Question

If G and H are groups, prove that $\mathbf{G}\mathbf{\times }\mathbf{H}\mathbf{\cong }\mathbf{H}\mathbf{\times }\mathbf{G}$.

Short answer

Answer:

It is proved that, $G\times H\cong H\times G$.

Step-by-step solution

Isomorphic Groups

Two groups $\mathbf{G}$and $\mathbf{H}$are said to be isomorphic if there exists an isomorphism $\mathbf{f}$from $\mathbf{G}$onto $\mathbf{H}$.

Prove that G×H≅H×G.

To prove $G\times H\cong H\times G$, that is to prove $G\times H$is isomorphic to $H\times G$.

Define a map $f:G\times H\to H\times G$as, $f\left(g,h\right)=\left(h,g\right)$where, $g\in G$and $h\in H$.

To show $G\times H$is isomorphic to $H\times G$.

It is enough to show that the map is one-one, onto, and homomorphism.

Check $f:G\times H\to H\times G$is one-one.

Let $f\left(g_1,h_1\right)=f\left(g_2,h_2\right)$where, $g_1,g_2\in G$and $h_1,h_2\in H$.

Then, $f\left(g_1,h_1\right)=f\left(g_2,h_2\right)$.

This implies that, $\left(h_1,g_1\right)=\left(h_2,g_2\right)$.

Further simplify $\left(h_1,g_1\right)=\left(h_2,g_2\right)$as: $$\begin{gathered} h_1=h_2 \\ {g}_1=g_2 \end{gathered}$$

Therefore, $h_1=h_2$and $g_1=g_2$.

Hence, $\left(g_1,h_1\right)=\left(g_2,h_2\right)$.

Therefore, $f:G\times H\to H\times G$mapping is one-one.

Check $f:G\times H\to H\times G$is onto.

For every $\left(h,g\right)\in H\times G$, there exists a unique $\left(g,h\right)\in G\times H$such that $f\left(g,h\right)=\left(h,g\right)$.

Therefore, $f:G\times H\to H\times G$mapping is onto.

Check $f:G\times H\to H\times G$is homomorphism in the following manner as:

$$\begin{gathered} f\left(\left(g_1,h_1\right)+\left(g_2,h_2\right)\right)=f\left(g_1,h_1\right)+f\left(g_2,h_2\right) \\ f\left(\alpha \left(g,h\right)\right)=\alpha f\left(g,h\right) \end{gathered}$$

Find $f\left(\left(g_1,h_1\right)+\left(g_2,h_2\right)\right)$as:

$$\begin{gathered} f\left(\left(g_1,h_1\right)+\left(g_2,h_2\right)\right)=f\left(g_1+g_2,h_1+h_2\right) \\ =\left(g_1+g_2,h_1+h_2\right) \\ =\left(h_1,g_1\right)+\left(h_2,g_2\right) \\ =f\left(g_1,h_1\right)+f\left(g_2,h_2\right) \end{gathered}$$

Find $f\left(\alpha \left(g,h\right)\right)$as:

$$\begin{gathered} f\left(\alpha \left(g,h\right)\right)=f\left(\alpha g,\alpha h\right) \\ =\left(\alpha h,\alpha g\right) \\ =\alpha \left(h,g\right) \\ =\alpha f\left(h,g\right) \end{gathered}$$

Thus, $f:G\times H\to H\times G$mapping is homomophism.

So, $f:G\times H\to H\times G$mapping is one-one, onto and homomorphism.

Hence, it is proved that $G\times H\cong H\times G$.