Question

If $\mathit{H}$ and$\overline{\mathbf{H}}$ are the groups in Example 5. Show that$\mathit{H}\mathbf{\cong }\overline{\mathbf{H}}$ .

Short answer

It is proved that,$H$ is isomorphic to$\overline{H}$ i.e., $H\cong \overline{H}$.

Step-by-step solution

To find kernel of f

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 ofrole="math" localid="1654599922530" $f$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$H$ and$K$ be two groups.

Consider the function $f:H\times K\to H$defined as,$f(h,k)=h$ .

Thenby Exercise 9 of Section 7.4, role="math" localid="1654600107559" $f$is a homomorphism.

Then, simplify ker $f$as follows:

$$\begin{gathered} \overline{H}=\text{Ker}f \\ =\{(h,k)\in H\times K:f(h,k)=e_H\} \\ =\{(h,k)\in H\times K:h=e_H\} \end{gathered}$$

$\therefore \overline{H}=\{(e_H,k):k\in K\}$

To show f is bijective

Now, we have to show that $H\cong \overline{H}$.

Define a function $f:\overline{H}\to H$by $f(e_H,k)=k$.

Now, for any $k\in H$ take$(e_H,k)\in \overline{H}$ such that$(e_H,k)=k$ .

Hence,$f$ is surjective.

Now, for any two elements$(e_H,k),(e_H,m)\in \overline{H}$ with $f(e_H,k)=f(e_H,m)\Rightarrow k=m$.

Thus, $f$is injective.

Hence, $f$ is bijective.

To show f is a homomorphism and H≅H¯

Now, in order to prove homomorphism, take $(e_H,k),(e_H,m)\in \overline{H}$.

Then, prove $f((e_H,k)\cdot (e_H,m))=f(e_H,k)f(e_H,m)$as:

$$\begin{gathered} f((e_H,k)\cdot (e_H,m))=f(e_H,km) \\ =km \\ =f(e_H,k)f(e_H,m) \end{gathered}$$

Therefore,$f((e_H,k)\cdot (e_H,m))=f(e_H,k)f(e_H,m)$.

Hence,$f$ is a homomorphism.

Thus,$f$ is bijective and homomorphism.

Hence,$H$ is isomorphic to$\overline{H}$ i.e.,$H\cong \overline{H}$ .