Show H × G₁ ≅ H × G₂ when G₁ ≅ G₂

[Parmenides]

Im asked to show that, given the groups $H, G_1$, and $G_2$ in which $G_1 \cong G_2$, that $$H\times{G_1} \cong H\times{G_2}$$


Because of the isomorphism between $G_1$ and $G_2$, their cardinalities (order) are equal, which i think will be of good use when considering their Cartesian product with $H$. So conceptually, it seems intuitive to believe that, almost vacuously, $$|H\times{G_1}| = |H\times{G_2}|$$ But I'm not sure how to explicitly show this. Since $G_1 \cong{G_2}$, there exists an isomorphism between the two such that: $$f: G_1 \rightarrow G_2$$ and that it is necessary to find an $F$ such that $$F: H\times{G_1} \rightarrow H \times G_2$$ Also, $f(a) = b \forall a \in G_1, b \in G_2$. I think these are some pieces to the puzzle, but how to stitch them together?

 

[Office_Shredder]

It is true that |HxG₁| = |HxG₂| and you should be able to prove it very easily, but it's not of much help in showing the groups are isomorphic. Remember, an element of HxG₁ looks like (h,g) with h in H and g in G₁. So F(h,g) = (h',g') where h' is in H and g' is in G₂ is going to be the function you're interested in... can you think of what F should look like? There aren't too many choices about how you can pick h' and g'.

 

[Parmenides]

Do you mean that since every element of $H \times G_1$ is unique and that every element of $H \times G_2$ can be defined as $$(c, f(a)) = (c, b), \forall c \in H, a \in G_1, b \in G_2$$ such that it is also unique, we have $$F: (c, a) \rightarrow (c, f(a)) = (c, b)$$ And thus, there is a one-to-one correspondence?