Groups

Abigail Huettig

23 Groups

Most if not all topics discussed in modern algebra are connected to groups in some way. However, before we can understand what a group is, we must understand what a binary operation is. Fraleigh and Katz provide the following definition [7].

Definition V.1
A binary operation $\ast$ on a set $S$ is a function mapping $S\times S$ into $S$ . For each $(a,b)\in S\times S$ , we will denote the element $\ast((a,b)))$ of $S$ by $a\ast b$ .

When a binary operation is a mapping from $S\times S$ to $S$ , we say that $S$ is closed under the binary operation.

A binary operation is represented by $\ast$ , just as addition is represented by + and multiplication by $\times$ . Because addition and multiplication are themselves binary operations, $\ast$ will sometimes be used in place of + or $\times$ .

Now that we have a basic understanding of what a binary operation is, we are ready to look at the definition of a group as given by Fraleigh and Katz [7].

Definition V.2
A group $\langle G, \ast\rangle$ is a set $G$ , closed under a binary operation $\ast$ , such that the following axioms are satisfied:

For all $a,b,c\in G$ , we have
$(a \ast b)\ast c=a\ast(b\ast c).$
There is an element $e$ in $G$ such that for all $x\in G$ ,
$e\ast x=x\ast e=x.$
Corresponding to each $a\in G$ , there is an element $a'$ in $G$ such that
$a\ast a'=a'\ast a=e.$

A group $G$ is a specific type of set (see Part III: Chapter 13) that is closed under a binary operation $\ast$ and must satisfy a few criteria. First, the binary operation must be associative. Second, the identity element of the binary operation is an element of $G$ . Third, every element in $G$ must have an inverse element that is also in $G$ .

So, to prove that a set $G$ is a group under a binary operation $\ast$ , we need to show that the group is closed under the operation and that the three conditions presented in the definition are satisfied. Let’s now look at an exercise from Fraleigh and Katz that I completed for this paper [7].

Example 49
Let $n$ be a positive integer and let $n\mathbb{Z}=\{nm|m\in\mathbb{Z}\}$ . Is $\langle n\mathbb{Z}, +\rangle$ a group?

Solution

Because every element in $n\mathbb{Z}$ is the product of a positive integer and an integer, every element in $n\mathbb{Z}$ is an integer because $\mathbb{Z}$ is closed under multiplication. So, for all $na,nb\in n\mathbb{Z}$ , $na+nb=n(a+b)$ because the distributive property holds in the integers. Because $\mathbb{Z}$ is closed under addition, $a+b\in \mathbb{Z}.$ Thus,

$na+nb=n(a+b)\in n\mathbb{Z},$

and $n\mathbb{Z}$ is closed under addition.
For all integers $a,b,c$ ,

$\begin{align*} (a+b)+c &=a+(b+c) &&\text{Associativity holds in integers.}\\ n[(a+b)+c]&=n[a+(b+c)] &&\text{Multiply both sides by the integer $n$.} \\ n(a+b)+nc&=na+n(b+c) &&\text{Distributive property holds in integers.} \\ (na+nb)+nc&=na+(nb+nc) &&\text{Distributive property holds in integers.} \end{align*}$

Thus, addition is associative for all elements $nm\in n\mathbb{Z}$ .
For all integers $a$ ,

$\begin{align*} a&=a+0 &&\text{0 is additive identity in integers.}\\ na&=n(a+0) &&\text{Multiply both sides by the integer $n$.}\\ &=na+n0 &&\text{Distributive property holds in integers.}\\ &=na+0 &&\text{Any integer multiplied by 0 is 0.} \end{align*}$

So, 0 is the identity element for $n\mathbb{Z}$ under addition, and $0=n0\in n\mathbb{Z}$ .
For all integers $a$ ,

$\begin{align*} a+(-a)&=0 &&\text{The integer $-a$ is the additive inverse of $a$}\\ n(a+(-a))&=n0 &&\text{Multiply both sides by the integer $n$}\\ na+(-na)&=0 &&\text{Distributive property holds in integers}\\ \end{align*}$

For each $na\in n\mathbb{Z}$ , $-na$ is the additive inverse. Additionally, $-na=n(-a)\in n\mathbb{Z}$ . Therefore, $\langle n\mathbb{Z}, +\rangle$ is a group.

