Question

Let G and H be finite cyclic groups. Prove that$\mathit{G}\times \mathit{H}$ is cyclic if and only if$\left(\left|\mathit{G}\right|,\left|\mathit{H}\right|\right)\mathbf{=}\mathbf{1}$ .

Short answer

It has been proved that,$G\times H$ is cyclic if and only if$\left(\left|G\right|,\left|H\right|\right)=1$ .

Step-by-step solution

Step 1:Statement of Lagrange Theorem

Lagrange Theorem states that, for any finite group G, the order of subgroup Hof group G divides the order of G.

Given that G and H are finite cyclic groups.

Let$g\in G$and$h\in H$are generators of these groups.

Let$\left|g\right|=n$and $\left|h\right|=m$.

Prove that G×H is cyclic if and only if G,H=1

According to Lagrange’s theorem, n divides $\left|G\right|$ and m divides $\left|H\right|$.

So, order of $\left(g,h\right)$in $G\times H$is least common multiple of $\left|g\right|$and $\left|h\right|$.

Thus, $\left|\left(g,h\right)\right|=lcm\left\{m,n\right\}$

Now, $lcm\left\{m,n\right\}=mn$if and only if $\left(m,n\right)=1$.

Hence, $\left(g,h\right)$is the generator of $G\times H$if and only if $\left(m,n\right)=1$.

Thus, it can be concluded that$G\times H$ is cyclic if and only if$\left(\left|G\right|,\left|H\right|\right)=1$ .