# 5.2: Abstract Algebra - Commutative Groups

- Page ID
- 62292

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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Every schoolchild learns about addition (\(+\)), subtraction (\(−\)), and multiplication (\(\times\)). Each of these is a “binary operation” on the set of real numbers, which means that it takes two numbers, and gives back some other number. In this section, we discuss binary operations on an arbitrary set; that is, we consider various ways of taking two elements of the set and giving back some other element of the set. (The official definition of the term “binary operation” is in Example \(6.3.6\), but it suffices to have an informal understanding for the present purposes.)

Let \(A\) be a set. We say that \(+\) is a **binary operation** on \(A\) if, for every \(a, b \in A\), we have a corresponding element \(a + b\) of \(A\).

(The element \(a + b\) must exist for *all* \(a, b \in A\). Furthermore, the sum \(a + b\) must depend only on the values of \(a\) and \(b\), not on any other information.)

- Addition (\(+\)), subtraction (\(−\)), and multiplication (\(\times\)) are examples of binary operations on \(\mathbb{R}\). They also provide binary operations on \(\mathbb{Q}\) and \(\mathbb{Z}\). However, subtraction (\(−\)) does not provide a binary operation on \(\mathbb{N}\), because \(x − y\) is not in \(\mathbb{N}\) when \(x < y\) (whereas the values of a binary operation on a set must all belong to the given set).
- Division (\(\div\)) is
*not*a binary operation on \(\mathbb{R}\). This is because \(x \div y\) does not exist when \(y = 0\) (whereas a binary operation on a set needs to be defined for*all*pairs of elements of the set). (On the other hand, division is a binary operation on the set \(\mathbb{R} \backslash\{0\}\) of all nonzero real numbers.) - Union (\(\cup\)), intersection (\(\cap\)), and set difference (\(\backslash\)) are binary operations on the collection of all sets.
- If a set does not have too many elements, then a binary operation on it can be specified by providing an “addition table.” For example, the following table defines a binary operation \(+\) on \(\{\mathrm{a}, \mathrm{b}, \mathrm{c}, \mathrm{d}, \mathrm{e}, \mathrm{f}\}\): \[\begin{array}{c|cccccc}

+ & \mathrm{a} & \mathrm{b} & \mathrm{c} & \mathrm{d} & \mathrm{e} & \mathrm{f} \\

\hline \mathrm{a} & \mathrm{a} & \mathrm{b} & \mathrm{c} & \mathrm{d} & \mathrm{e} & \mathrm{f} \\

\mathrm{b} & \mathrm{b} & \mathrm{c} & \mathrm{a} & \mathrm{e} & \mathrm{f} & \mathrm{d} \\

\mathrm{c} & \mathrm{c} & \mathrm{a} & \mathrm{b} & \mathrm{f} & \mathrm{d} & \mathrm{e} \\

\mathrm{d} & \mathrm{d} & \mathrm{e} & \mathrm{f} & \mathrm{a} & \mathrm{b} & \mathrm{c} \\

\mathrm{e} & \mathrm{e} & \mathrm{f} & \mathrm{d} & \mathrm{b} & \mathrm{c} & \mathrm{a} \\

\mathrm{f} & \mathrm{f} & \mathrm{d} & \mathrm{e} & \mathrm{c} & \mathrm{a} & \mathrm{b}

\end{array}\]

To calculate \(x + y\), find the row that has \(x\) at its left end, and find the column that has \(y\) at the top. The value \(x + y\) is the entry of the table that is in that row and that column. For example, \(f\) is at the left of the bottom row and \(e\) is at the top of the second-to-last column, so \(f + e = a\), because a is the second-to-last entry of the bottom row.

Let \(+\) be a binary operation on a set \(G\).

- \(+\) is
**commutative**iff \(g + h = h + g\) for all \(g, h \in G\). - \(+\) is
**associative**iff \(g + (h + k) = (g + h) + k\) for all \(g, h, k \in G\). - An element 0 of \(G\) is an
**identity element**iff \(g + 0 = g\), for all \(g \in G\). - For \(g \in G\), a
**negative**of \(g\) is an element \(−g\) of \(G\), such that \(g + (−g) = 0\), where \(0\) is an identity element of \(G\). - We say (\(G, +\)) is a
**commutative group**iff all three of the following conditions (or “axioms”) are satisfied:- \(+\) is commutative and associative,
- there is an identity element, and
- every element of \(G\) has a negative.

For historical reasons, most mathematicians use the term “abelian group,” rather than “commutative group,” but they would agree that “commutative group” is also acceptable.