There are many different types of groups. One is $n\mathbb{Z}$ which we just saw is the set of all integers that are multiples of $n$ . Another common group is $\mathbb{Z}_n=\{0,1,2,3,...,n-1\}$ . Other types include cyclic, abelian, dihedral, alternating, and symmetric.

We are now going to look at subgroups, which are specific groups within a group. In particular, we will examine cyclic subgroups.

Cyclic Subgroups

Before we discuss what cyclic subgroups are, we need to understand what a subgroup is. It seems that a subgroup would be group within a group, similar to how a subset is a a set within a set. However, it is a little more complex than that. Fraleigh and Katz provide the following definition [7].

Definition V.3
If a subset $H$ of a group $G$ is closed under the binary operation of $G$ and if $H$ with the induced operation from $G$ is itself a group, then $H$ is a subgroup of $G.$

If $H$ is a subgroup of a group $G$ under $\ast$ , it is also a subset of $G$ . That is, every element in $H$ must also be in $G$ . Additionally, $H$ must be closed under the binary operation $\ast$ and fulfill the three criteria given in definition V.2 for $\ast$ . We denote a subgroup as $H\leq G$ . If we know that $H$ is not equal to $G$ , $H$ is a proper subgroup denoted as $H<G$ . Notice how the notation and terminology is very similar to that of sets (see Part III: Chapter 13).

Now we are ready to discuss cyclic subgroups. Fraleigh and Katz provide the following definition [7].

Definition V.4
Let $G$ be a group and let $a\in G$ . Then the subgroup $\{a^n|n\in \mathbb{Z}\}$ of $G$ is called the cyclic subgroup of $G$ generated by $a$ , and denoted by $\langle a \rangle$ .

The notation of $a^n$ is a little misleading. We are not taking $a$ to the power of $n$ unless the binary operation is multiplication. For an element $a$ of a group $G$ , $a^n$ is actually equal to $a_1 \ast a_2\ast ... \ast a_n$ , where $a=a_i$ for all $1\leq i\leq n$ and $\ast$ is the binary operation of $G$ .

There is another interesting aspect of these subgroups that Fraleigh and Katz give in the following theorem [7].

Theorem V.5

Let $G$ be a group and let $a\in G$ . Then

$H=\{a^n|n\in\mathbb{Z}\}$

is a subgroup of $G$ and is the smallest subgroup of $G$ that contains $a$ , that is, every subgroup containing $a$ contains $H$ .

Any subgroup $H$ of a group $G$ , defined under $\ast$ , generated by an element $a$ in $G$ , is the smallest subgroup containing $a$ . That is, every subgroup containing $a$ contains $H=\langle a\rangle$ . Let’s take a moment to think through why this is true. Let $I$ be a subgroup of $G$ that contains $a$ . We have that $(a,a)$ is mapped to $a^2$ by $\ast$ , $(a,a^2)$ is mapped to $a^3$ by $\ast$ , and so on. Eventually, we will have $(a,a^{n-1})$ is mapped to $(a,a^n)$ . Because $I$ is a subgroup, $I$ is closed. Because $I$ contains $a$ and is closed, $a^2$ is in $I$ which implies $a^3$ is in $I$ and so on. Thus, $a^n$ is in $I$ for all integers $n$ . Because $H=\langle a\rangle=\{a^n|n\in\mathbb{Z}\}$ , $I$ must contain $H=\langle a \rangle$ .

Cyclic subgroups are helpful in that they make finding the smallest subgroup containing a specific element quite easy. Furthermore, we are guaranteed that the set of elements generated by an element $a\in G$ is a subgroup of $G$ . Let’s look at an example that I created and completed for this paper. Recall that $a+b(\text{mod } n)$ is equal to the remainder when $(a+b)$ is divided by $n$ .

Example 50
Consider the group

$\mathbb{Z}_9=\{0,1,2,...,7,8\}$

with binary operation $a\ast b=a+b=(a+b)$ mod 9. Find the smallest subgroup that contains 3, denoted as $\langle 3\rangle$ .

Solution
By Theorem V.5, the smallest subgroup containing 3 would be the cyclic subgroup generated by 3.

