Question

Let ${\mathbf{N}}_{\mathbf{1}}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{,}{\mathbf{N}}_{\mathbf{k}}$be normal subgroups of a group $\mathbf{G}$. Let ${\mathbf{N}}_{\mathbf{1}}{\mathbf{N}}_{\mathbf{2}}\mathbf{.}\mathbf{.}\mathbf{.}{\mathbf{N}}_{\mathbf{k}}$denote the set of all elements of the form ${\mathbf{a}}_{\mathbf{1}}{\mathbf{a}}_{\mathbf{2}}\mathbf{.}\mathbf{.}\mathbf{.}{\mathbf{a}}_{\mathbf{k}}$with ${\mathbf{a}}_{\mathbf{j}}\mathbf{\in }{\mathbf{N}}_{\mathbf{j}}$. Assume that $\mathbf{G}\mathbf{=}{\mathbf{N}}_{\mathbf{1}}{\mathbf{N}}_{\mathbf{2}}\mathbf{.}\mathbf{.}\mathbf{.}{\mathbf{N}}_{\mathbf{k}}$and that ${\mathbf{N}}_{\mathbf{i}}\mathbf{\cap }\left(N_1...N_{i+1}...N_k\right)\mathbf{=}<e>$for each $\mathbf{i}\left(1\le i\le n\right)$. Prove that $\mathbf{G}\mathbf{=}{\mathbf{N}}_{\mathbf{1}}\mathbf{\times }{\mathbf{N}}_{\mathbf{2}}\mathbf{\times }\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\times }{\mathbf{N}}_{\mathbf{k}}$.

Short answer

Answer:

It has been proved that, $G=N_1\times N_2\times ...\times N_k$.

Step-by-step solution

Given that

Given that $N_1,...N_k$are normal subgroups of a group $G$.

Also $N_1,N_2...N_k$denote the set of all elements of the form $a_1{a}_2...a_k$with $a_j\in N_j$.

It is assumed that $G=N_1{N}_2...N_k$and that for each $i\left(1\le i\le n\right)$.

Prove by induction

Let $H=N_1...N_{k-1}$.

Since $N_1,...,N_k$are normal subgroups, so $H$is also a normal subgroup.

Also by assumption, .

Thus, $G\cong H\times N_k$.

By induction on $k$, it is clear that $G=N_1\times N_2\times ...N_k$.

Thus, it can be concluded that $G=N_1\times N_2\times ...\times N_k$.