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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 $.

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You probably meant $\mathbb{Z} \subseteq HK$ the first time. – Adeel Nov 16 '12 at 21:35
How did you show that the intersection of $<3>,<7>$ equals {0}? – Amr Nov 16 '12 at 21:38
up vote 2 down vote accepted

The second part is wrong: consider the isomorphism $\mathbb{Z} \to \left<3\right>$ given by $n \mapsto 3n$ and the analogous one for $\left<7\right>$; it follows that $H \times K$ is isomorphic to $\mathbb Z \times \mathbb Z$ which is easily shown to be not isomorphic to $\mathbb{Z}$.

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21 belongs to the intersection of $<3>,<7>$. It is not true that $Z=<3>$x$<7>$. Note that <3>,<7> are isomorphic to $Z$. Thus $Z$ is isomorphic to the direct sum of $Z$ and $Z$. However, the direct sum of $Z$ and $Z$ is not cyclic (a contradiction).

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Didn't think the last part I wrote down was correct. Thanks. I will use your proof by contradiction. – mkeachie Nov 16 '12 at 22:34

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