Show Isomorphism btwn H x G_1 & H x G_2

[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?