\magnification=\magstep1
\centerline{MATH 351 -- Modern Algebra I}
\centerline{Selected Solutions for Problem Set 9}
\centerline{November 30, 2018}
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\noindent
6.26.  By Theorem 6.6 in Lee, we know that every 
$\sigma \in S_n$ is a product of transpositions, but
those transpositions can contain any of the $(a b)$,
with $1 \le a < b \le n$.  On the other hand, by 
Lemma 6.2, we know that 
$$(1 a)(1 b)(1 a) = (1 a)(1 b)(1 a)^{-1} = (a b).$$
Hence in any factorization of $\sigma$ into transpositions,
we can replace any transposition $(a b)$ with $a,b \ne 1$
by the product $(1 a)(1 b)(1 a)$ as above.  This shows
that every $\sigma \in S_n$ can be written as a product
of the ``special'' transpositions $(1 i)$ with $2 \le i \le n$.  
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\noindent
6.28.  There are many different proofs of this fact.  Let's look
at two, both of which are based on Exercise 6.26 above.  
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\noindent
{\it Proof 1:} (This one might be the ``slickest''.)  Letting $\sigma = (1 2)$
and $\tau = (1 2 \cdots n)$ as in the problem statement, note that
Lemma 6.2 implies:
$$\eqalign{
\tau \sigma \tau^{-1} &= (2 3)\cr
\tau (2 3) \tau^{-1} &= \tau^2 \sigma \tau^{-2} = (3 4)\cr
\tau (3 4) \tau^{-1} &= \tau^3 \sigma \tau^{-3} = (4 5),\cr}$$
and hence, continuing in the same way every {\it ``consecutive''}
transposition $(k\ k+1)$ with $1 \le k \le n -1$ can be written 
as required by the problem.  Then, using Lemma 6.2 again, note
that 
$$\eqalign{ 
(2 3)(1 2)(2 3) &= (1 3) = (\tau\sigma\tau^{-1})\sigma(\tau\sigma\tau^{-1})\cr
(3 4)(1 3)(3 4) &= (1 4) = 
(\tau^2\sigma\tau^{-2})(\tau\sigma\tau^{-1})\sigma(\tau\sigma\tau^{-1})(\tau^2\sigma\tau^{-2})\cr}$$
and similarly, all of the ``special transpositions'' $(1 i)$ from Exercise 6.26 can
be written as products of powers of $\sigma$ and $\tau$.  It follows from Exercise 6.26, then,
that every $\rho \in S_n$ can be written in this way too.  
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\noindent
{\it Proof 2a:} A second way to derive the same conclusion (using Exercise 6.26 in
a similar way), is to alternate conjugating $2$-cycles and $n$-cycles, like the following:
We have, by Lemma 6.2,
$$\eqalign{\sigma\tau\sigma^{-1} &= (2 1 3 4 \cdots n) = \tau_2\cr
\tau_2\sigma\tau_2^{-1} &= (3 1) = (1 3)\cr
(1 3)\tau_2 (1 3) &= (2 3 1 4 \cdots n) = \tau_3\cr
\tau_3(1 3)\tau_3^{-1} &= (4 1) = (1 4)\cr}$$
and continuing in the same way, we get all the $(1 i)$ with $2 \le i \le n$ as products
of powers of $\sigma$ and $\tau$.  
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\noindent
{\it Proof 2b:}
Proof 2a can also be phrased as an argument {\it by induction} as follows.  We want to 
show that in $S_n$, all of the ``special'' transpositions $(1 i)$ for $2 \le i \le n$
can be obtained as products of powers of $\sigma$ and $\tau$.  The base case is $i = 2$
and there is nothing to prove since $\sigma = (1 2)$.  So now let's assume that we 
have shown $(1 2), (1 3), \ldots, (1 k)$ can be written this way.  Consider $(1\ k+1)$.  
Then, following what we did in Proof 2a, we consider the product giving the element we
called $\tau_k$ above:
$$\tau_k = (1 k)(1\ k-1)\cdots(1 3)(1 2)\tau(1 2)(1 3) \cdots (1\ k-1)(1 k)$$
By Lemma 6.2, since the product on the right of $\tau$ is the inverse of the product
on the left of $\tau$, this equals
$$\tau_k = (2\ 3\ 4\ \cdots\ k\ 1\ k+1\ k+2\ \cdots\ n).$$
(In the $n$-cycle $\tau$, we're transposing the 1 and the 2, then the 1 and the 3, etc. up to the 1 and the $k$, 
so $2,3,\ldots,k$ all end up to the left of $1$, and the remaining numbers $k + 1, \ldots, n$ have
not moved.)  By Lemma 6.2 yet again, then 
$$\tau_k(1 k)\tau_k^{-1} = (k+1\ 1) = (1\ k+1)$$
is a product of powers of $\tau$ and $\sigma$, since $\tau_k$ and $(1 k)$ are 
such products (that's the induction hypothesis) and this finishes the proof. 
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7.8.  We want to show that if $a$ is odd order, then $C(a) = C(a^4)$.  We'll set this
up as a proof showing each of these sets is contained in the other.
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$\subseteq:$ Let $g \in C(a)$.  Then $ga = ag$, so 
$$ga^4 = (ga)a^3 = (ag)a^3 = a(ga)a^2 = a(ag)a^2 = a^2(ga)a = a^3ga = a^4g.$$
Hence $g \in C(a^4)$, so $C(a) \subseteq C(a^4)$.  (Note: We don't need the hypothesis 
that $a$ has odd order for this inclusion.)
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$\supseteq:$  Now let $g \in C(a^4)$.  By definition this means $ga^4 = a^4g$.  We have
to show this implies $ga = ag$ when $|a|$ is odd.  Let $k = |a|$.  Then by integer
division we have $k = 4q + r$ with $0 \le r < 4$.  But note $r = 0,2$ are not 
possible, since then $k$ would be even.  This says there are two possible values for
$r$, namely $r = 1$ and $r = 3$.  We handle each of those cases separately.
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\noindent
If $r = 1$, then $a^{4q + 1} = e$, so $a = a^{-4q}$.  Note that $a^{-4q} = (a^4)^{-q}$.  
Since we have $ga^4 = a^4g$, we also have $a^{-4}g = ga^{-4}$ by multiplying both 
sides of the previous equation by $a^{-4}$ on the left and the right. It follows that
$ga^{-4q} = a^{-4q}g$ for all integers $q$.  Hence  
$$ga = ga^{-4q} = a^{-4q}g = ag$$
and $g \in C(a)$.  This shows $C(a^4) \subseteq C(a)$ when $|a| = 4q + 1$.  
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If $r = 3$, then $a^{4q + 3} = e$, so $a = a^{4(q+1)}$.  Note that $a^{4(q+1)} = (a^4)^{q+1}$.  
If $g \in C(a^4)$, then it follows that
$ga^{4(q+1)} = a^{4(q+1)}g$.  Hence  
$$ga = ga^{4(q+1)} = a^{4(q+1)}g = ag$$
and $g \in C(a)$.  This shows $C(a^4) \subseteq C(a)$ when $|a| = 4q + 3$ as well.
\bye  
