Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
4  
Yes product refers to the law of the group, even when it's the usual addition. –  Davide Giraudo May 31 '12 at 16:46
1  
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
1  
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

2 Answers 2

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 $$G=\{x_1^{\varepsilon_1}\star\ldots\star x_n^{\varepsilon_n},n\in\Bbb N,x_i\in S,\varepsilon_i\in\{-1,1\}\}.$$

share|improve this answer

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: $$ -4 = (-1) + (-1) + (-1) + (-1). $$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.