Could someone please verify whether my solution is okay? The problem with my answer is that it isn't rigorous, meaning I just assumed a lot of properties to be true, since the professor did not prove these in class. Additionally, I am not sure what the question is asking and so this is the best idea I came up with.

Show that $S=\left \{ 2^{m}3^{n}:m,n\in \Bbb{Z} \right \}\leq \Bbb{Q}\setminus \left \{0\right \}$ is the internal direct product isomorphic to $\Bbb{Z}\times \Bbb{Z}$.

Let $H=\left \{ 2^{m}:m\in \Bbb{Z} \right \}$ and $K=\left \{ 3^{n}:n\in \Bbb{Z} \right \}$.

Claim: $S$ is the internal direct product of $H$ and $K$.

$S=HK=\left \{ 2^{m}3^{n}:2^{m}\in H, 3^{n}\in K \right \}$, which is true.

If $m=n=0$, then $2^{0}=3^{0}=1$. Since $2^{m}\neq 3^{n}$ for any $m,n>0$, it must be that $H\cap K=\left \{1 \right \}$.

I am honestly not even certain whether the above statement is correct. I just wrote it to make the condition hold in my favor.

Let $2^{m}\in H$ and $3^{n}\in K$. Then $2^{m}3^{n}=3^{n}2^{m}$, since integers are abelian.

Claim: $S\cong \Bbb{Z}\times \Bbb{Z}$

Define $\phi:\Bbb{Z}\times \Bbb{Z}\to S$ by $\phi(m,n)=2^{m}3^{n}$.

Well-defined: If $(m,n)=(x,y)$, then $\phi(m,n)=2^{m}3^{n}=2^{x}3^{y}=\phi(x,y)$ by coordinate equality.

Operation-preserving: $\phi((m,n)+(x,y))=\phi(m+x,n+y) = 2^{m+x}3^{n+y}=2^{m}2^{x}3^{n}3^{y}=2^{m}3^{n}2^{x}3^{y}=\phi(m,n)\phi(x,y)$

Injective: If $\phi(m,n)=\phi(x,y)$, then $2^{m}3^{n}=2^{x}3^{y}$. Then $m=x$ and $n=y$ by countability of $\Bbb{Z}\times \Bbb{Z}$ and laws of exponents.

Surjective: $\phi$ is onto iff for all $2^{m}3^{n}\in S$, there exists an $(m,n)\in \Bbb{Z}\times \Bbb{Z}$ such that $\phi(m,n)=2^{m}3^{n}$. By construction of $\phi$, it is surjective.


Your proof seems fine to me.

If you want to justify that $2^m\neq 3^n$ for $m, n\ge 1$, note that the LHS is even whereas the RHS is odd. It also follows by uniqueness of factorisation.

  • 1
    $\begingroup$ Thank you for taking the time to look at my long solution. And thanks, that is actually very helpful! $\endgroup$ – numericalorange Mar 27 '18 at 0:54
  • $\begingroup$ You're welcome! :) $\endgroup$ – Shaun Mar 27 '18 at 0:55

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