External Direct Product isomorphism

• Sep 30th 2007, 02:12 PM
External Direct Product isomorphism
Suppose that $\displaystyle G_{1} \approx G_{2}, H_{1} \approx H_{2}$, prove that $\displaystyle G_{1} \oplus H_{1} \approx G_{2} \oplus H_{2}$

For this problem, I really don't have much clue, I feel I'm not too comfortable with the whole direct product thingy.

Thanks.
• Sep 30th 2007, 02:42 PM
Plato
• Sep 30th 2007, 06:25 PM
No, I do not yet know how to solve that other problem as well. But I was hoping that by understanding how to solve this one, I might be able to solve the other one.
• Oct 1st 2007, 11:10 AM
ThePerfectHacker
Quote:

Suppose that $\displaystyle G_{1} \approx G_{2}, H_{1} \approx H_{2}$, prove that $\displaystyle G_{1} \oplus H_{1} \approx G_{2} \oplus H_{2}$
There exists a bijective homomorphism $\displaystyle \phi: G_1\mapsto G_2$ and $\displaystyle \psi: H_1\mapsto H_2$ by definition.
Now create $\displaystyle G_1\times H_1 = \{(g_1,h_1)|g_1\in G_1 \mbox{ and }h_1\in H_1$ and similarly $\displaystyle G_2\times H_2$.
For $\displaystyle x\in G_1\times H_1$, i.e. $\displaystyle x = (g_1,h_1)$ for some $\displaystyle g_1\in G_1$ and $\displaystyle h_1\in H_1$. Define $\displaystyle \tau (x) = (\phi (g_1),\psi (h_1))$.
Show that $\displaystyle \tau$ is a mapping from $\displaystyle G_1\times H_1$ into $\displaystyle G_2\times H_2$. Now show it is one-to-one and onto. Finally conclude by showing it is a homomorphism between these groups. It is a straightforward computation.