Show that the group ordering in the direct product is doesn't matter to the group structure.

Question

Show that the group ordering in the direct product is doesn't matter to the group structure.

For instance, for $A,B,C \ne \emptyset$, $A×B×C$ and $A×C×B$ is same (That is, the group structure of $A×B×C$ is same with $A×C×B$).

Any idea?

Answer 1

First, $A,B,C$ are not only nonempty, they must be groups. This is because empty set can never be a group.
Second, in group theory, we always say that two groups have same structure if they are isomorphic.

Hint:
To show that $A\times B\times C$ is isomorphic to $A\times C\times B$, we need to construct an isomorphism between them. The obvious one should be $\phi:A\times B\times C \rightarrow A\times C\times B$ defined by $$(a,b,c)\mapsto (a,c,b)$$ where $(a,b,c)\in A\times C\times B.$

In general, let $A_1,\dots,A_n$ be groups and $\sigma\in S_n$, then $$A_1\times\dots\times A_n \cong A_{\sigma(1)}\times\dots\times A_{\sigma(n)}$$ and this can be proven in a similar way.

Answer 2

You could use group presentations.

If $X\cong\langle G_X\mid R_X\rangle$, then, by this result, we have, assuming without loss of generality the presentations are given by disjoint symbols, that

$$\begin{align} A\times B\times C&\cong\langle G_A\cup G_B\cup G_C\mid R_A\cup R_B\cup R_C\cup\{ xy=yx, xz=zx, yz=zy\mid x\in G_A, y\in G_B, z\in G_C\}\rangle\\ & \cong\langle G_A\cup G_C\cup G_B\mid R_A\cup R_C\cup R_B\cup\{ xy=yx, xz=zx, yz=zy\mid x\in G_A, z\in G_C, y\in G_B\}\rangle\\ &\cong A\times C\times B. \end{align}$$