Question

A group $\mathbf{G}$ is said to be indecomposable if it is not the direct product of two of its proper normal subgroups. Prove that each of these groups is indecomposable:${\mathbf{D}}_{\mathbf{4}}$

Short answer

It has been proved that, ${\mathrm{D}}_4$ is an indecomposable group.

Step-by-step solution

Step-by-Step SolutionStep 1: Definition of Indecomposable

A group$G$ is said to be indecomposable if it is not the direct product of its two proper normal subgroups.

Prove that D4 is indecomposable

Consider the symmetric group, $D_4=\{1,r,r^2,r^3,t,rt,r^{2}t,r^{3}t\}$.

Claim: If $G$is non-abelian group with $\left|G\right|<12$then, $G$is indecomposable.

Consider that role="math" localid="1652848333970" $G$is not indecomposable.

Then, $G=A\times B$for proper subgroups $A,B$.

Hence,$\left|A\right|,\left|B\right|\le \left|G\right|$, which implies order should be less than 6.

This implies $A,B$are abelian. But then, $G$would also be abelian.

This is the contradiction since $G$is non-abelian group with $\left|G\right|<12$.

Therefore,$G$ is indecomposable.

This implies $D_4$is also a non abelian group with order 8(<12).

Hence,$D_4$ is indecomposable.

Thus, it can be concluded that $D_4$is indecomposable.