Suppose that we are given the groups $ A_1, A_2, B_1, B_2 $.
Qe are also given that the functions $ a: A_1 \to A_2$ and $b: B_1 \to B_2 $ are isomorphisms.
Now this new "c" which is what I'm trying to prove is an isomorphism let's define as $ c: A_1 \times B_1 \to A_2 \times B_2 $ by setting:
$$\forall x \in A_1, y \in B_2: c(x,y) = (a(x) , b(y)) $$
- So I see that $c$ is built from the two isomorphisms $a$ and $b$ which leads me to believe that $c$ itself is also an isomorphism, but proving this via a formal proof has been more difficult that expected.
My idea: this is how i've been trying to show that $c$ respects the operation of composition (but let me know the flaws)
$ c(x_1x_2, y_1y_2) = (a(x_1x_2) , b(y_1y_2)) $ – definition of c
$ = (x_1x_2 , y_1y_2) $ – now these $x_1x_2$ and $y_1y_2$ are in $A_2$ and $B_2$
$ = (x_1, y_1) (x_2, y_2) $ – Cartesian products
$ = (a(x_1), b(y_1))(a(x_2), b(y_2)) $ – rewriting $x_1$ in $A_2$ as "$a(x_1)$"
$ = c(x_1, y_1) c(x_2, y_2) $ – definition of $c$
- but then when it comes to showing that it respects inverses I have been having some problems… I know if it's bijective that it's also invertible, and since it is built up of isomorphic groups that are themselves invertible that this should be invertible as well, but showing that with a proof is where I am getting tripped up..
any and all input is appreciated for I'm doing my best to fully understand these concepts but it has proved more difficult than expected
Best Answer
If $a\colon A_1\to A_2$ and $b\colon B_1\to B_2$ are isomorphisms, then $c=a\times b\colon A_1\times B_1\to A_2\times B_2$ is also an isomorphism.
You already showed it’s a morphism (the argument is essentially correct, but you should start from $c\bigl( (x_1,y_1)(x_2,y_2)\bigr)$, not from $c(x_1x_2,y_1y_2)$).
(Also, you should use notation that is suggestive! If your domain is called $A_1\times B_1$, then why not refer to the elements of $A_1$ by lower case $a$-s and those of $B_1$ by lower case $b$-s?)
Anyhow... two different ways of showing it is an isomorphism:
It has an inverse: we know $a$ has an inverse $a^{-1}\colon A_2\to A_1$ and that $b$ has an inverse $b^{-1}\colon B_2\to B_1$. Show that $d=a^{-1}\times b^{-1}\colon A_2\times B_2\to A_1\times B_1$ is the inverse of $c$.
Show $c$ is bijective. To show that $c$ is one-to-one, consider the kernel: if $c(x,y) = (e,e)$, show that then $a(x)=e$ and $b(y)=e$, and use the fact that $a$ and $b$ are both one-to-one to conclude $c$ is one-to-one. And to show $c$ is surjective, take $(z,w)\in A_2\times B_2$; use the fact that $a$ and $b$ are both surjective to show that you can find an element in $A_1\times B_1$ whose image under $c$ is $(z,w)$.
Pick your favorite method.