Question

Let ${\mathbf{N}}_{\mathbf{1}}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{,}{\mathbf{N}}_{\mathbf{k}}$be normal subgroups of a finite group $\mathbf{G}$. If $\mathbf{G}\mathbf{=}{\mathbf{N}}_{\mathbf{1}}\mathbf{,}{\mathbf{N}}_{\mathbf{2}}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}{\mathbf{N}}_{\mathbf{k}}$(notation as in Exercise 25) and $\left|G\right|\mathbf{=}\left|N_1\right|\mathbf{·}\left|N_2\right|\mathbf{.}\mathbf{.}\mathbf{.}\left|N_k\right|$, 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

Modify the statement

Second Isomorphism Theorem.

Let$\mathbf{K}$and$\mathbf{N}$be subgroups of a group$\mathbf{G}$, with$\mathbf{N}$ normal in$\mathbf{G}$.

Then$\mathbf{NK}\mathbf{=}\mathbf{\{}\mathbf{nk}\mathbf{}\overline{)\mathbf{n}}\mathbf{\in }\mathbf{N}\mathbf{,}\mathbf{}\mathbf{k}\mathbf{\in }\mathbf{K}\mathbf{\}}$ is a subgroup of$\mathbf{G}$that contains both$\mathbf{K}$and $\mathbf{N}$.

Claim: Let $N_i$be the normal subgroups of a finite group $G$. Then, $\left|N_1...N_k\right|$divides$\left|N_1\right|...\left|N_k\right|$with equality if and only if $N_1...N_k\cong N_1\times ...\times N_k$.

Prove by induction

Suppose $k\ge 2$and let $H=N_1...N_{k-1}$.

Then, $H$is normal and bySecond Isomorphism Theorem, $\left|N_1...N_k\right|=\left|H\right|·\left|N_k\right|/\left|H\cap N_k\right|$

By the induction hypothesis, this divides $\left|N_1\right|...\left|N_{k-1}\right|·\left|N_k\right|/\left|H\cap N_k\right|$, which divides $\left|N_1\right|...\left|N_k\right|$.

The equality holds if and only if $\left|H\right|=\left|N_1\right|...\left|N_{k-1}\right|$and .

By induction, this occurs if and only if $H\cong N_1\times ...\times N_{k-1}$and .

Thus, it can be concluded that $\left|N_1...N_k\right|$divides $\left|N_1\right|...\left|N_k\right|$with equality if and only if $N_1...N_k\cong N_1\times ...\times N_k$.

Hence, $G=N_1\times N_2\times ...\times N_k$.