$\begin{align*} 3^2&=3+3=6\\ 3^3&=3+3+3=6+3=0\\ 3^4&=3+3+3+3=0+3=3\\ 3^5&=3+3+3+3+3=3+3=6\\ \end{align*}$

Since $3^2$ and $3^5$ both result in 6, we have found all of the elements of the set. If we keep going we will have a pattern of 0,3,6,0,3,6,… as $n$ increases. Therefore, the smallest subgroup of $\mathbb{Z}_9$ containing 3 is $H=\langle 3\rangle=\{0,3,6\}.$

We have seen how a subgroup, specifically a cyclic subgroup, relates to a group. Now, we are going to look at how groups relate to other groups through isomorphisms.

Isomorphisms

Let’s begin our discussion of isomorphishms by looking at the formal definition Fraleigh and Katz provide for us [7].

Definition V.6
Let $\langle S, \ast \rangle$ and $\langle S', \ast ' \rangle$ be binary algebraic structures. An isomorphism of $S$ with $S'$ is a one-to-one function $\phi$ mapping $S$ onto $S'$ such that

$\phi(x \ast y)=\phi(x)\ast'\phi (y) \text{for all}\, x,y\in S.$

Recall that in Part III: Chapter 15, we discussed one-to-one and onto functions. Now, suppose that we have such a function from one binary algebraic structure $S$ to another binary algebraic structure $S'$ . The fact that the function is one-to-one guarantees that each element in $S$ will be mapped to only one element in $S'$ . Because the function is onto, we are guaranteed that each element in $S'$ will be mapped to only one element in $S$ . If we let $s$ be the number of elements in $S$ and $s'$ be the number of elements in $S'$ , we have that $s\leq s'$ because the function is one-to-one and $s'\leq s$ because the function is onto. Thus, $S$ and $S'$ must have the same number of elements. This gives us a glimpse into why isomorphisms are so important.

When two binary algebraic structures $S$ and $S'$ are isomorphic ( $S\simeq S'$ ), we say that they have the same structure, or structural properties. As we have seen, isomorphic binary algebraic structures have the same number of elements, though not necessarily the same elements [7]. Their binary operations would also have the same properties, such as being commutative, even if the binary operations themselves are not the same [7]. There are many more structural properties, but this provides the basic idea. If we can find an isomorphism from $S$ to $S'$ , then we know all the structural properties are the same for $S$ and $S'$ .

Isomorphisms are of great importance because they help us study complex groups that are not very common and are more difficult to work with. Because two groups that are isomorphic have the exact same structural properties, isomorphisms allow us to find a simpler group with the same structural properties as a more complicated group [11]. Thus, structural properties of the complicated group can be discovered by way of the simpler one that otherwise may have gone undiscovered or taken much more effort to find.

To prove that two groups $\langle G, \ast\rangle$ and $\langle H, \ast'\rangle$ are isomorphic, we must find a one-to-one mapping $\phi$ from $G$ onto $H$ that satisfies

$\phi(g\ast i)=\phi(g)\ast'\phi(i)$

for all $g, i \in G$ . We already discussed how to prove a mapping is one-to-one and onto in Part III: Chapter 15, and we can then use a direct proof (see Part II: Chapter 11) to show that

$\phi(g\ast i)=\phi(g)\ast'\phi(i)$

for all $g, i\in G$ .

Before we complete the following example, we need to introduce one more definition so that what is done in the problem makes sense. This definition comes from Macauley [17].

Definition V.7
The direct product of groups $A$ and $B$ consists of the set $A\times B$ , and the group operation is done component-wise: if $(a,b),(c,d)\in A\times B$ , then

$(a,b)\ast (c,d)=(a\ast c,b\ast d).$

Now we are ready to begin our example which is a homework problem I completed for Modern Algebra [18].

Example 51
Let $\langle\mathbb{Q^{\ast}},\cdot\rangle$ be the group of non-zero rational numbers under multiplication and $\langle\mathbb{Q^+},\cdot\rangle$ be the group of positive rational numbers under multiplication.
Show that

$\langle\mathbb{Q^{\ast}},\cdot\rangle\simeq\langle\mathbb{Q^+},\cdot\rangle\times C_2$

where $C_2=\{e^{0\pi i},e^{\pi i}\}=\{1,-1\}$ with multiplication as the binary operation.

Proof.
Let $\phi:\langle\mathbb{Q^+},\cdot\rangle\times C_2\rightarrow\langle\mathbb{Q^{\ast}},\cdot\rangle$ be defined as $\phi((a,b))= a\cdot b.$

To prove that $\phi$ is one-to-one, we need to show that if $\phi((a,b))=\phi((c,d))$ then $(a,b)=(c,d)$ . We will show this by proving two cases.

