Problem Statement: Let $H$, $K$, $N$ be nontrivial normal subgroups of a group $G$ and suppose $G=H\times K$. Prove that $N$ is in the center of $G$ or $N$ intersects one of $H$, $K$ non-trivially.
This current section discusses internal and external weak direct products. I discussed this problem with my professor, and I believe I understand (almost) every part of this problem now, but I am still having one missed connection.
We wanted to prove this problem by assuming that $N\cap H=N\cap K=\langle e\rangle$, and show that $N\subset Z(G)$. She said that $G$ is the internal weak direct product of $H$ and $K$, i.e. $G=H\vee K$, but I am not seeing that based on the definitions. (Also, I understand that Hungerford is kind of abusing notation when he uses "$=$" rather than "$\cong$", but essentially $G$ and $H\times K$ are the same thing.)
The definition of weak direct product says that if $G=H\vee K$ and $H\cap K=\langle e\rangle$, then $G\cong H\times K$. But the reverse implication isn't necessarily true, correct? We cannot assume anything about $H\cap K$ either, other than normality, correct?
Now, if I can somehow show that $G=H\vee K=\langle H\cup K\rangle$. Then by another theorem in the book, since $H$ and $K$ are normal in $G$, then $G=H\vee K=HK$, and by the same theorem, it tells us that since $N\cap H=N\cap K=\langle e\rangle$, then $nk=kn$ and $nh=hn$ for $n\in N$, $k\in K$, $h\in H$. From there, we can show that since $G=HK$, then any $g\in G=HK$ can be expressed as $g=hk$ for some $h\in H$ and $k\in K$. So then it must be that $$nhk=(nh)k=(hn)k=h(nk)=h(kn)=hkn.$$ So then $n$ commutes with anything in $G$, and thus, $N$ is in the center.
Can someone explain to me why $G=H\times K$ implies that $G=H\vee K$? I appreciate your help!
I'll denote $\times_i$ (internal direct product) and $\times_e$ (external direct product).
If $G = H \times_i K$, this means that $H \times_e K \cong H \times_i K=G$ where our isomorphism is $(h,k) \mapsto hk$.
This means that for each $g \in G$ there exists some (unique choice of) $h \in H$, $k \in K$ such that $g=hk$. This implies (among other things that $H \times_i K = HK = G$. Notice that $(h,1)(1,k) = (h1,1k) = (h,k) = (1h,k1) = (1,k)(h,1)$ in $H \times_e K$ therefore $hk=kh$ in $G=H \times_i K$.
Consider $w \in H \vee K = \langle H \cup K \rangle$. Then $w=h_1k_1\cdots h_\ell k_\ell$ for some finite string of $h$'s and $k$'s (anything in $\langle H \cup K \rangle$ can be written as a word in the alphabet $H \cup K$ -- we don't need inverses because $H$ and $K$ are subgroups and can start and/or end on either an $h$ or $k$ because $1$ belongs to both $H$ and $K$). Well, we know that $hk=kh$ for any $h$ and $k$. Therefore, $w=h_1k_1\cdots h_\ell k_\ell = h_1\cdots h_\ell \, k_1\cdots k_\ell \in HK=G$. In other words, $H \vee K = H \times_i K = G$.
Hopefully, something in this discussion will help with your hangup! :)