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In mathematics, a group $G$ is called free if there is a subset $S$ of $G$ such that any element of $G$ can be written in one and only one way as a product of finitely many elements of $S$ and their inverses.

The group $(\mathbb{Z},+)$ of integers is free; we can take $S = \{1\}$. ("Free group", Wikipedia)

By saying "product of finitely many elements of $S$ and their inverses", in case of $(\mathbb{Z},+)$, does product refer to $+$?

If not, can anyone correct misunderstanding?

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Yes product refers to the law of the group, even when it's the usual addition. – Davide Giraudo May 31 '12 at 16:46
Yes; the product just refers to the group operation. For abelian groups, $+$ is often used instead of $\cdot$ or mere juxtaposition. – Dylan Moreland May 31 '12 at 16:46
Note: When we talk about a "free group on X", then $X$ is a set whose elements correspond to the $S$ of the definition. So a "free group on $\mathbb{Z}$" would be a free group of countable rank, rather than the free group $\mathbb{Z}$. – Arturo Magidin May 31 '12 at 19:31

The term "product" refers to the law of the group, even when it's the usual addition. If we denote $\star$ the law, $G$ is free with generator $S$ if $$G=\{x_1^{\varepsilon_1}\star\ldots\star x_n^{\varepsilon_n},n\in\Bbb N,x_i\in S,\varepsilon_i\in\{-1,1\}\}.$$

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In a group we are considering only one operation. So when we talk about $(\mathbb{Z}, +)$ then the operation is addition. If we talked about for example $(\mathbb{R}^{\times}, \cdot)$, then the operation is multiplication. When starting out learning about groups this can often confuse, so it is sometimes recommended that when you come across a group, you also stop and think what the operation is.

With $(\mathbb{Z}, +)$ we have the operation being addition $+$. And we can take $S = \{1\}$, because given any integer $n$, we can write $n$ as a sum of $1$ in a unique way. For for example $$ 7 = 1 + 1 + 1 + 1 + 1 + 1 + 1. $$ The inverse of $1$ in $(\mathbb{Z},+)$ is $-1$, so it also works for negative integers. For example: $$ -4 = (-1) + (-1) + (-1) + (-1). $$

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