We were given the following statement to prove: "in $ \mathbb Z $, let $ H= \langle 3 \rangle $ and $ K = \langle 7 \rangle $. Prove that $ \mathbb Z = HK $. Does $ \mathbb Z = H \times K $?" I have come up with the following answer but I'm very unsure of it. Just hoping someone could help me correct it:
$ H= \langle 3 \rangle = \{0, \pm 3, \pm6, ...\}$ and $ K = \langle 7 \rangle =\{0, \pm 7, \pm14,...\}$. As $ HK = \{hk:h \in H, k \in K\}$, we have $ -6+7 = 1 \in HK $, & so $\langle 1 \rangle \subseteq HK$. The set of integers $\mathbb Z $ under addition is cyclic and 1 is a generator. So $ HK \subseteq \mathbb Z $. Now suppose $ x \in HK $. Then $ x= hk = (\pm n3)(\pm n7) $, and so clearly, $ x \in \mathbb Z $, and $ HK \subseteq \mathbb Z $. Thus $ HK = \mathbb Z$.
If $ x \in \mathbb Z $, and $ h \in H $, then $ x + h -x=h \in H$, and so $ xHx^{-1} \subseteq H $. The situation with $ K $ is identical. So $ H $ and $ K $ are both normal subgroups of $ \mathbb Z $. Furthermore as 3 and 7 are relatively prime we have $ \langle 3 \rangle \cap \langle 7 \rangle = \{ e \}$. So $ H \times K = \mathbb Z $.