- (\(\mathbb{R}, +\)) is a commutative group: we all learned in elementary school that addition is commutative and associative, that \(g + 0 = g\), and that \(g + (−g) = 0\). (The same is true for (\(\mathbb{Q}, +\)) and (\(\mathbb{Z}, +\)).)
- (\(\mathbb{N}, +\)) is not a commutative group (even though addition is commutative and associative, and 0 is an identity element), because no nonzero element has a negative in \(\mathbb{N}\).
- (\(\mathbb{R}, −\)) is
*not*a commutative group, because subtraction is not commutative: \(g − h\) is usually not equal to \(h − g\). (Another reason (\(\mathbb{R}, −\)) is not a commutative group is that subtraction is not associative: \((g − h) − k\) is usually not equal to \(g − (h − k)\).) - (\(\mathbb{R}, \times\)) is
*not*a commutative group (even though addition is commutative and associative, and 1 is an identity element for multiplication), because 0 does not have a “negative” or “multiplicative inverse”: there does not exist \(h \in \mathbb{R}\), such that \(0 \times h = 1\).

Our condition to be an identity element is what is usually called a “right identity element.” For 0 to be an identity element, it should be true not only that \(g + 0 = g\), but also that \(0 + g = g\). However, the only binary operations of interest to us are commutative groups, where the second condition is also satisfied (see Exercise \(5.2.11(1)\) below), so there is no need for us to make this distinction.

For the binary operation on \(\{\mathrm{a}, \mathrm{b}, \mathrm{c}, \mathrm{d}, \mathrm{e}, \mathrm{f}\}\) in Example \(5.2.2(4)\), verify that:

- \(a\) is an identity element, and
- every element has a negative, namely: \(-\mathrm{a}=\mathrm{a},-\mathrm{b}=\mathrm{c},-\mathrm{c}=\mathrm{b},-\mathrm{d}=\mathrm{d},-\mathrm{e}=\mathrm{f}, -\mathrm{f}=\mathrm{e}\).

The following result tells us that, instead of an identity element of a commutative group, we may speak of *the* identity element.

*The identity element of any commutative group is unique.*

**Proof**-
Suppose 0 and \(\theta\) are any identity elements of a commutative group (\(G, +\)). Then \[\begin{aligned}

0 &=0+\theta & &(\theta \text { is an identity element) }\\

&=\theta+0 & &(+\text { is commutative) }\\

&=\theta & &(0 \text { is an identity element). }

\end{aligned}\]

Since 0 and \(\theta\) are arbitrary identity elements of (\(G, +\)), this implies that all identity elements are equal to each other, so the identity element is unique (there is only one of them).

In this section, the symbol 0 will always represent the identity element of whatever commutative group is under consideration.

*Warning:* This means that 0 will usually not represent the number zero. For example, in Example \(5.2.2(4)\), we have \(0 = a\).

Similarly, instead of a negative, we may speak of *the* negative of an element of \(G\):

*Let* (\(G, +\)) *be a commutative group. For each* \(g \in G\)*, the negative of* \(g\) *is unique.*

**Proof**-
Suppose \(−g\) and \(h\) are any negatives of \(g\). Then \[\begin{aligned}

-g &=-g+0 & &(0 \text { is the identity element) }\\

&=-g+(g+h) & &(h \text { is a negative of } g) \\

&=(-g+g)+h & &(+\text { is associative }) \\

&=h+(g+(-g)) & &(+\text { is commutative) }\\

&=h+0 & &(-g \text { is a negative of } g) \\

&=h & &(0 \text { is the identity element). }

\end{aligned}\]

Therefore, all negatives of \(g\) are equal, so the negative is unique.

*For every commutative group* (\(G, +\))*, we have* \(−0 = 0\).

**Proof**-
Let \(g = 0\). Then \(g\) is the identity element, so \(0 + g = 0\). By definition of the negative, this means that \(g = −0\). Since \(g = 0\), we conclude that \(0 = −0\).

Assume (\(G, +\)) is a commutative group. For \(g, h \in G\), we use \(g − h\) as an abbreviation for \(g + (−h)\).

Assume (\(G, +\)) is a commutative group, and \(g, h \in G\).

*Carefully justify each step of your proofs, using only the axioms stated in Definition* \(5.2.3\). *Do not assume any other properties of addition that you were taught in school*.

- Show \(0 + g = g\).
- Show \((−g) + g = 0\).
- Show \(g − g = 0\).
- Show \(−(−g) = g\).
- Show \((−g) + h = h − g\).
- Show \((g − h) + h = g\).

The associative law tells us that \((g +h)+k = g + (h+k)\). Therefore, we learn in elementary school to simply write \(g+h+k\), because it does not matter where the parentheses go. Officially, the associative law In fact (as we also learn in elementary school), there is no need to include parentheses in any sum (even if it has more than three terms). Also, the commutative law tells us that \(g + h = h + g\). We learn in elementary school that this allows us to rearrange the terms in a sum of any length, and the same is true for commutative groups. For example: \[g_{1}+g_{2}+g_{3}+g_{4}+g_{5}=g_{4}+g_{3}+g_{1}+g_{5}+g_{2} .\]

Here is an official statement of these observations:

*If* (\(G, +\)) *is a commutative group, *\(n \in \mathbb{N}^{+}\)*, and *\(g_{1}, g_{2}, \ldots, g_{n} \in G\)*, then:*

