If $G_1\cong G_2$ and $H_1\cong H_2$ then $G_1 \times H_1 \cong G_2 \times H_2$


$f_G:G_1\rightarrow G_2$ and $f_H:H_1\rightarrow H_2$.

Question 1: Is the following statement valid? Does it belong in this proof?

$$f_G:G_1\times A \rightarrow G_2 \times A\textrm{ for any group }A$$

Question 2: Can I finish the proof like this? Do I need to show any steps in between?

$$\left[\ f_H \circ f_G\right] :G_1\times H_1 \rightarrow G_2 \times H_2$$

$f_G$ and $f_H$ are both bijective. Thus $f_H\circ f_G$ is bijective. Therefore, $G_1 \times H_1 \cong G_2 \times H_2$.

Does this work? I was considering breaking it down further and manipulating individual elements in the groups $G_1,G_2,H_1,H_2.$ I would appreciate your advice. Thanks

  • $\begingroup$ Where does your $f_G$ take an arbitrary element $(g,h)$? $\endgroup$
    – Mathmo123
    Sep 29, 2014 at 19:51
  • $\begingroup$ I suppose $f_G\left[(g_1,h)\right]=(g_2,h)$ where $g_1\in G_1, g_2\in G_2$. Edit: Although maybe I did this improperly--can $f_G$ take a two-element coordinate and simply not change the 2nd element of the coordinate? This assumption is what I based my entire proof on. $\endgroup$ Sep 29, 2014 at 19:54
  • $\begingroup$ @Mathmo123 forgot to tag your username in that last post. $\endgroup$ Sep 29, 2014 at 20:00
  • $\begingroup$ yes it can and in which case your proof works (you still need to verify that that $f_G$ acts as a homomorphism in the way you've constructed it, not just a bijection, but that isn't tricky). The easier way to write it is to let $f$ be the function that takes $(g,h) \mapsto (f_G(g), f_H(h))$ $\endgroup$
    – Mathmo123
    Sep 29, 2014 at 20:08
  • 1
    $\begingroup$ Yes exactly - and then you can use that $f_G$ and $f_H$ are homomorphisms $\endgroup$
    – Mathmo123
    Sep 29, 2014 at 20:23

1 Answer 1


For the sake of completeness. Take $G_1 \cong G_2,H_1 \cong H_2$ with isomorphisms $f_1,f_2$ respectively. Then take $g((x,y)) = (f_1(x),f_2(y))$.


$g((x_1,y_1)) = g((x_2,y_2)) \implies (f_1(x_1),f_2(y_1)) = (f_1(x_2),f_2(y_2))$

$\implies f_1(x_1) = f_1(x_2) \land f_2(y_1) = f_2(y_2) \implies x_1 = x_2 \land y_1 = y_2$


Take $(x,y) \in G_2 \times H_2$, by surjectivity of $f_1$ there is an $x_1$ such that $f_1(x_1) = x$ and by surjectivity of $f_2$ there is an $x_2$ such that $f_2(x_2) = y$. Thus, $g(x_1,x_2) = (x,y)$


$g((x_1,y_1)(x_2,y_2)) = g(x_1x_2,y_1y_2) = (f_1(x_1x_2),f_2(y_1y_2)) = (f_1(x_1),f_2(y_1))(f_1(x_2),f_2(y_2))$ since $f_1,f_2$ are homomorphisms.


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