After learning how to verify a group, we now apply the method to the most familiar class of examples: additive groups of number systems. These examples are important because they provide the model for many later algebraic structures. In an additive group, the identity element is usually $0$, and the inverse of an element $a$ is usually $-a$. The main work is to check that the set is closed under addition and that additive inverses remain inside the set. In this lesson, students will verify additive groups such as $(\mathbb{Z},+)$, $(m\mathbb{Z},+)$, $(\mathbb{Q},+)$, $(\mathbb{R},+)$, and $(\mathbb{C},+)$. :::definition[Additive Group] Let $G$ be a non-empty set with addition $+$. Then $(G,+)$ is called an $additive group$ if $(G,+)$ is a group. In additive notation, the identity element is denoted by $0$, and the inverse of $a\in G$ is denoted by $-a$. ::: The notation changes slightly in additive groups. Instead of writing the identity as $e$, we usually write it as $0$. Instead of writing the inverse of $a$ as $a^{-1}$, we write it as $-a$. This is only a change of notation, not a change of meaning. The inverse condition becomes $$\begin{equation*} a+(-a)=(-a)+a=0. \end{equation*}$$ Students should not write $a^{-1}$ for an additive inverse unless the context has explicitly chosen multiplicative notation. For additive groups, the identity is $0$ and the inverse of $a$ is $-a$. Choose a multiple set $m\mathbb{Z}$ and an element $a$. The preview checks whether $a$ belongs to $m\mathbb{Z}$, displays its additive inverse, and verifies that the inverse remains in the same set. This is the key membership step in additive group proofs. :::scientific-preview[Additive Group Pattern Explorer] ::: :::theorem Let $(\mathbb{Z},+)$ be the set of integers under ordinary addition. Then $(\mathbb{Z},+)$ is a group. ::: :::proof Given that $\mathbb{Z}$ is the set of integers and $+$ is ordinary addition. To prove that $(\mathbb{Z},+)$ is a group. [1] To prove closure. Let $a,b\in\mathbb{Z}$. Then $$\begin{equation*} a+b\in\mathbb{Z}. \end{equation*}$$ Therefore $\mathbb{Z}$ is closed under addition. [2] To prove associativity. Let $a,b,c\in\mathbb{Z}$. Then $$\begin{equation*} (a+b)+c=a+(b+c). \end{equation*}$$ Therefore addition is associative on $\mathbb{Z}$. [3] To prove the existence of identity element. There exists $0\in\mathbb{Z}$ such that $$\begin{equation*} a+0=0+a=a \quad \forall\, a\in\mathbb{Z}. \end{equation*}$$ Therefore $0$ is the identity element of $(\mathbb{Z},+)$. [4] To prove the existence of inverse elements. Let $a\in\mathbb{Z}$. Then $-a\in\mathbb{Z}$ and $$\begin{equation*} a+(-a)=(-a)+a=0. \end{equation*}$$ Therefore every element of $\mathbb{Z}$ has an additive inverse in $\mathbb{Z}$. Hence, $(\mathbb{Z},+)$ is a group. $\square$ ::: :::theorem Let $m\in\mathbb{Z}$. If $m\mathbb{Z}=\{mk:k\in\mathbb{Z}\}$, then $(m\mathbb{Z},+)$ is a group under ordinary addition. ::: :::proof Given that $m\in\mathbb{Z}$ and $m\mathbb{Z}=\{mk:k\in\mathbb{Z}\}$. To prove that $(m\mathbb{Z},+)$ is a group. [1] To prove closure. Let $a,b\in m\mathbb{Z}$. Then there exist $r,s\in\mathbb{Z}$ such that $$\begin{equation*} a=mr,\quad b=ms. \end{equation*}$$ Now $$\begin{align*} a+b&=mr+ms\\ &=m(r+s). \end{align*}$$ Since $r+s\in\mathbb{Z}$, we get $a+b\in m\mathbb{Z}$. [2] To prove associativity. Let $a,b,c\in m\mathbb{Z}$. Since addition of integers is associative, $$\begin{equation*} (a+b)+c=a+(b+c). \end{equation*}$$ [3] To prove the existence of identity element. Since $0=m\cdot0$, we get $0\in m\mathbb{Z}$. Also $$\begin{equation*} a+0=0+a=a \quad \forall\, a\in m\mathbb{Z}. \end{equation*}$$ Therefore $0$ is the identity element. [4] To prove the existence of inverse elements. Let $a\in m\mathbb{Z}$. Then there exists $r\in\mathbb{Z}$ such that $a=mr$. Now $$\begin{align*} -a&=-(mr)\\ &=m(-r). \end{align*}$$ Since $-r\in\mathbb{Z}$, we get $-a\in m\mathbb{Z}$. Also $$\begin{equation*} a+(-a)=(-a)+a=0. \end{equation*}$$ Therefore every element of $m\mathbb{Z}$ has an additive inverse in $m\mathbb{Z}$. Hence, $(m\mathbb{Z},+)$ is a group. $\square$ ::: The proof for $m\mathbb{Z}$ is a useful model because it shows how to use the internal form of the elements. If $a\in