- (\(+\)
*is***associative**)*The expression*\(g_{1}+g_{2}+\cdots+g_{n}\)*represents a well-defined element of*\(G\)*, which does not depend on how the expression is parenthesized.* - (\(+\)
*is***commutative**)*If*\(h_{1}, h_{2}, \ldots, h_{n}\)*is a list of the same elements of*\(G\)*, but perhaps in a different order, then*\[h_{1}+h_{2}+\cdots+h_{n}=g_{1}+g_{2}+\cdots+g_{n} .\]

Assume (\(G, +\)) is a commutative group, and \(g, h \in G\). Show \[-(g+h)=(-g)+(-h) .\]

**Proof**-
We have \[\begin{aligned}

(g+h)+((-g)+(-h)) &=g+h+(-g)+(-h) & &(+\text { is associative }) \\

&=g+(-g)+h+(-h) & &(+\text { is commutative }) \\

&=(g+(-g))+(h+(-h)) & & \text { (+ is associative) } \\

&=0+0 & &\text { (definition of }-g \text { and }-h) \\

&=0 & &(0 \text { is the identity element) }

\end{aligned}\]

So \((−g) + (−h)\) is the negative of \(g + h\). In other words, \((−g) + (−h) = −(g + h)\).

Assume (\(G, +\)) is a commutative group, and \(g, h, a \in G\).

- Show \(−(g − h) = h − g\).
- Show that if \(g + a = h + a\), then \(g = h\).

Let (\(G, +\)) be a commutative group. A subset \(H\) of \(G\) is a **subgroup** of (\(G, +\)) iff

- \(H \neq \varnothing\),
- (
**closed under negatives**) \(−h \in H\), for all \(h \in H\), and - (
**closed under addition**) \(h_{1} + h_{2} \in H\), for all \(h_{1}, h_{2} \in H\).

\(\mathbb{Z}\) and \(\mathbb{Q}\) are subgroups of (\(\mathbb{R}, +\)), but \(\mathbb{N}\) is **not** a subgroup (because it is not closed under negatives).

*If* \(H\) *is a subgroup of a commutative group* (\(G, +\))*, then* \(0 \in H\) (*where, as usual,* 0 *is the identity element of* (\(G, +\))).

**Proof**-
We know that \(H \neq \varnothing\) (from the definition of subgroup), so there is some \(h \in H\). Since \(H\) is closed under negatives (because it is a subgroup) this implies \(−h \in H\). Then, since \(H\) is closed under addition (because it is a subgroup), we have \(h + (−h) \in H\). Since \(h + (−h) = 0\) (by the definition of \(−h\)), this means \(0 \in H\).

Assume \(H\) is a subgroup of a commutative group (\(G, +\)), and \(h, k \in H\).

- Show \(h − k \in H\).
- For all \(a \in G\), show that if \(a \notin H\), then \(a+h \notin H\).

Assume \(H\) is a subgroup of a commutative group (\(G, +\)), and \(a, b \in G\). Let \(a+H=\{a+h \mid h \in H\}\) and \(b+H=\{b+h \mid h \in H\}\).

- Show that if \(a + H\) is a subgroup of \(G\), then \(a \in H\). [
*Hint:*Every subgroup contains 0.] - Show that if \(a \in H\), then \(a + H = H\).
- Show that \(a + H = b + H\) iff \(a − b \in H\).
- Show that if \((a+H) \cap(b+H) \neq \varnothing\), then \(a + H = b + H\).

*Assume* (\(G, +\)) *is a commutative group, and let* \(T=\{t \in G \mid t+t=0\}\)*. Then* \(T\) *is a subgroup of* \(G\).

**Proof**-
It suffices to show that \(T\) is: nonempty, closed under negatives, and closed under addition.

(nonempty) We have \(0 \in T\) (because it is immediate that \(0 + 0 = 0\)), so \(T \neq \varnothing\).

(closed under negatives) Given \(t \in T\), we have \[\begin{aligned}

(-t)+(-t) &=-(t+t) \\

&=-0 \\

&=0

\end{aligned}\]

(First Line: Example \(5.2.13\), Second Line: \(t \in T\), Third Line: Proposition \(5.2.9\))

This means \(−t \in T\), so \(T\) is closed under negatives.

(closed under addition) Given \(s, t \in T\), we have \(s + s = 0\) and \(t + t = 0\). Therefore \[\begin{aligned}

(s+t)+(s+t) &=(s+t)+(t+s) & &(+\text { is commutative }) \\

&=s+(t+(t+s)) & & \text { ( }+\text { is associative) } \\

&=s+((t+t)+s) & & \text { (+ is associative) } \\

&=s+(0+s) & &(t \in T) \\

&=s+s & &(0 \text { is the identity element) }\\

&=0 & &(s \in T)

\end{aligned}\]

so \(s + t \in T\(). Therefore \(T\) is closed under addition.

Assume \(H\) and \(K\) are subgroups of a commutative group (\(G, +\)).

- Show that \(H \cap K\) is a subgroup of (\(G, +\)).
- Let \(H+K=\{h+k \mid h \in H, k \in K\}\). Show that \(H + K\) is a subgroup of (\(G, +\)).