\magnification=\magstep1
\centerline{MATH 351 -- Modern Algebra 1}
\centerline{Selected Solutions -- Problem Set 8}
\centerline{November 10, 2018}
\bigskip
\noindent
5.10.  We are given that $N_1, \ldots, N_k$ are finite normal subgroups of some group
$G$ such that $\gcd(|N_i|,|N_j|) = 1$ whenever $i \ne j$ and we want to show
that 
$$N_1 \times \cdots \times N_k \cong N_1 N_2\cdots N_k.$$
It is true that this is intimately connected with Lemmas 5.1 and 5.2 and Theorem 
5.2.  However there are some differences, too, that we must take into account.
First, note that we are not assuming $G = N_1\cdots N_k$, so it might not be true
that $G$ is the internal direct product of the subgroups $N_i$.  The product
$N_1\cdots N_k$ could be some proper subgroup of $G$.  Second, the hypothesis
$\gcd(|N_i|,|N_j|) = 1$ here is different from the condition that appears in the
definition of the internal direct product.  Hence there are several different 
ways to proceed here.  
\bigskip
\noindent
{\it Solution 1:} (The ``slickest'' way.) We argue by induction on $k$.  If
$k = 1$, there is nothing to show.  If this seems unsatisfying as a base case,
consider the case $k = 2$.  Then note that $\gcd(|N_1|,|N_2|) = 1$ implies
that $N_1 \cap N_2 = \{e\}$ by Lagrange's theorem, since $N_1 \cap N_2$ is a subgroup
of both $N_1$ and $N_2$. Then $N_1 \times N_2 \cong N_1N_2 = H$ follows from Theorem 5.2
applied to the group $H = N_1N_2$ (not $G$), since $H$ is the internal direct product of 
its subgroups $N_1$ and $N_2$.  Now assume the statement holds for all 
collections of $k$ subgroups satisfying the given conditions, and consider
$N_1,\ldots,N_k, N_{k+1}$.  By the induction hypothesis, $N_1N_2 \cdots N_k$
is isomorphic to $N_1 \times N_2 \times \cdots \times N_k$.  In particular,
$$|N_1N_2 \cdots N_k| = |N_1||N_2| \cdots |N_k|.$$
It follows that $\gcd(|N_1N_2\cdots N_k|, |N_{k+1}|) = 1$, since 
$\gcd(|N_i|,|N_{k+1}|) = 1$ for all $1 \le i \le k$.  But this implies 
$$(N_1N_2 \cdots N_k) \cap N_{k+1} = \{e\},$$ 
as above.  Since $N_1N_2 \cdots N_k$ is normal in $G$,
the proof that 
$$(N_1\times N_2 \times \cdots \times N_k) \times N_{k+1} \cong 
(N_1N_2 \cdots N_k) N_{k+1}$$
will follow exactly as in the case $k = 2$ done above.  
\bigskip
\noindent
{\it Solution 2:} (This one is essentially repeating the proof of Theorem 5.2, 
which is OK, but less ``elegant.'')
Consider the mapping 
$$\eqalign{
\alpha : N_1 \times N_2 \times \cdots \times N_k &\to N_1\cdots N_k\cr
(n_1,n_2,\ldots,n_k) &\mapsto n_1 n_2 \cdots n_k\cr}$$
Then 
\item{$\bullet$} $\alpha$ is surjective by definition.  
\item{$\bullet$} $\alpha$ is a group homomorphism by Lemma 5.1 and the
same argument as in the text in the proof of Theorem 5.2:
$$\eqalign{
\alpha(n_1n_1',n_2n_2',\ldots,n_kn_k') &= (n_1n_1')(n_2n_2') \cdots (n_kn_k')\cr
&= (n_1n_2\cdots n_k)(n_1'n_2'\cdots n_k')\cr
&= \alpha(n_1,n_2,\ldots,n_k)\alpha(n_1',n_2',\ldots,n_k'),\cr}
$$
since Lemma 5.1 implies that the $n_i'$ and the $n_j$ all commute with each other.
\item{$\bullet$}  Finally, we show that $\alpha$ is injective, using induction on $k$.
If $k = 1$, there is nothing to prove.  So assume that the result is true whenever
$k = \ell$ and consider products of $\ell + 1$ subgroups.  To show $\alpha$ is injective, 
it suffices to show that $\ker(\alpha) = (e,e,\ldots,e)$.  
So consider $(n_1,n_2,\ldots,n_{\ell+1})$ such that 
$$\alpha(n_1,n_2,\ldots,n_\ell,n_{\ell+1}) = n_1n_2\cdots n_\ell n_{\ell + 1} = e.$$
This implies 
$$n_{\ell+1}^{-1} = n_1 n_2 \cdots n_\ell \in (N_1N_2\cdots N_\ell) \cap N_{\ell+1}.$$
In the ``slick'' solution, we saw that $(N_1N_2\cdots N_\ell) \cap N_{\ell + 1} = \{e\}$
because $N_1 N_2 \cdots N_\ell$ is isomorphic to $N_1 \times N_2 \times \cdots \times N_\ell$
so the order of $N_1 N_2 \cdots N_\ell$ is the product of the orders of the $N_i$, hence
is relatively prime to $|N_{\ell+1}|$.  The same is true here.  Hence
that implies $n_{\ell + 1} = e$ and then by induction $n_1 n_2 \cdots n_{\ell} = e$
implies $n_1 = n_2 = \cdots = n_\ell = e$ as well.  
\bigskip
\noindent
5.18.  In the cyclic group $G = \langle a \rangle$ of order $90 = 2 \cdot 3^2 \cdot 5$, the 
subgroup of $2$-elements is $H_2 = \{e,a^{45}\} = \langle a^{45} \rangle $.  
The subgroup of $3$-elements is $\langle a^{10}\rangle$.  (Note that $a^{10}$, $a^{20}$,
$a^{40}$, $a^{50}$, $a^{70}$, $a^{80}$ have order $9$, while $a^{30}$ and $a^{60}$
have order $3$, and $e$ has order $1$.  So all the elements of $\langle a^{10}\rangle$
are $3$-elements and they the only ones.)  Similarly, the subgroup of $5$-elements
is $\langle a^{18}\rangle$.  Since $G$ is the internal direct product of its subgroups
$H_2,H_3,H_5$ by Lemma 5.5, we know there is exactly one way to write 
$a = b c d$ with $b \in H_2$, $c \in H_3$ and $d \in H_5$.  A little experimentation
gives 
$$b = a^{45}, \quad c = a^{10}, \quad d = a^{36}.$$
(Note, it would also be possible to solve this by setting up a congruence on the 
exponents writing $b = a^{45 i}, c = a^{10 j},  d = a^{18 k}$.  We want to find
$i,j,k$ with $45i + 10j + 18k \equiv 1 \bmod 90$, and $i = 1, j = 1, k = 2$ is
clearly one solution.)
\bye
