# A condition involving internal direct products.

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 Khan 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

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 $\mathbb{Z}$.
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).