Understanding an example of the product of two finite cyclic groups in MacLane/Birkhoff's Algebra I am working through MacLane/Birkhoff's abstract algebra textbook "Algebra" (3rd edition), but I am having trouble understanding one of their examples.
On p.46, they present their definition of the product of two groups:

If $G$ and $G'$ are given (multiplicative) groups, their product $G\times G'$ is the product set consisting of all pairs $(a,a')$ for $a\in G$, $a'\in G'$ with the binary operation
  $$(a,a')(b,b')=(ab,a'b')$$
  of "termwise multiplication" of these pairs.

However, on p.61 in a section discussing the defining relations for the generators of certain types of groups (e.g. cyclic groups, dihedral group $D_{2n}$, etc.) they give the following example:

Consider the product $E=B\times C$ of two finite cyclic groups $B$ and $C$ with generators $b$ and $c$ of orders $m$ and $n$, respectively. Every element in $E$ has the form of a product $b^{i}c^{j}$ with integral exponents $i$ and $j$. Since $b^{m}=1$ and $c^{n}=1$, one can reduce $i$ modulo $m$ and $j$ modulo $n$, and $E$ has just $mn$ elements. Thus $E$ is generated by two elements $b$ and $c$ and its multiplication table may be computed directly from the following (defining) relations:
  $$b^{m}=1,\,\,\,\,\,\, c^{n}=1,\,\,\,\,\,\, bc=cb.$$
  Indeed,from these relations one may calculate any product of two elements of $E$ as $(b^{i}c^{j})(b^{k}c^{l})=b^{i+k}c^{j+l}$ where the exponents are to be reduced modulo $m$ or modulo $n$, as the case may be. Thus the "defining relations" on the generators $b$ and $c$ of $E$ describe this group up to isomorphism. 

Their two uses of the term "product of two groups" doesn't seem to match up to me. In their second use of the term, the elements of the group $E$ are not ordered pairs, they are the pairwise products of the elements of the groups $B$ and $C$. This, to me, seems more akin to the definition of the direct product of subgroups of a group, but they have not mentioned that $B$ and $C$ are subgroups of another group.
Are they using the term "product of two groups" loosely and the group in the example is really something other than the product of the two groups $B$ and $C$? And if so, what exactly is this group?
I also do not see why, in their "defining relations" every element $b$ of $B$ must commute with every element $c$ of $C$. But I think this may be due to the fact that I am not completely sure how the group was constructed. (Maybe it has to do with the fact that every cyclic group is abelian?)
Thanks in advance.
 A: If $A$ and $B$ are groups then $A$ and $B$ may be interpreted as subgroups of $A\times B$, where $a\in A$ is identified with $(a,e)\in A\times B$, and similarly $b\in B$ identified with $(e,b)\in A\times B$. Therefore it makes sense to speak of the elements $a,b\in A\times B$. Moreover, it is easily checked that elements of these forms commute: $(a,e)(e,b)=(a,b)=(e,b)(a,e)$ (i.e. $ab=ba$).
Often one thinks of groups according to their isomorphism type instead of how they are actually constructed in this or that definition. If your only understanding of direct products is "the elements look like tuples," you will be unable to recognize a direct product just by its structure $-$ this is where the notion of internal direct product comes in. It is easily checked that $A\times B$ is an internal direct product of the subgroups $A\times\{e\}$ and $\{e\}\times B$, and conversely any group which has subgroups (isomorphic to) $A$ and $B$ which is an internal direct product of them, is isomorphic to $A\times B$.
