Question

If ${\mathit{G}}_i\cong {\mathit{H}}_i$for $i=1,2,\dots ,n,$prove that

${\mathit{G}}_1\times \cdots \times {\mathit{G}}_n\cong {\mathit{H}}_1\times \cdots {\mathit{H}}_n$.

Short answer

It is proved that,$G_1\times \cdots \times G_n\cong H_1\times \cdots H_n$ .

Step-by-step solution

Isomorphic Groups

Two groupsGandHare said to be isomorphic if there exists an isomorphism ffromG ontoH .

Prove that G1×⋯×Gn≅H1×⋯Hn

To show isomorphism just show mapping is one-one, onto, and homomorphism.

Check f is one-one as:

Let $\left(g_1,g_2,\dots ,g_n\right)\in$ker f.

This implies that $f\left(g_1,g_2,\dots ,g_n\right)=0\mathrm{or}\left(f_1\left(g_1\right),f_2\left(g_2\right),\dots ,f_n\left(g_n\right)\right)=\left(0,0,\dots ,0\right)$.

This implies each $f_i\left(g_i\right)=0$for $i=1,2,\dots ,n$.

Since each map $f_i:G_i\to H_i$is an isomorphism, this implies each $g_i=0$for $i=1,2,\dots ,n,$which means $\left(g_1,g_2,\dots ,g_n\right)=0$.

This further implies ker $f=\left\{\left(0,0,\dots ,0\right)\right\}$.

Therefore, the map f is one-one.

Check f is onto as:

Let$\left(h_1,h_2,\dots ,h_n\right)\in H_1\times H_2\times \dots \times H_n$where$h_i\in H_i$for $i=1,2,\dots ,n$.

Since the map $f_i:G_i\to H_i$is an isomorphism, there exists$g_i\in G_i$such that $f_i\left(g_i\right)=h_i$.

Therefore, $\left(h_1,h_2,\dots ,h_n\right)=\left(f_1\left(g_1\right),f_2\left(g_2\right),\dots ,f_n\left(g_n\right)\right)$.

Thus, there exists$\left(g_1,g_2,\dots ,g_n\right)\in G_1\times G_2\times \dots \times G_n$such that $f\left(g_1,g_2,\dots ,g_n\right)=\left(f_1\left(g_1\right),f_2\left(g_2\right),\dots ,f_n\left(f_n\right)\right)$.

Hence, f is onto.

Check f is homomorphism as:

Show that$f\left(\alpha \left(a_1,a_2,\dots ,a_n\right)+\left(b_1,b_2,\dots ,b_n\right)\right)=\alpha f\left(a_1,a_2,\dots ,a_n\right)+f\left(b_1,b_2,\dots ,b_n\right)$ where$\left(a_1,a_2,\dots ,a_n\right),\left(b_1,b_2,\dots ,b_n\right)\in G_1\times G_2\times \dots \times G_n$ as follows:

$\begin{array}{rcl}f\left(\alpha \left(a_1,a_2,\dots ,a_n\right)+\left(b_1,b_2,\dots ,b_n\right)\right)& =& f\left(\left(\alpha a_1,\alpha a_2,\dots ,\alpha a_n\right)+\left(b_1,b_2,\dots ,{b_n}_{}\right)\right)\\ & =& f\left(\left(\alpha a_1+b_1,\alpha a_2+b_2,\dots ,\alpha a_n+b_n\right)\right)\\ & =& \left(f\left(\alpha a_1+b_1\right),f_2\left(\alpha a_2+b_2\right),\dots ,f_n\left(\alpha a_n+b_n\right)\right)\end{array}$

Since, the map $f_i:G_i\to H_i$is an isomorphism, further simplify as:

$\begin{array}{rcl}f\left(\alpha \left(a_1,a_2,\dots ,a_n\right)+\left(b_1,b_2,\dots ,b_n\right)\right)& =& \left(\alpha f_1\left(a_1\right)+f_1\left(b_1\right),\alpha f_2\left(a_2\right)+f_2\left(b_2\right),\dots ,\alpha f_n\left(a_n\right)+f_n\left(b_n\right)\right)\\ & =& \alpha \left(f_1\left(a_1\right),f_2\left(a_2\right),\dots ,f_n\left(a_n\right)+\left(f_1\left(b_1\right),f_2\left(b_2\right),\dots ,f_n\left(b_n\right)\right)\right)\\ & =& \alpha f\left(a_1,a_2,\dots ,a_n\right)+f\left(b_1,b_2,\dots ,b_n\right)\end{array}$

Thus, the map f is a homomorphism.

Therefore, the map $f:G_1\times G_2\times \dots \times G_n\to H_1\times H_2\times \dots \times H_n$is an isomorphism.

Hence, it is proved that$G_1\times \dots \times G_n\cong H_1\times \dots H_n$ .