Question

If $\left(m,n\right)\mathbf{\ne }\mathbf{1}$, prove that role="math" localid="1659418227168" ${\mathbf{\square }}_{\mathbf{m}\mathbf{n}}$is not isomorphic to role="math" localid="1659418258581" ${\mathbf{\square }}_{\mathbf{m}}\mathbf{\times }{\mathbf{\square }}_{\mathbf{n}}$.

Short answer

It is proved that ${\square }_{\mathrm{mn}}$ is not isomorphic to ${\square }_{\mathrm{m}}\times {\square }_{\mathrm{n}}$.

Step-by-step solution

General concept

In contrast, assume that ${\mathbf{\square }}_{\mathbf{m}\mathbf{n}}$is isomorphic to role="math" localid="1659418470322" ${\mathbf{\square }}_{\mathbf{m}}\mathbf{\times }{\mathbf{\square }}_{\mathbf{n}}$. It implies that ${\mathbf{\square }}_{\mathbf{m}}\mathbf{\times }{\mathbf{\square }}_{\mathbf{n}}$is cyclic. It implies that there exist $\left(a,b\right)\mathbf{\in }{\mathbf{\square }}_{\mathbf{m}}\mathbf{\times }{\mathbf{\square }}_{\mathbf{n}}\mathbf{}\mathit{s}\mathit{u}\mathit{c}\mathit{h}\mathbf{}\mathit{t}\mathit{h}\mathit{a}\mathit{t}\mathbf{}\mathit{O}\left(\left(a,b\right)\right)\mathbf{=}\mathit{m}\mathit{n}$.

Show that that □mn is not isomorphic to □m×□n.

Consider integers m and n, such that $\left(m,n\right)\ne 1$.

Recall that if $\left(a,b\right)\in {\square }_m\times {\square }_n$, then $O\left(\left(a,b\right)\right)=lcm\left(o\left(a\right),O\left(b\right)\right)$. Note that role="math" localid="1659419357979" $O\left(a\right)\overline{)m}$and $O\left(a\right)\overline{)n}$. This implies that:

$\begin{array}{rcl}lcm\left(O\left(a\right),O\left(b\right)\right)& \le & lcm\left(m,n\right)\\ mn& \le & lcm\left(m,n\right)......\left(1\right)\end{array}$

Since $lcm\left(m,n\right)\le mn$, equation (1) implies that $lcm\left(m,n\right)=mn$. Then:

$\begin{array}{rcl}lcm\left(m,n\right)& =& \frac{mn}{lcm\left(m,n\right)}\\ & =& \frac{mn}{mn}\\ & =& 1\end{array}$

This is a contradiction to the condition $ged\left(m,n\right)\ne 1$. Therefore, the assumption is wrong.

Hence, it is proved that ${\square }_{mn}$is not isomorphism to ${\square }_m\times {\square }_n$.