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\centerline{MATH 351 -- Modern Algebra 1}
\centerline{Selected Solutions for Problem Set 7}
\centerline{November 3, 2018}
\bigskip
\noindent
4.46.  (1) We are given that $N$ and $K$ are subgroups of index $2$ of $G$ with
$N \ne K$.  By Theorem 4.1 in Lee, note that both $N$ an $K$ are normal
subgroups of $G$.  By the result of Exercise 4.5, $N\cap K$ is also a 
normal subgroup of $G$.  This follows because if $g \in G$ is arbitrary, and
$a \in N\cap K$, then $a \in N$, so $gag^{-1} \in N$ and similarly $a \in K$,
so $gag^{-1} \in K$.  Hence $gag^{-1} \in N \cap K$, so $N \cap K$ is normal in
$G$.  The same argument just given also works if we only take $g \in N$.
Hence $N\cap K$ is also a normal subgroup of $N$.  Now by the {\it Second Isomorphism
Theorem}, we have 
$$N/(N \cap K) \cong NK/K.$$
We claim that $NK = G$.  This is true because $K$ has index $2$ in $G$.  In
the set 
$$NK = \{nk \mid n \in N, k \in K\},$$
if $n \in K$ as well, then $nK = K$, so we get $K \subset NK$.
On the other hand, we are assuming $N \ne K$, so there are $n \in N$ that are
in $G \setminus K$ as well.  If $n \notin K$, then the coset $nK = G \setminus K$, the
other left coset of $K$ in $G$.  So $G \setminus K \subset NK$ as well.  
Hence $K \cup (G \setminus K) = G \subseteq NK \subseteq G$.  It follows that
$NK = G$. Now we have proved what was required since 
$$[N : N\cap K] = \left|N/(N \cap K)\right| = |NK/K| = |G/K| = 2.$$
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\noindent
(2)  Now consider the mapping
$$\eqalign{\alpha : G &\to G/N \times G/K\cr
                    g &\mapsto (gN,gK)\cr}$$
$G/N$ and $G/K$ have order $2$ so each is isomorphic to ${\bf Z}_2$ by Corollary 4.2.  We want
to show that $\alpha$ is a surjective group homomorphism and identify the kernel as $N \cap K$
to show that $G/(N\cap K) \cong {\bf Z}_2 \times {\bf Z}_2$.  Here are the individual steps:
\medskip
\item{$\bullet$}  We claim first that $\alpha$ is a group homomorphism.  This is true because 
$$\alpha(g\cdot h) = ((g\cdot h)N,(g\cdot h)K) = (gN\cdot hN, gK\cdot hK) = (gN,gK)\cdot (hN,hK) = \alpha(g)\cdot \alpha(h),$$
using the definition of the coset product in $G/N$ and $G/K$.  
\item{$\bullet$} The kernel of $\alpha$
consists of the $g\in G$ such that $(gN,gK) = (N,K)$, the identity element of $G/N\times G/K$.  
But $(gN,gK) = (N,K)$ if and only if $g \in N$ and $g\in K$, so $g\in N \cap K$.  Hence
$\ker(\alpha) = N\cap K$.  
\item{$\bullet$} Next, we claim that $\alpha(G) = (G/N,G/K)$ (that is, that $\alpha$
is surjective).  This follows by thinking about what part (1) of the problem says.  We
have $[N:N\cap K] = 2$.  The same argument, but reversing the roles of $N,K$ shows that
$[K:N\cap K] = 2$ as well.  Moreover $[G:N] = [G:K] = 2$ by assumption.  It follows
that $G$ decomposes into four disjoint subsets:
$$N \cap K, (G \setminus N) \cap K, N \cap (G \setminus K),\ {\rm and}\ (G \setminus N) \cap (G\setminus K).$$
All of these are nonempty since $[N:N\cap K] = [K:N \cap K] = 2$. 
The $g$ in the first map to $(N,K)$ under $\alpha$, and the $g$ in the other three subsets
map to the other three elements in $G/N \times G/K$.  Therefore $\alpha$ is surjective. 
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\noindent
{\it Comment:}  We actually saw an example of the pattern described here when we considered
$G = D_8$, $N = \langle R_{90}\rangle$ and $K = \{R_0,R_{180},F_1,F_2\}$.  These are
two unequal subgroups of order 4 in a group of order 8, so we're exactly in the situation
of the problem. In this case, 
$N \cap K = \{R_0,R_{180}\} = Z(D_8)$ and we showed directly that 
$$D_8/Z(D_8) \cong {\bf Z}_2 \times {\bf Z}_2.$$
\bye