Case 1: Let $b=d=1.$

$\begin{align*} \phi((a,1))&=\phi((c,1))\\ a\cdot 1&=c\cdot 1\\ a&=c \end{align*}$

Case 2: Let $b=d=-1$ .

$\begin{align*} \phi((a,-1))&=\phi((c,-1))\\ a\cdot -1&=c\cdot -1\\ a&=c \end{align*}$

Thus, $\phi$ is one-to-one. To prove that $\phi$ is onto, we need to show that for each $c\in\langle\mathbb{Q^{\ast}},\cdot\rangle$ , there exists $(a,b)\in\langle\mathbb{Q^+},\cdot\rangle\times C_2$ such that $\phi((a,b))=c$ . We will show this by proving two cases as well.

Case 1: Let $c$ be in $\mathbb{Q}^+.$

$\begin{align*} c&=\phi(a,1)\\ &=a\cdot 1\\ &=a\\ \end{align*}$

Case 2: Let $c$ be in $\mathbb{Q}^-$ .

$\begin{align*} c&=\phi(a,-1)\\ &=a\cdot -1\\ &=-a\\ -c&=a \end{align*}$

Thus, $\phi$ is onto.

Finally, for all $a,c\in \langle\mathbb{Q^+},\cdot\rangle$ and $b,d\in C_2$ ,

$\begin{align*} \phi((a,b)\ast (c,d))&=\phi(a\ast c,b\ast d) &&\text{Definition of direct product}\\ &=\phi(a\cdot c,b\cdot d)\\ &=a\cdot c \cdot b\cdot d\\ &=a\cdot b \cdot c\cdot d &&\text{Rational multiplication is commutative.}\\ &=(a\cdot b) \cdot (c\cdot d) &&\text{Associativity of Group Operations}\\ &=\phi(a,b)\ast \phi(c,d) \end{align*}$

Because $\phi$ is a one-to-one mapping from $\langle\mathbb{Q^+},\cdot\rangle\times C_2$ onto $\langle\mathbb{Q^{\ast}},\cdot\rangle$ and

$\phi((a,b)\ast (c,d))=\phi(a,b)\ast \phi(c,d)$

holds for all $a,c\in \langle\mathbb{Q^+},\cdot\rangle$ and $b,d\in C_2$ ,

$\langle\mathbb{Q^{\ast}},\cdot\rangle\simeq\langle\mathbb{Q^+},\cdot\rangle\times C_2.$

Isomorphisms will come up again later when we discuss factor groups, which are groups made up of cosets. So, let’s find out what cosets are.

Cosets

Cosets are derived from an element of a group $G$ and a subgroup of $G$ . Let’s examine the following definition provided by Fraleigh and Katz [7].

Definition V.8
Let $H$ be a subgroup of a group $G$ . The subset $aH=\{ah|h\in H\}$ of $G$ is the left coset of $H$ containing $a$ , while the subset $Ha=\{ha|h\in H\}$ is the right coset of $H$ containing $a$ .

For a group $G$ under a binary operation $\ast$ and a subgroup $H$ of $G$ , if we take an element $a$ of $G$ and perform the binary operation on $a$ and every element in $H$ , the resulting set of elements is known as a coset. As not every binary operation is commutative, $a\ast h$ is not necessarily equal to $h\ast a$ , where $h$ is an element of $H$ . The elements of the form $a\ast h$ make up the left coset of $H$ containing $a$ , denoted as $aH$ , because $a$ is to the left of $h$ . Similarly, the elements $h\ast a$ make up the right coset $Ha$ because $a$ is to the right of $h$ .

Another interesting fact of left cosets is that they partition the group $G$ [7]. That is, if we take the union of all of the left cosets, we will get the entire group $G$ . Additionally, none of the left cosets have an element in common. That is, the intersection of the left cosets is the empty set.

We could also replace “left” with “right” in the preceding paragraph and still get true statements. (Refer back to Part III: Chapter 13 for a review of the terms union, intersection, and empty set.)

Let’s now try to find all of the cosets of a subgroup. The following example is an exercise from Fraleigh and Katz that I completed for this paper [7].

Example 52
Find all cosets of the subgroup $\langle4\rangle$ of

$\mathbb{Z}_{12}=\{0,1,2,...,10,11\}$

with the binary operation $a\ast b=a+b=(a+b)\,$ mod(12).

