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