Question

If $\mathbf{N}\mathbf{,}\mathbf{K}$are subgroups of a group $\mathbf{G}$such that $\mathbf{G}\mathbf{=}\mathbf{N}\mathbf{\times }\mathbf{K}$and $\mathbf{M}$is a normal subgroup of $\mathbf{N}$, prove that $\mathbf{M}$is a normal subgroup of $\mathbf{G}$.

[Compare this with Exercise 14 in Section 8.2.]

Short answer

Answer:

It is proved that, $M$is a normal subgroup of $G$.

Step-by-step solution

Define isomorphic map with normal subgroup

Definition of subgroup:

Let $\left(G,*\right)$be a group under binary operation, say $*$. A non-empty subset $H$of $G$is said to be a subgroup of $G$, if $\left(H,*\right)$is itself a group.

In other words, if $e_G\in H$and $a,b\in H$then, $a{b}^{-1}\in H$.

Definition of Normal Subgroup:

A subgroup $N$of a group $G$is called a normal subgroup of $G$if $Na=aN,\forall a\in G$.

Note that, if $\phi :G\to H$is an isomorphism then, any normal subgroup goes to normal subgroup as if $N\subset G$is normal.

Now consider, $a\in \phi \left(N\right)$and $b\in H$.

Let $\phi \left(a\text{'}\right)=a$and $\phi \left(b\text{'}\right)=b$.

Then, find $ba{b}^{-1}$as:

$$\begin{gathered} ba{b}^{-1}=\phi \left(b\text{'}\right)\phi \left(a\text{'}\right)\phi \left(b{\text{'}}^{-1}\right) \\ =\phi \left(b\text{'}a\text{'}b{\text{'}}^{-1}\right) \\ \in \phi \left(N\right) \end{gathered}$$

Therefore, $b\text{'}a\text{'}b{\text{'}}^{-1}$$\in N$since $N$is normal.

To prove M is a normal subgroup of G

Let $N,K$are subgroups of $G$.

Also, $M$is a subgroup of $G$.

Now, consider $M$as a subgroup of $G=N\times K$, which implies $M\times \left\{e_G\right\}$.

Then, $\left(m,e_G\right)\in M\times \left\{e_G\right\}$and $n,k\in G$gives as:

$$\begin{gathered} \left(n,k\right)\left(m,e_G\right){\left(n,k\right)}^{-1}=\left(n,k\right)\left(m,e_G\right)\left(n^{-1},k^{-1}\right) \\ =\left(nm{n}^{-1},k{k}^{-1}\right) \\ =\left(nm{n}^{-1},e_G\right) \\ \in M\times \left\{e_G\right\} \end{gathered}$$

Hence, $nm{n}^{-1}\in M$since $M$is normal in $N$.

Thus, $M$is a normal subgroup of $G$.