Solution
The subgroup $\langle4\rangle$ is the cyclic subgroup generated by 4.

$\begin{align*} 4^2&=4+4=8\\ 4^3&=4+4+4=8+4=0\\ 4^4&=4+4+4+4=0+4=4\\ 4^5&=4+4+4+4+4=4+4=8 \end{align*}$

If we continue on, we will have a pattern of 0,4,8,0,4,8,…. So,

$\langle4\rangle=\{0,4,8\}.$

The left cosets are then:

$\begin{align*} 0+\langle4\rangle&=0+\{0,4,8\}=\{0,4,8\}=4+\langle4\rangle=8+\langle4\rangle\\ 1+\langle4\rangle&=1+\{0,4,8\}=\{1,5,9\}=5+\langle4\rangle=9+\langle4\rangle\\ 2+\langle4\rangle&=2+\{0,4,8\}=\{2,6,10\}=6+\langle4\rangle=10+\langle4\rangle\\ 3+\langle4\rangle&=3+\{0,4,8\}=\{3,7,11\}=7+\langle4\rangle=11+\langle4\rangle\\ \end{align*}$

and the right cosets are:

$\begin{align*} \langle4\rangle+0&=\{0,4,8\}+0=\{0,4,8\}=\langle4\rangle+4=\langle4\rangle+8\\ \langle4\rangle+1&=\{0,4,8\}+1=\{1,5,9\}=\langle4\rangle+5=\langle4\rangle+9\\ \langle4\rangle+2&=\{0,4,8\}+2=\{2,6,10\}=\langle4\rangle+6=\langle4\rangle+10\\ \langle4\rangle+3&=\{0,4,8\}+3=\{3,7,11\}=\langle4\rangle+7=\langle4\rangle+11\\ \end{align*}$

Because the right cosets are identical to the left cosets, the cosets of the subgroup $\langle4\rangle$ of $\mathbb{Z}_{12}$ are $\{0,4,8\}, \{1,5,9\}, \{2,6,10\},$ and $\{3,7,11\}$ .

Notice how none of the left cosets have an element in common. Additionally, every element of $\mathbb{Z}_{12}$ appears in one of these left cosets. The same is true for the right cosets. This is what it means to say that the left or right cosets partition the group that they are subsets of. This now brings us to the Theorem of Lagrange which is developed from this idea that cosets partition the group.

Theorem of Lagrange

This theorem is stated by Fraleigh and Katz as follows, and the proof is one I wrote based off of the proof given by Fraleigh and Katz [7].

Theorem V.9 (Theorem of Lagrange)

Let $H$ be a subgroup of a finite group $G$ . Then, the order of $H$ is a divisor of the order of $G$ .

Proof.

The order of a group is simply the number of elements in the group. So, what we need to show is that the number of elements in $H$ is a divisor of the number of elements in $G$ .

We have that every left (or right) coset of $H$ has the same number of elements as $H$ . Because the union of the left cosets is $G$ and their intersection is the empty set, we have

$x+x+...+x=y$

where $x$ and $y$ are the orders of $H$ and $G$ respectively. Thus, $nx=y$ for some $n\in \mathbb{N}$ . Therefore, $x$ is a divisor of $y$ .

The contrapositive (see Part II: Chapter 11) of this theorem is true. That is, any subset $H$ of a group $G$ with an order that is not a divisor of the order of $G$ is not a subgroup of $G$ . However, the converse of this theorem is not true. If the order of $H$ is a divisor of the order of $G$ , it does not necessarily follow that $H$ is a subgroup of $G$ .

Let’s look at an example that I wrote that puts these ideas into practice.

Example 53
Consider the group

$G=\mathbb{Z}_{16}=\{0,1,2,...,14,15\}$

under addition modulo 16. Are the sets

$A=\{2z:z<7 \text{ is a positive integer}\}$

and

$B=\{2z+1:z<8 \text{ is a nonnegative integer}\}$

subgroups of $G$ ?

Solution
We have that

$A=\{2z:z<7 \text{ is a positive integer}\}=\{2,4,6,8,10,12\}.$

Because the order of $A$ is 6 and the order of $G$ is 16, $A$ is not a subgroup of $G$ by the Theorem of Lagrange.

Additionally, we have that

$B=\{2z+1:z<8 \text{ is a nonnegative integer}\}=\{1,3,5,7,9,11,13,15\}.$

Because the order of $B$ is 8 and the order of $G$ is 16, the Theorem of Lagrange is inconclusive.

