# Isomorphism Problem

1. Oct 24, 2013

### 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?

2. Oct 24, 2013

### Office_Shredder

Staff Emeritus
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'.

3. Oct 24, 2013

### 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?