Question

$\mathit{\phi }\mathbf{:}{\mathit{S}}_{\mathbf{n}}\mathbf{\to }{\mathit{S}}_{\mathbf{n}\mathbf{+}\mathbf{1}}$, where for eachrole="math" localid="1654592036136" $\mathit{f}\mathbf{\in }{\mathit{S}}_{\mathbf{n}}$ ,$\mathit{\phi }\mathbf{\left(}\mathbf{f}\mathbf{\right)}\mathbf{\in }{\mathit{S}}_{\mathbf{n}\mathbf{+}\mathbf{1}}$ is given by,$\mathit{\phi }\mathbf{\left(}\mathbf{f}\mathbf{\right)}\mathbf{\left(}\mathbf{k}\mathbf{\right)}\mathbf{=}\{\begin{array}{l}f\left(k\right)if1\le k\le n\\ n+1ifk=n+1\end{array}$

Short answer

It is proved that,$\phi$ is a homomorphism and Ker $\phi =\left\{e\right\}$.

Step-by-step solution

To show f is a homomorphism

Definition of Group Homomorphism

Let $(G,\ast )$and $(G\text{'},\ast \text{'})$be any two groups. A function $f:G\to G\text{'}$is said to be a group homomorphism if $f(a\ast b)=f\left(a\right)\ast \text{'}f\left(b\right),\forall a,b\in G$.

Definition of Kernel of a Function

Let $f:G\to H$be a homomorphism of groups. Then the kernel of $f$is defined by the set $\{a\in G:f\left(a\right)=e_H\}$, where $e_H$is an identity element.

It is the mapping from elements in $G$onto an identity element in $H$by the homomorphism $f$.

Let $\phi :S_n\to S_{n+1}$be the function defined as follows:

$\phi \left(f\right)\left(k\right)=\left\{\begin{array}{l}f\left(k\right)if1\le k\le n\\ n+1ifk=n+1\end{array}\right.$

We have to show that $\phi$is a homomorphism.

Let$f,g\in S_n$.

Then to prove homomorphism we will show that $\phi \left(f{g}^{-1}\right)=\phi \left(f\right)\phi {\left(g\right)}^{-1}$.

First, for $1\le k\le n$, we have:

$$\begin{gathered} \phi \left(f{g}^{-1}\right)\left(k\right)=f{g}^{-1}\left(k\right) \\ =f\left(g^{-1}\left(k\right)\right) \\ =f\left(\phi {\left(g\right)}^{-1}\left(k\right)\right) \\ =\phi \left(f\right)\left(\phi {\left(g\right)}^{-1}\left(k\right)\right) \end{gathered}$$

$\therefore \phi \left(f{g}^{-1}\right)\left(k\right)=\phi \left(f\right)\phi \left(g^{-1}\right)\left(k\right)$

Thus,$\phi \left(f{g}^{-1}\right)=\phi \left(f\right)\phi {\left(g\right)}^{-1}$.

Now, for $k=n+1$, we have:

$$\begin{gathered} \phi \left(f{g}^{-1}\right)\left(k\right)=\phi \left(f{g}^{-1}\right)(n+1) \\ =n+1 \\ =\phi \left(f\right)(n+1)\phi \left(g^{-1}\right)(n+1) \\ =\phi \left(f\right)\left(k\right)\phi \left(g^{-1}\right)\left(k\right) \end{gathered}$$

$\therefore \phi \left(f{g}^{-1}\right)\left(k\right)=\phi \left(f\right)\phi \left(g^{-1}\right)\left(k\right)$

Thus,$\phi \left(f{g}^{-1}\right)=\phi \left(f\right)\phi {\left(g\right)}^{-1}$ .

Hence,$\phi$ is a homomorphism.

To find kernel of φ

We find kernel of$f$as:

Ker$f=\{f\in S_n:\phi \left(f\right)\left(k\right)=k,1\le k\le n+1\}$

Since for all$1\le k\le n,$we have$\phi \left(f\right)\left(k\right)=k$and$\phi \left(f\right)(n+1)=n+1$.

Then, it follows that, for some $f\in S_n$, we have $\phi \left(f\right)\left(k\right)=k$, for all $1\le k\le n+1$.

Then,$\phi \left(f\right)\left(k\right)=k$ , for all $1\le k\le n$.

Hence, $\phi$is an identity permutation.

Therefore, ker $\phi =\left\{e\right\}$.

Thus,$\phi$ is a homomorphism and Ker$\phi =\left\{e\right\}$ .