By definition V.3, $B$ is a subgroup of $G$ if and only if it is a subset of $G$ that is a group under the binary operation of $G$ . Because every element in $B$ is in $G$ , it is a subset. However, the identity element for addition modulo 16 is 0, which is not an element of $B$ . Thus, $B$ is not a group under the binary operation of $G$ . Therefore, $B$ is not a subgroup of $G$ .

We have seen how the Theorem of Lagrange grew from the idea of cosets. Now, we are going to discuss another topic that would not be possible without cosets, factor groups.

Factor Groups

A factor group is $G/H$ where $H$ is a subgroup of $G$ [7]. Because the elements of $G/H$ are just the left cosets of $H$ , we are more concerned with finding what group a factor group is isomorphic to than we are with finding what the actual elements of a factor group are.

To compute a factor group is to find a group that the factor group is isomorphic to [7]. The best way to understand how to compute a factor group is to see an example so we will work through one now. This is a homework problem that I completed for this paper that comes from Fraleigh and Katz [7].

Example 54
Compute the factor group $2\mathbb{Z}/8\mathbb{Z}$ , where $2\mathbb{Z}$ and $8\mathbb{Z}$ are defined under addition.

Solution
We first need to show that $8\mathbb{Z}$ is a subgroup of $2\mathbb{Z}$ . By definition V.3, $8\mathbb{Z}$ must be a group under the binary operation of $2\mathbb{Z}$ . We are given that both
$8\mathbb{Z}$ and $2\mathbb{Z}$ are defined under addition. Moreover, they are both groups under addition by example 49. All that remains left to show is that every element of $8\mathbb{Z}$ is in $2\mathbb{Z}$ . That is, we need to show that any integer that is a multiple of 8 is also a multiple of 2. We have that for an element $a\in 8\mathbb{Z}$ ,

$\begin{align*} a&=8n &&\text{where $n$ is an integer}\\ &=2\cdot 4\cdot n\\ &=2\cdot 4n &&\text{Associativity holds in the integers} \end{align*}$

Because $n$ and 4 are integers and the integers are closed under multiplication, $4n$ is an integer. Thus, if an integer is a multiple of 8, it is also a multiple of 2. Therefore, we can conclude that $8\mathbb{Z}$ is a subgroup of $2\mathbb{Z}$ .

Now, we need to find all left cosets of the subgroup $8\mathbb{Z}$ of $2\mathbb{Z}$ . Because $2\mathbb{Z}$ is the group of all integers that are multiples of 2, we need to add $8\mathbb{Z}$ to every even integer to get the left cosets. We have then

$\begin{align*} 0+8\mathbb{Z}&=\{...,-16,-8,0,8,16,...\}=8n+8\mathbb{Z}\\ 2+8\mathbb{Z}&=\{...,-14,-6,2,10,18,...\}=(8n+2)+8\mathbb{Z}\\ 4+8\mathbb{Z}&=\{...,-12,-4,4,12,20,...\}=(8n+4)+8\mathbb{Z}\\ 6+8\mathbb{Z}&=\{...,-10,-2,6,14,22,...\}=(8n+6)+8\mathbb{Z}\\ \end{align*}$

where $n\in\mathbb{Z}$ . So, we have that

$\frac{2\mathbb{Z}}{8\mathbb{Z}}=\{0+8\mathbb{Z}, 2+8\mathbb{Z}, 4+8\mathbb{Z},6+8\mathbb{Z}\},$

which is a group under the binary operation

$(a+8\mathbb{Z})\ast(b+8\mathbb{Z})=(a+b)\text{mod(8)}+8\mathbb{Z}.$

Because $2\mathbb{Z}/8\mathbb{Z}$ has four elements, we know that any group isomorphic to it must have four elements as well. This is because an isomorphism is a one-to-one, onto mapping, which means that there is a one-to-one correspondence between the elements of the domain and codomain (see Part III: Chapter 15). Thus, the domain and codomain (when they are finite) have the same number of elements. One group that has four elements is $\mathbb{Z}_4$ with the binary operation

$(a\ast b)=(a+b)\text{mod(4)}.$

Now, we need to see if we can find an isomorphism $\phi$ from $\mathbb{Z}_4$ to $2\mathbb{Z}/8\mathbb{Z}$ . Let $\phi(z)=2z+8\mathbb{Z}.$ It is easy to show that $\phi$ is one-to-one and onto.

