Question

Let $\mathbf{N}$be a normal subgroup of a group $\mathbf{G}$and let $\mathbf{f}\mathbf{:}\mathbf{G}\mathbf{\to }\mathbf{H}$be a homomorphism of groups such that the restriction of $\mathbf{f}$to $\mathbf{N}$is an isomorphism $\mathbf{N}\mathbf{\cong }\mathbf{H}$. Prove that $\mathbf{G}\mathbf{\cong }\mathbf{N}\mathbf{\times }\mathbf{K}$, where K is the kernel of $\mathbf{f}$.

Short answer

Answer:

It is proved that, $G$is isomorphic to $N\times K$, where $K$is the kernel of $f$.

Step-by-step solution

First Isomorphism Theorem

Let $f:G\to H$be a surjective homomorphism of groups with kernel $K$. Then, the quotient group $G/K$ is isomorphic to H.

g is one to one

Define a map $g:N\times K\to G$by $g\left(n.k\right)=nk$.

Let $\left(n,k\right),\left(n\text{'},k\text{'}\right)\in N\times K$.

Then, $g\left(n,k\right)=g\left(n\text{'},k\text{'}\right)$implies that $nk=n\text{'}k\text{'}$. Similarly, $f\left(nk\right)=f\left(n\text{'}k\text{'}\right)$ implies $nk=n\text{'}k\text{'}$.

Since f is an isomorphism $n=n\text{'}$, implies $k=k\text{'}$.

Therefore, g is one to one.

g is onto

Let $g\in G$.

Then, $f\left(g\right)\in H$.

Since $f\left(g\right)\in H$and f is an isomorphism, there exists $n\in N$such that $f\left(n\right)=f\left(g\right)$or $f\left(n^{-1}g\right)=e$.

Since $n^{-1}g\in K,n{n}^{-1}g=g$implies $g\left(n,n^{-1}g\right)=g$.

Therefore, g is onto.

g is a homomorphism

Let $\left(n,k\right),\left(n\text{'},k\text{'}\right)\in N\times K$then,

$$\begin{gathered} g\left(\left(n,k\right)\left(n\text{'},k\text{'}\right)\right)=g\left(nn\text{'},kk\text{'}\right) \\ =nn\text{'}kk\text{'} \\ =nkn\text{'}k\text{'} \\ =g\left(n,k\right)g\left(n\text{'},k\text{'}\right) \end{gathered}$$

Thus, g is a homomorphism.

Therefore, $G$is isomorphic to $N\times K$where $K$is the kernel of $f$.