# internal direct product groups

Let $G = \{ 3^a 6^b 10^c: a, b, c \in \mathbb{Z}\}$ under multiplication and $H = \{ 3^a 6^b 12^c: a, b,c \in \mathbb{Z}\}$ under multiplication. Prove that $G = \langle3\rangle \times \langle6\rangle \times \langle10\rangle$ whereas $H$ is not the IDP of $\langle3\rangle \times \langle6\rangle \times \langle12\rangle$.

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Hints: $3^16^012^1 =3^06^212^0$ Also, $3^a6^b10^c=2^{b+c}3^{a+b}5^c$, and from unique factorization the values of $a+b$, $b+c$ and $c$ are uniquely determined. – Jyrki Lahtonen Sep 23 '11 at 12:49

Hint. Use prime factorisation to show that any element of $G$ can be written as $3^a 6^b 10^c$ for a unique triple $(a, b, c)$. Conclude that $G$ is the internal direct product $\langle 3 \rangle \times \langle 6 \rangle \times \langle 10 \rangle$. On the other hand, find two distinct triples $(a, b, c)$ and $(a', b', c')$ such that $3^a 6^b 12^c = 3^{a'} 6^{b'} 12^{c'}$. Conclude that $(\langle 3 \rangle \times \langle 6 \rangle) \cap \langle 12 \rangle$ is non-trivial and therefore $H$ is not an internal direct product of $\langle 3 \rangle \times \langle 6 \rangle$ and $\langle 12 \rangle$.