One-to-One

$\begin{align*} \phi(a)&=\phi(b)\\ 2a+8\mathbb{Z}&=2b+8\mathbb{Z}\\ 2a&=2b\\ a&=b \end{align*}$

Onto

$\begin{align*} b+8\mathbb{Z}&=\phi(a)\\ &=2a+8\mathbb{Z}\\ b&=2a\\ \frac{b}{2}&=a \end{align*}$

Additionally, we have

$\begin{align*} \phi(a\ast b)&=\phi((a+b)\text{mod(4)})\\ &=2((a+b)\text{mod(4)})+8\mathbb{Z}\\ &=2(a+b-4d)+8\mathbb{Z} &&\text{where $d$ is an integer}\\ &=2a+2b-8d+8\mathbb{Z}\\ &=(2a+2b)\text{mod(8)}+8\mathbb{Z}\\ &=(2a+8\mathbb{Z})\ast(2b+8\mathbb{Z})\\ &=\phi(a)\ast \phi(b) \end{align*}$

Therefore, $\displaystyle\frac{2\mathbb{Z}}{8\mathbb{Z}}$ is isomorphic to $\mathbb{Z}_4$ .

We have now seen two examples of how to find an isomorphism, each one demonstrating how long this process can be. Thankfully, we do have some shortcuts, such as the FTFGAG.

FTFGAG

FTFGAG stands for the Fundamental Theorem of Finitely Generated Abelian Groups. This theorem is given by Macauley as follows [17].

Theorem V.10 (FTFGAG)

Every finite abelian group $G$ is isomorphic to a direct product of cyclic groups, i.e., for some integers $n_1,n_2,...,n_m,$

$A\simeq\mathbb{Z}_{n_1}\times\mathbb{Z}_{n_2}\times...\times\mathbb{Z}_{n_m},$

where each $n_i$ is a prime power, i.e., $n_i=p_i^{d_i}$ , where $p_i$ is prime and $d_i\in\mathbb{N}$ .

There are a couple terms in this theorem that we have not yet covered. A group $G$ under $\ast$ is abelian if and only if $\ast$ is commutative for all elements in $G$ . A group $G$ is cyclic if and only if there exists an element in $G$ that generates all of $G$ [17]. For example, $\mathbb{Z}_n$ is cyclic for all natural numbers $n>1$ because

$\mathbb{Z}_n=\langle 1 \rangle=\{0,1,2,3,...,n-1\}.$

So, every finite abelian group is isomorphic to a direct product of cyclic groups. More specifically, the cyclic groups are of the form $\mathbb{Z}_n$ where $n$ is a prime number raised to some power. Let’s demonstrate this using an example. This exercise is a homework problem I completed for Modern Algebra [18].

Example 55
List every direct product of cyclic groups that is isomorphic to an abelian group of order 54.

Solution
Finite isomorphic groups have the same order because an isomorphism is a one-to-one, onto mapping (see Part III: Chapter 15). Because the order of $\mathbb{Z}_{n_1}\times \mathbb{Z}_{n_2}\times...\times\mathbb{Z}_{n_m}$ is equal to $n_1\cdot n_2\cdot ...\cdot n_m$ , we need to find every combination of products of primes to some power that equal 54. Because $54=2\cdot 3^3$ , we have the following combinations:

$\begin{align*} \mathbb{Z}_2\times&\mathbb{Z}_3\times\mathbb{Z}_3\times\mathbb{Z}_3\\ \mathbb{Z}_2&\times\mathbb{Z}_3\times\mathbb{Z}_9\\ &\mathbb{Z}_2\times\mathbb{Z}_{27}.\\ \end{align*}$

Therefore, every finite abelian group of order 54 is isomorphic to one of these direct products of cyclic groups by the FTFGAG.

Listing out all of the possible direct products becomes a long task when an abelian group is of an order that is the product of many primes. But if we only need to find one group that is isomorphic to a particular abelian group, this theorem is extremely useful.

We have covered a good amount of information relating to the algebraic binary structure known as a group. We covered cyclic subgroups, isomorphisms, cosets, and factor groups. We also discussed a couple of important and useful theorems, the Theorem of Lagrange and the FTFGAG. We are now going to turn our attention to other types of algebraic binary structures. These are rings, integral domains, and fields.