Introduction to Abstract Algebra/Lecture 3

Definition of a Group

In this lesson, the basic structure of abstract algebra is introduced: the group. A group is defined by the pairing of a set and an operation between elements of that set. Let $G$ be a set and let $\cdot$ be an operation on that set. We say that $G$ is a group with the operation $\cdot$ if and only if:

• The operation is closed with respect to the set: $\forall a,b\in G$, it is the case that $a\cdot b\in G$.
• There is an identity element: $\exists e\in G$ such that $\forall a\in G$, it is the case that $a\cdot e=e\cdot a=a$.
• Each element has an inverse in the set: $\forall a\in G$, it is the case that $\exists a^{-1}\in G$ such that $a\cdot a^{-1}=a^{-1}\cdot a=e$.
• The operation is associative with respect to the elements of the group: $\forall a,b,c\in G$, it is the case that $a\cdot (b\cdot c)=(a\cdot b)\cdot c$.

These are precise definitions which enable us to say which sets and operations form groups and what additional properties groups have. Because we can derive properties beyond the mere definition of groups and the relations between its elements, we can begin to develop an abstracted form of algebra which allows us to work with and manipulate equations which may involve ideas beyond simple numbers. Also of interest with groups is exploring the questions that merely being a group cannot answer. For instance, in the definition for groups, you should note that there is no rule which says that $a\cdot b=b\cdot a$ (that is, it is not necessarily commutative, like matrix multiplication). In essence, a group is free to have or not have any property which is not forbidden by the rules above. This aspect of abstract algebra is what gives the topic its depth.

In the following sections, additional properties will be derived for simple groups, those possessing only the properties described above. The assignment for this lecture extends the set of properties by prompting for additional fundamental proofs.

Properties of a Group

The Uniqueness of the Identity

Suppose that both $e$ and $f$ are identity elements of a group. Then it is the case that $e\cdot f=e$ and $e\cdot f=f$. Therefore, $e=f$.

The Inverse of the Identity is the Identity

Let $e$ be the identity element of a group so that $e^{-1}$ is its inverse. $e\cdot e^{-1}=e^{-1}$ by the definition of the identity, but it is also the case that $e\cdot e^{-1}=e$ by the definition of the inverse. Therefore, $e=e^{-1}$.

Cancellation Law

Suppose $a\cdot b=a\cdot c$. By the definition of the operation, there exists $x$ such that $a\cdot b=a\cdot c=x$. Additionally, $a^{-1}\cdot x$ equals itself so that it is the case that $a^{-1}\cdot a\cdot b=a^{-1}\cdot a\cdot c$. Therefore, using the definition of the inverse, $e\cdot b=e\cdot c$ so that, by the definition of the identity, $b=c$. A similar argument holds that $b\cdot a=c\cdot a$ requires $b=c$.

The Uniqueness of the Inverse

Suppose $a^{-1}$ and $b$ are inverses of $a$ so that $a\cdot a^{-1}=e$ and $a\cdot b=e$ by the definition of the identity. Therefore, $a\cdot b=a\cdot a^{-1}$, so $b=a^{-1}$ by the cancellation law.

Strategies for Proof

You will see groups defined by sets of logical statements which often involve traditional operations. Inherently, this means you can work with those definitions to establish that the group conditions hold and to search for additional properties. One pitfall in this process is ambiguity of notation which can arise when the desired symbol for the group operation matches closely to the normal algebraic symbol, but is not exactly equivalent (or perhaps the operation is so similar to the traditional variation that it is best to use the same symbol). For instance, say we create the operation $+$ for a potential group which involves the operation $+$ for standard real addition; when you say $a+b=a+b-10$ to mean that the application of your operation is equivalent to 10 less than the application of real addition, the meaning of the definition is ambiguous outside of this context. This raises the point that it's extremely important to remember context when working with groups. Sometimes, authors will help by using non-standard operators like $a\diamond b$ or $a\star b$ to mean the application of the operation. Other times, the operation is presented as a function $f:G\times G\rightarrow G$ where $G$ is the group's set. Indeed, any logical representation can be used to define potential groups. In terms of the proof process, this means it may be helpful or necessary to switch between the various representations of the definitions.

Homomoprhisms

One strategy used to prove various properties of groups is the idea of creating a homomorphism between the group you are working with and some other group. Say you have a group $G$ under the $\diamond$ operation and there exists another group $H$ under the $\star$ operation and you create a function $f:G\rightarrow H$ in which the following holds: $f(a\diamond b)=f(a)\star f(b)$. Conceptually, $f$ acts as a translator between the two sets. The implication of having a homomorphism from $G$ to $H$ is that whatever algebraic properties are applicable to $G$ are also applicable to some or all of $H$. In other words, if dealing with $H$ is difficult, you may construct a $G$ such that $f$ exists and you may show properties of $Rng(f)\subseteq H$ by showing features of $G$. For example, if the $\diamond$ operation is commutative, then you can argue that $f(a\diamond b)=f(b\diamond a)$ therefore $f(a)\star f(b)=f(b)\star f(a)$ which means that for the elements of the range of $f$, the $\star$ operation is commutative.