m\mathbb{Z}$, then $a$ must be written as $mr$ for some integer $r$. This representation is the key to proving closure and inverse membership. For example, the inverse of $mr$ under addition is $-mr=m(-r)$, and this again belongs to $m\mathbb{Z}$. This type of argument will appear repeatedly when proving that subsets form groups. :::example The following are additive groups: [1] $(\mathbb{Q},+)$ is a group because the sum and negative of rational numbers are rational, and $0\in\mathbb{Q}$. [2] $(\mathbb{R},+)$ is a group because the sum and negative of real numbers are real, and $0\in\mathbb{R}$. [3] $(\mathbb{C},+)$ is a group because the sum and negative of complex numbers are complex, and $0\in\mathbb{C}$. Hence, these standard number systems are groups under addition. ::: :::solved-problem Prove that $(3\mathbb{Z},+)$ is a group. ::: :::solution Let $3\mathbb{Z}=\{3k:k\in\mathbb{Z}\}$. To prove that $(3\mathbb{Z},+)$ is a group. [1] To prove closure. Let $a,b\in3\mathbb{Z}$. Then there exist $r,s\in\mathbb{Z}$ such that $$\begin{equation*} a=3r,\quad b=3s. \end{equation*}$$ Now $$\begin{align*} a+b&=3r+3s\\ &=3(r+s). \end{align*}$$ Since $r+s\in\mathbb{Z}$, we get $a+b\in3\mathbb{Z}$. [2] To prove associativity. Let $a,b,c\in3\mathbb{Z}$. Since integer addition is associative, $$\begin{equation*} (a+b)+c=a+(b+c). \end{equation*}$$ [3] To prove identity element. Since $0=3\cdot0$, we get $0\in3\mathbb{Z}$. Also $$\begin{equation*} a+0=0+a=a \quad \forall\, a\in3\mathbb{Z}. \end{equation*}$$ [4] To prove inverse elements. Let $a\in3\mathbb{Z}$. Then $a=3r$ for some $r\in\mathbb{Z}$. Now $$\begin{align*} -a&=-3r\\ &=3(-r). \end{align*}$$ Since $-r\in\mathbb{Z}$, we get $-a\in3\mathbb{Z}$. Also $$\begin{equation*} a+(-a)=(-a)+a=0. \end{equation*}$$ Hence, $(3\mathbb{Z},+)$ is a group. $\square$ ::: :::remark Not every subset of a number system forms an additive group. For example, $\mathbb{N}$ under addition is not a group if $\mathbb{N}=\{1,2,3,\ldots\}$ because it does not contain the additive identity $0$. Also, $\mathbb{R}-\{0\}$ under addition is not a group because $1+(-1)=0$ and $0\notin\mathbb{R}-\{0\}$. Thus additive group examples require careful membership checks. ::: Use the calculator below to test additive inverse membership for sets of the form $m\mathbb{Z}$. Enter an integer $m$ and an element $a$. If $a$ belongs to $m\mathbb{Z}$, the calculator checks whether $-a$ also belongs to $m\mathbb{Z}$. This illustrates the inverse step in the proof of additive groups. :::calculator[Additive Inverse Membership Calculator] ::: :::exercise [1] Prove that $(4\mathbb{Z},+)$ is a group. [2] Determine whether $(\mathbb{N}\cup\{0\},+)$ is a group. [3] Determine whether $(\mathbb{Q},+)$ is a group. [4] Find the additive inverse of $\dfrac{7}{5}$ in $(\mathbb{Q},+)$. [5] Determine whether $(\mathbb{R}-\{0\},+)$ is a group. ::: :::answer [1] If $a=4r$ and $b=4s$, then $a+b=4(r+s)\in4\mathbb{Z}$. The identity is $0$, and the inverse of $4r$ is $4(-r)$. [2] No. The element $1$ has no additive inverse in $\mathbb{N}\cup\{0\}$. [3] Yes. Rational numbers are closed under addition, addition is associative, $0$ is the identity, and $-a$ is rational for every rational $a$. [4] The additive inverse is $-\dfrac{7}{5}$. [5] No. Closure fails because $1+(-1)=0$ and $0\notin\mathbb{R}-\{0\}$. ::: :::faq Q: What is the identity element in an additive group? A: The identity element is usually denoted by $0$. Q: What is the inverse of $a$ in an additive group? A: The inverse of $a$ is denoted by $-a$. Q: Is every subset of $\mathbb{Z}$ an additive group? A: No. The subset must contain $0$, be closed under addition, and contain additive inverses. Q: Why is $m\mathbb{Z}$ closed under addition? A: If $a=mr$ and $b=ms$, then $a+b=m(r+s)$, and $r+s\in\mathbb{Z}$. ::: :::call-to-action[Multiplicative Groups] subtitle: Continue to multiplicative groups of nonzero numbers, where identity and inverses are written in multiplicative notation. button: Next Lesson | /abstract-algebra/introduction-to-groups/definition-and-examples-of-groups/multiplicative-groups-nonzero-numbers button-ghost: Previous Lesson | /abstract-algebra/introduction-to-groups/definition-and-examples-of-groups/verify-group-axioms :::