# If $G_1 \rightarrow G_2$ and $H_1 \rightarrow H_2$, then $G_1\times H_1 \rightarrow G_2\times H_2$

Dr. Pinter's "A Book of Abstract Algebra" presents the exercise:

If $G_1 \rightarrow G_2$ and $H_1 \rightarrow H_2$, then $G_1\times H_1 \rightarrow G_2\times H_2$.

where $\rightarrow$ means "is isomorphic to."

Taking that $f: G_1 \rightarrow G_2$ and $g: H_1 \rightarrow H_2$, my solution is

$h: G_1\times H_1 \rightarrow G_2\times H_2 = (x, y) = (f(x), g(y))$

Do I need to prove that $h$ is an isomorphism? Or can I rely on the isomorphisms of $f$ and $g$?

• $G \times H$ means $GH$ ?? Jun 4, 2015 at 16:15
• Since $f$ and $g$ are isomorphisms, there are $f^{-1}$ and $g^{-1}$. Using $f^{-1}$ and $g^{-1}$, construct $h'\colon G_2\times H_2\to G_1\times G_1$. Easy to check $h\circ h'=\operatorname{id}$ and $h'\circ h=\operatorname{id}$. Jun 4, 2015 at 16:15

You've correctly identified the most common way to establish a bijection between $G_1 \times H_1$ and $G_2 \times H_2$. Now you just have to go through the work of proving that $h$ is indeed a bijection. Luckily, showing that $h$ is one-to-one and onto are very short proofs relying on $f$ and $g$ being bijections.
As for the "morphism" part, I'll give names to the operations. Suppose $(G_1, *_1)$, $(G_2, *_2)$, $(H_1, +_1)$, and $(H_2, +_2)$ are the operational systems in question.
Let $\cdot_1$ be defined for all $(x,y),(z,w) \in G_1 \times H_1$ by $(x,y) \cdot_1 (z,w) = (x *_1 z, y +_1 w)$, and similarly for $\cdot_2$ with elements in $G_2 \times H_2$. (That is, $\cdot_1$ and $\cdot_2$ are the traditional componentwise operations.)
Let $(x,y),(z,w) \in G_1 \times H_1$. Then
\begin{align} h((x,y) \cdot_1 (z,w)) &= h((x *_1 z, y +_1 w)) \\ &= (f(x *_1 z), g(y +_1 w)) \\ &= (f(x) *_2 f(z), g(y) +_2 g(w)) \\ &= (f(x),g(y)) \cdot_2 (f(z), g(w)) \\ &= h((x,y)) \cdot_2 h((z,w)) \end{align}
and thus $h$ is a homomorphism. Together with it being a bijection, you get that $h$ is an isomoprhism.