Abstract Algebra/Group Theory/Group

In this section we will begin to make use of the definitions we made in the section about binary operations. In the next few sections, we will study a specific type of binary structure called a group. First, however, we need some preliminary work involving a less restrictive type of binary structure.

Monoids

Definition 1: A monoid is a binary structure $(M,\ast )$ satisfying the following properties:

(i) $(a\ast b)\ast c=a\ast (b\ast c)$ for all $a,b,c\in M$. This is defined as associativity.

(ii) There exists an identity element $e\in M$ such that $a\ast e=a=e\ast a$ for all $a\in M$.

Now we have our axioms in place, we are faced with a pressing question; what is our first theorem going to be? Since the first few theorems are not dependent on one another, we simply have to make an arbitrary choice. We choose the following:

Theorem 2: The identity element of $M$ is unique.

Proof: Assume $e$ and $e^{\prime }$ are both identity elements of $M$. Then both satisfy condition (ii) in the definition above. In particular, $e=e\ast e^{\prime }=e^{\prime }$, proving the theorem. ∎

This theorem will turn out to be of fundamental importance later when we define groups.

Theorem 3: If $a_1,a_2,\dots ,a_n$ are elements of $M$ for some $n\in \mathbb{N}$, then the product $a_1\ast a_2\ast \cdots \ast a_n$ is unambiguous.

Proof: We can prove this by induction. The cases for $n=1$ and $n=2$ are trivially true. Assume that the statement is true for all $n<k$. For $n=k$, the product $a_1\ast \cdots \ast a_k$, inserting parentheses, can be "partitioned" into $(a_1\ast \cdots \ast a_i)\ast (a_{i+1}\ast \cdots \ast a_k)$. Both parts of the product have a number of elements less than $k$ and are thus unambiguous. The same is true if we consider a different "partition", $(a_1\ast \cdots \ast a_j)\ast (a_{j+1}\ast \cdots \ast a_k)$, where $j>i$. Thus, we can unambiguously compute the products $(a_1\ast \cdots \ast a_i)$, $(a_{i+1}\ast \cdots \ast a_j)$, and $(a_{j+1}\ast \cdots \ast a_k)$, and rewrite the two "partitions" as $b_1\ast (b_2\ast b_3)$ and $(b_1\ast b_2)\ast b_3$. These equal each other by the definition of a monoid.∎

This is about as far as we are going to take the idea of a monoid. We now proceed to groups.

Groups

Definition 4: A group is a monoid $(G,\ast )$ that also satisfies the property

(iii) For each $a\in G$, there exists an element $a^{\prime }\in G$ such that $a\ast a^{\prime }=a^{\prime }\ast a=e$.

Such an element $a^{\prime }$ is called an inverse of $a$. When the operation on the group is understood, we will conveniently refer to $(G,\ast )$ as $G$. In addition, we will gradually stop using the symbol $\ast$ for multiplication when we are dealing with only one group, or when it is understood which operation is meant, instead writing products by juxtaposition, $a\ast b\equiv ab$.

Remark 5: Notice how this definition depends on Theorem 2 to be well defined. Therefore, we could not state this definition before at least proving uniqueness of the identity element. Alternatively, we could have included the existence of a distinguished identity element in the definition. In the end, the two approaches are logically equivalent.

Also note that to show that a monoid is a group, it is sufficient to show that each element has either a left-inverse or a right-inverse. Let $a\in G$, let $b$ be a right-inverse of $a$, and let $c$ be a right-inverse of $b$. Then, $a=a(bc)=(ab)c=c$. Thus, any right-inverse is also a left-inverse, or $ab=ba$. A similar argument can be made for left-inverses.

Theorem 6: The inverse of any element is unique.

Proof: Let $g\in G$ and let $g^{\prime }$ and $g^{″}$ be inverses of $g$. Then, $g^{\prime }=g^{\prime }\ast e=g^{\prime }\ast g\ast g^{″}=e\ast g^{″}=g^{″}$. ∎

Thus, we can speak of the inverse of an element, and we will denote this element by $a^{-1}$. We also observe this nice property:

Corollary 7: $(a^{-1}{)}^{-1}=a$.

Proof: This follows immediately since $a\ast a^{-1}=a^{-1}\ast a=e$.

The next couple of theorems may appear obvious, but in the interest of keeping matters fairly rigorous, we allow ourselves to state and prove seemingly trivial statements.

Theorem 8: Let $G$ be a group and $a,b\in G$. Then $(a\ast b{)}^{-1}=b^{-1}\ast a^{-1}$.

Proof: The result follows by direct computation: $(a\ast b)\ast (b^{-1}\ast a^{-1})=a\ast b\ast b^{-1}\ast a^{-1}=a\ast e\ast a^{-1}=a\ast a^{-1}=e$. ∎

Theorem 9: Let $a,b,c\in G$. Then, $a\ast b=a\ast c$ if and only if $b=c$. Also, $a\ast c=b\ast c$ if and only if $a=b$.

Proof: We will prove the first assertion. The second is identical. Assume $a\ast b=a\ast c$. Then, multiply on the left by $a^{-1}$ to obtain $b=c$. Secondly, assume $b=c$. Then, multiply on the left by $a$ to obtain $a\ast b=a\ast c$. ∎

Theorem 10: The equation $a\ast x=b$ has a unique solution in $G$ for any $a,b\in G$.

