Proof:

We shall merely provide a sketch of the argument. Let $G_1$ and $G_2$ be two given groups. A word is by definition a finite sequence $(x_1, x_2, \dots x_n)$ such that each $x_i$ ( $i = 1, 2\dots, n$) belongs to one of the groups, no pair of adjacent terms of the sequence belong to the same group and none of the $x_i$ is the identity element of either of the groups. We call the integer $n$ the length of the word and also include the empty word of length zero. Denoting by $W$ is the set of all words, the idea is to define a binary operation of juxtaposition of words. The empty word would serve as the identity and the inverse of a word $(x_1, x_2,\dots, x_n)$ would be the word $(x_n^{-1}, x_{n-1}^{-1}, \dots, x_1^{-1})$. One would hope that the operation of juxtaposition would make $W$ a group. This however would not quite suffice. The juxtaposition of two words $(x_1, x_2,\dots, x_n)$ and $(y_1, y_2, \dots, y_m)$ may result in a sequence that does not qualify to be called a word for the simple reason that $x_n$ and $y_1$ may belong to the same group. When this happens we may try to replace the juxtaposed string by the smaller string

$$(x_1, x_2, \dots, x_{n-1}, z, y_2, \dots, y_m)$$

where $z = x_ny_1$. If $z$ is not the unit element we do get a legitimate word but if $z$ is the unit element of one of the groups we must drop it altogether obtaining instead the still smaller string

$$(x_1, x_2, \dots, x_{n-1}, y_2, \dots, y_m)$$

If $x_{n-1}$ and $y_2$ belong to the same group the above process must continue and thus in finitely many steps we obtain a legitimate word that ought to be the product of the two given words. To check that we do get a group that qualifies as the coproduct of the given groups can be tedious. The reader may consult [11], pp 72-73.

We now introduce the notion of a direct sum of abelian groups which will play a crucial role in the second part of the course.