Example Groups

Positive Reals Under Multiplication

Consider the set $\mathbb {R} ^{+}$, that is, the set of positive, real numbers (noting that zero is not an element of this set). Let the operation be the usual multiplication for real numbers. First, we must demonstrate that the operation is closed to the set. Since each element of the set is both positive and real and the product of any two such numbers has the same properties, it is the case that the operation is closed under the set. Next, we must find an identity element in the group. It is known that the product of any real number and 1 is the same number, so we can choose 1 to be the identity, noting that it is in our set and $1\cdot a=a\cdot 1$. Next, we must demonstrate that each element of the set has an inverse. Since the product of a positive real and its reciprocal (also a positive real) is equal to the identity, 1, the existence of an inverse is established for each element of the set. We take note that the product of a positive real and its reciprocal has the same effect regardless of the order, $a\cdot a^{-1}=a^{-1}\cdot a=1$. Next, we acknowledge that fact that real multiplication is associative, therefore the operation is associative for our purposes too. We have established that $\mathbb {R} ^{+}$ with real multiplication $\cdot$ forms a group.

Integers Modulo N Under Addition

Consider the set of integers containing zero and all positive integers less than some other positive integer, $N$. These are the "integers modulo N" or "integers mod N" which is denoted as $\mathbb {Z} /N\mathbb {Z}$. Let an operation, $\diamond$, between elements of the set be defined conditionally as if $a+b<N$, then $a\diamond b=a+b$; otherwise, $a\diamond b=a+b-N$. Conceptually, the set is easily understood as $\left\{0,1,...,N-1\right\}$ while the operation easily understood as addition which "wraps around" if necessary to keep the sum inside the set. It must be shown that the operation is closed to the group. If $a+b<N$, then $a\diamond b=a+b$ and $0\leq a+b<N$ so $0\leq a\diamond b<N$ (and thus is in the set). For the case that $a+b\geq N$, let's consider the case where both $a$ and $b$ are the largest element in $\mathbb {Z} /N\mathbb {Z}$, that is $a=b=N-1$. Only the sum of the two largest numbers in the set yields the highest sum since if either were less, then the sum would be less. With those maximum elements, the maximum sum of any two elements of the set is produced, $a+b=2N-2<2N$. Therefore, $a+b<2N$ for any $a$ and $b$ so it is the case that $N\leq a+b<2N$, so $0\leq a+b-N<N$ and $0\leq a\diamond b<N$. Therefore, for both conditions, $a\diamond b\in \mathbb {Z} /N\mathbb {Z}$. Therefore it is established that the operation is closed with respect to the set. Next, an identity element must be established. Let $0$ be the proposed identity. For all $a$ in the set, $a\diamond 0=a+0=0+a=0\diamond a=a$. Since $0$ is in the set, $0$ is the identity. Next, it must be demonstrated that every element has an inverse. Let $a$ be some element in the set so that $a^{-1}$ is its inverse. If we define the inverse to be $a^{-1}=N-a$ if $a\neq 0$, otherwise $a^{-1}=0=a\diamond a^{-1}$, then we can explore the non-identity condition by noting that $0\leq N-a<N$ so that $0\leq a^{-1}<N$ for all $a$. Additionally, note that the sum of $a$ and its inverse is $a+(N-a)=N$, which meets or exceeds $N$, therefore we use the definition $a\diamond a^{-1}=a+a^{-1}-N=a+(N-a)-N=0$. Therefore, it is established that the inverse of each element exists in the set and that the application of the operation between any item and its inverse yields the identity. Finally, it must be demonstrated that the operation is associative. Suppose that $a$, $b$, and $c$ are members of the set, then $(a\diamond b)\diamond c=(a+b-pN)+c-qN=a+b+c-kN$ for some $p,q\in \left\{0,1\right\}$ with $k=p+q$. Alternatively, $a\diamond (b\diamond c)=a+(b+c-xN)-yN=a+b+c-hN$ for some $x,y\in \left\{0,1\right\}$ with $h=x+y$. If $h\neq k$ then, either $(a\diamond b)\diamond c$ is not in the set or $a\diamond (b\diamond c)$ is not in the set, which is impossible from the definition of the operation. Therefore, the operation is associative for members of the set. Therefore, the integers modulo N form a group under addition.

Example Non-groups

Negative Reals Under Multiplication

Consider the set, $\mathbb {R} ^{-}$, of negative real numbers (zero is not included in the set) and the typical multiplication operation for reals. It is sufficient to show that any one of the properties of a group is false in order to show that the set and the operation do not form a group. Suppose $e$ is the identity of the group. $e<0$ by the definition of the set, but $e\cdot e>0$ because of typical multiplication. So, not only is the operation not closed to the set, but there is no identity since $e\cdot a\notin \mathbb {R} ^{-}$ for all $e,a\in \mathbb {R} ^{-}$.

Further Notation

As you study further into abstract algebra and group theory, shorthand notation will be used in various cases since the context of groups and algebra is assumed. When speaking about groups in general, the addition and multiplication operations are used frequently. You may see $a+b$ or $ab$ to mean the application of the group operation. Exponents are used to mean the repeated application of the multiplication operation, you may see $a^{3}$ which means $aaa$. Additionally, you might see $a^{-3}$ to mean $a^{-1}a^{-1}a^{-1}$. When the a addition symbol is used for the operation, you may also come across $3a$ to mean $a+a+a$. Many times, you might study two groups with nearly the same set of elements. For example, the set $\mathbb {R}$ forms a group under addition, but also forms a group under multiplication if you take away the zero element (or more generally, identity under addition). You will see the multiplicative group denoted $\mathbb {R} ^{*}$ or $\mathbb {R} -\left\{0\right\}$ or $\mathbb {R} \setminus \left\{0\right\}$ depending on the author's background and preference.

In some contexts, you will see groups labeled by the pairing of its set with the operation; for example, $(G,+)$ means the group having the set $G$ under the operation $+$.

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