# Does an isomorphism of groups that can be written as a direct product induce isomorphisms on the factors?

An answer to question Isomorphism of Direct Product of Groups says if you have two (or more) group isomorphisms $\phi_1:A_1 \rightarrow X_1$ and $\phi_2:A_2 \rightarrow X_2$ then it follows that $A_1 \times A_2 \cong X_1 \times X_2$ under the isomorphism $\phi(a_1,a_2)=(\phi_1(a_1),\phi_2 (a_2) )$

I am interested in whether the converse of this statement is true.

If $\phi: A_1 \times A_2 \rightarrow X_1 \times X_2$ is an isomorphism, is it true that $A_1 \cong X_1$ under an isomorphism $\phi_1$ and $A_2 \cong X_2$ under an isomorphism $\phi_2$ such that $\phi(a_1,a_2)= (\phi_1 (a_1), \phi (a_2))$?

• Do you have any reason to believe it is true? The first one is true because you define the groups up from isomorphic groups (in the language of groups without set theory they are the same damn object). – Jacob Wakem Aug 11 '16 at 6:39

## 4 Answers

No, and this is always false for any group that can be written as a direct product in a non-trivial way. For example, if $G = A \times B$ with neither $A$ nor $B$ the trivial group, then

$$A \times B = G \equiv G \times \{e\}$$

• I can't believe how obvious this was. Got stuck in the trap of trying to prove something that was false... – Bernard W Dec 21 '15 at 7:42
• @BernardWojcik As a way to help build intuition, try asking the analogous question for numbers: If $xy = ab$, is $x = a$ or $x = b$? Obviously not. The analogy between integer factorization and group factorization (at least for finite abeliean groups) is strong and useful. – user296602 Dec 21 '15 at 7:44

Yet another example: let $A\not\cong B$. Then $A\times B\cong B\times A$ . . .

I would add to the answers already provided that even if $A_1\cong X_1$ and $A_2\cong X_2$, there may not exist isomorphisms $\phi_1:A_1\to X_1$ and $\phi_2:A_2\to X_2$ such that $\phi(a_1,a_2)=(\phi_1(a_1),\phi_2(a_2))$. For instance, let $A_1=A_2=X_1=X_2=\mathbb{Z}$ and consider $\phi:\mathbb{Z}\times\mathbb{Z}\to\mathbb{Z}\times\mathbb{Z}$ given by $\phi(a,b)=(a,b+a)$. Then $\phi$ is an isomorphism (its inverse is given by $(a,b)\mapsto(a,b-a)$), but it cannot come from a pair of isomorphisms $\phi_1$ and $\phi_2$ because the second coordinate of $\phi$ depends on both coordinates of the input.

This is not true: Let $A_1 = \mathbb{Z}$, and consider $A_2 = \mathbb{Z} \times \mathbb{Z} \times \mathbb{Z} \times \cdots$. Also let $X_1$ be the trivial group, and $X_2 = A_2$. Then $$A_1 \times A_2 \cong A_2 \cong X_1 \times X_2$$ and $A_2 \cong X_2$, but $A_1$ is not isomorphic to $X_1$.