $N(f)$ is the translates. In general
\[ g^{-1}fg(x) = g(f(g^{-1}(x)))=c\left(\frac{x-d}{c}+b\right)+d=x+bc \]
We are given $b\neq 0$ and $c\neq 0$ so $bc$ is an arbitrary nonzero
real and so
so the centralizer $C(f)$ is all the translates $h(x)=x+\gamma$ with $\gamma\neq 0$.
\item Describe $C(f)$ and $N(f)$ when $m=1$ and $b=0$.
\So This is the identity so $C(f)=G$ and $N(f)=\{e\}$.
\item Describe $Z[G]$, the center of the group.
\So Only the identity $f(x)=x$. Notice the relation $f\sim g$ does split $G$ into
equivalence classes. The identity is (as always) its own equivalence class. For
$m\neq 1$ the set of functions $h(x)=mx+b$ forms an equivalence class. For
$m=1,b\neq 0$ the set of ``nontrivial translates" $h(x)=x+b$ forms an equivalence class.
\een
\item Let $\sig\in S_n$ be (in cycle notation)
\[ \sig = (12\cdots n) \]
\ben
\item Describe in words the $\gamma\in C(\sig)$.
\So All $\gamma$ consisting of a single cycle of length $n$.
\item Find $o(C(\sig))$.
\So $(n-1)!$. Write any $(x_1\cdots x_n)$ except that the same
cycle is written precisely $n$ different ways.
\item Deduce $o(N(\sig))$.
\So $n!/(n-1)!=n$.
\item (*) Describe $N(\sig)$ explicitly. (Idea: Since you already
know that $N(\sig)$ has precisely woggle elements, you should look
for woggle distinct elements that are in $N(\sig)$ and then you have
them all.)
\So $\{e,\sig,\sig^2,\ldots,\sig^{n-1}\}$. These are all distinct,
they all commute with $\sig$ and there are $n$ of them.
\een
\item Let $o(G)=p^n$, $p$ prime. Prove that $o(Z[G])\neq p^{n-1}$.
(Idea: Examine the proof that groups of order $p^2$ are Abelian.)
\So Assume $o(Z[G])=p^{n-1}$ and pick $\ah\not\in Z[G]$. Its normalizer
contains $Z[G]$ but it also contains $\ah$ itself so it has strictly
more than $p^{n-1}$ elements so (using the powerful theorem that the
order of a subgroup (here the normalizer of $\ah$) must divide the
order of the group)
it has $p^n$ elements so its everything so $\ah\in Z[G]$,
a contradiction.
\item Let $G=Z_5\times Z_{25}\times Z_{125}$.
Find the order of $(3,10,12)$. Give a good description and a precise
count on those $(a,b,c)\in G$ order $25$.
\So $i(3,10,12)=(0,0,0)$ requires $5|i$, $5|i$ and $125|i$ so it need
$125|i$ so the order is $125$. (The order must be a power of $5$ as
$o(G)$ is a power of five. For the second: $25(a,b,c)=(0,0,0)$ iff
$5|c$ as the others hold automatically. However, we must exclude
those $(a,b,c)$ with $5(a,b,c)=(0,0,0)$ and this holds when $5|b$ and
$25|c$. So the desciption is: those $(a,b,c)$ with $5|c$ but not having
both $5|b$ and $25|c$. For the count: there are $3125=5^5$ choices with
$5|c$ ($5$ choices for $a$, $25$ for $b$ and $25$ for $c$) and
$125$ choices with $5|b$ and $25|c$ ($5$ for each of $a,b,c$) leaving
$3125-125=3000$ elements of order $25$.
\item Let $H$ be a normal subgroup of $G$ (all under multiplication).
Let $g\in G$ and let (as usual) $\ol{g}$ denote the corresponding
element in $G/H$. Suppose $\ol{g}$ has order $a$ (in $G/H$) and
$g^a$ has order $b$ (in $G$). Prove $g$ has order $ab$ in $G$.
[Note that, as a Corollary, $o(\ol{g})$ divides $o(g)$ whenever both
are finite.]
\So As $e=(g^a)^b=g^{ab}$, $g$ has order at most $ab$.
Now suppose $g^i=e$. Then $\ol{g}^i=\ol{g^i}=\ol{e}=e$. As $\ol{g}$
has order $a$ this means $a$ divides $i$ so we can write $i=aj$ where
$j$ is an integer. Now $e=g^{aj}=(g^a)^j$. As $g^a$ has order $b$
we must have $b$ dividing $j$ so that $j=bk$ for some integer $k$.
Thus $i=abk$ for some $k$. As $k\geq 1$, $ab$ is the order.
\item Let $H$ be the subgroup of $S_4$ which is the normalizer of
$(12)(34)$, as discussed in class. Is this group isomorphic to
the group of symmetries of the square, as given in the solutions
to Assignment 1. If they are not isomorphic, prove it. If they
are isomorphic, give an explicit isomorphism.
\So Hmmmm. Certainly $e\lra I$. That leaves
$7!=5040$ possibilities. But there are ways to make it simpler.
Both groups have two elements of order four so the isomorphism
must send these to each other. So lets associate $R\lra (1324)$.
Products go into products. So then $S \lra (12)(34)$ and
$T\lra (1423)$. How about $V$. Well, lets try $V\lra (12)$.
Now everything is forces. $A=VR$ so it must correspond to
the product of $(12)$ and $(1324)$, that is: $A\lra (14)(23)$.
$H=VS$ so $H \lra (12)[(12)(34)] = (34)$.
$D=VT$ so $D \lra (12)(1423) = (13)(24)$.
Checking out the other products (still lots of grunt work here) --
it works! The groups are indeed isomoprhic.
\\ {\tt Extra:} One student had a great approach. Label the
vertices of the square $1,3,2,4$ with $1$ in the upper right
in counterclockwise order. The eight permutations
of $H$ then correspond to the symmetries of the square.
For example: $(12)$ is a flip on the main diagonal, $(34)$
a flip on the other diagonal, $(14)(23)$ a flip on the
horozontal axis, $(1324)$ a rotation by $\pi/2$, etc.
\een
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\een
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