# Does external direct product preserve isomorphic relationships?

Problem: I am currently learning group theory and am not sure if the following is true: let $A \cong A'$ and $B \cong B'$, is $A \times B \cong A' \times B'$ (here "$\times$" denote the external direct product)?

Attempt: Since $A \cong A'$ and $B \cong B'$, there exist isomorphisms $f_A: A \rightarrow A'$ and $f_B: B \rightarrow B'$. Let $f: A \times B \rightarrow A' \times B'$ be defined by $f(a,b) = (f_A(a),f_B(b))$ for every $a \in A, b \in B$. Then $f$ is an isomorphism.

I know that for finite group $A$ and arbitrary group $B, C$, we have $A \times B \cong A \times C \iff B \cong C$, but I cannot find anything as general as stated in this problem, so where did I go wrong in my proof?

• Why, your proof is correct. Dec 16, 2017 at 4:28
• To prove that you are correct you could show 1) $f$ is a bijection and 2) $f$ is a homomorphism by using the corresponding properties of $f_A \text { and } f_B$ Dec 16, 2017 at 5:30

You need to show that $f$ is a bijection and a homomorphism. Let $x = (a,b)$ and $y = (c, d)$ be members of $A\times B$. Then,
$$f(xy) = f((a, b)(c, d)) = f((ac, bd)) = (f_A(ac), f_B(bd)) = (f_A(a)f_A(c), f_B(b)f_B(d)) = (f_A(a), f_B(b))(f_A(c), f_B(d)) = f(x)f(y)$$
Hence, $f$ is a homomorphism.
Also,$f$ is a bjection (see proof here).
This proves that $f$ is an isomorphism from $A \times B$ to $A' \times B'$. Thus, $A\times B \cong A' \times B'$.