Proof: We must show existence and uniqueness. For existence, observe that $x=a^{-1}\ast b$ is a solution in $G$. For uniqueness, multiply both sides of the equation on the left by $a^{-1}$ to show that this is the only solution. ∎

Notation: Let $G$ be a group and $a\in G$. We will often encounter a situation where we have a product $\underset{\mathrm{n}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{m}\mathrm{s}}{\underbrace{a\ast a\ast \cdots \ast a}}$. For these situations, we introduce the shorthand notation $a^n=\underset{\mathrm{n}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{m}\mathrm{s}}{\underbrace{a\ast a\ast \cdots \ast a}}$ if $n$ is positive, and $a^n=\underset{\mathrm{|}\mathrm{n}\mathrm{|}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{m}\mathrm{s}}{\underbrace{a^{-1}\ast a^{-1}\ast \cdots \ast a^{-1}}}$ if $n$ is negative. Under these rules, it is straightforward to show $g^n\ast g^m=g^{n+m}$ and $(a^n{)}^{-1}=a^{-n}$ and $a^0=e$ for all $a\in G$.

Definition 11: (i) The order of a group $G$, denoted $|G|$ or $o(G)$, is the number of elements of $G$ if $G$ is finite. Otherwise $|G|$ is said to be infinite.

(ii) The order of an element of $g\in G$, similarly denoted $|g|$ or $o(g)$, is defined as the lowest positive integer $n$ such that $g^n=e$ if such an integer exists. Otherwise $|g|$ is said to be infinite.

Theorem 12: Let $G$ be a group and $a,b\in G$. Then $|ab|=|ba|$.

Proof: Let the order of $ab$ be $n$. Then, $(ab{)}^n=abab\mathrm{...}ab=e$, $n$ being the smallest positive integer such that this is true. Now, multiply by $b$ on the left and $a$ on the right to obtain $(ba{)}^{n+1}=ba$ implying $(ba{)}^n=e$. Thus, we have shown that $|ba|\le |ab|$. A similar argument in the other direction shows that $|ab|\le |ba|$. Thus, we must have $|ab|=|ba|$, proving the theorem. ∎

Corollary 13: Let $G$ be a group with $a,b\in G$. Then, $|ab{a}^{-1}|=|b|$.

Proof: By Theorem 12, we have that $|ab{a}^{-1}|=|b{a}^{-1}a|=|be|=|b|$. ∎

Theorem 14: An element of a group not equal to the identity has order 2 if and only if it is its own inverse.

Proof: Let $g$ have order 2 in the group $G$. Then, $g^2=gg=e$, so by definition, $g^{-1}=g$. Now, assume $g^{-1}=g$ and $g\ne e$. Then $e=g{g}^{-1}=gg=g^2$. Since $g\ne e$, 2 is the smallest positive integer satisfying this property, so $g$ has order 2. ∎

Definition 15: Let $G$ be a group such that for all $a,b\in G$, $ab=ba$. Then, $G$ is said to be commutative or abelian.

When we are dealing with an abelian group, we will sometimes use so-called additive notation, writing $+$ for our binary operation and replacing $a^n$ with $na$. In such cases, we only need to keep track of the fact that $n$ is an integer while $a$ is a group element. We will also talk about the sum of elements rather than their product.

Abelian groups are in many ways nicer objects than general groups. They also admit more structure where ordinary groups do not. We will see more about this later when we talk about structure-preserving maps between groups.

Definition 16: Let $G$ be a group. A subset $S\subseteq G$ is called a generating set for $G$ if every element in $G$ can be written in terms of elements in $S$. We write $G=⟨S⟩=\{s_1^{e_1}\dots s_m^{e_m}\mid s_i\in S,e_i\in \{1,-1\}\}$.

Now that we have our definitions in place and have a small arsenal of theorems, let us look at three (really, two and a half) important families of groups.

Multiplication tables

We will now show a convenient way of representing a group structure, or more precisely, the multiplication rule on a set. This notion will not be limited to groups only, but can be used for any structure with any number of operations. As an example, we give the group multiplication table for the Klein 4-group $K_4$. The multiplication table is structured such that $g\ast h$ is represented by the element in the "$(g,h)$-position", that is, in the intersection of the $g$-row and the $h$-column.

$\begin{array}{ccccc}\ast & e& a& b& c\\ e& e& a& b& c\\ a& a& e& c& b\\ b& b& c& e& a\\ c& c& b& a& e\end{array}$

This next group is for the group of integers under addition modulo 4, called ${\mathbb{Z}}_4$. We will learn more about this group later.

$\begin{array}{ccccc}+& 0& 1& 2& 3\\ 0& 0& 1& 2& 3\\ 1& 1& 2& 3& 0\\ 2& 2& 3& 0& 1\\ 3& 3& 0& 1& 2\end{array}$

We can clearly see that $K_4$ and ${\mathbb{Z}}_4$ are "different" groups. There is no way to relabel the elements such that the multiplication tables coincide. There is a notion of "equality" of groups that we have not yet made precise. We will get back to this in the section about group homomorphisms.

The reader might have noticed that each row in the group table features each element of the group exactly once. Indeed, assume that an element $k\in G$ appeared twice in some row of the multiplication table for $G$. Then there would exist $g,h,h^{\prime }\in G$ such that $gh=g{h}^{\prime }$, implying $h=h^{\prime }$ and contradicting the assumption of $k$ appearing twice. We state this as a theorem:

Theorem 17: Let $G$ be a group and $a\in G$. Then $aG=\{a\ast g\mid g\in G\}=G$.

Using this, the reader can use a multiplication table to find all groups of order 3. He/she will find that there is only one possibility.

Problems

Problem 1: Show that $M_n(ℝ)$, the set of $n\times n$ matrices with real entries, forms a group under the operation of matrix addition.

Problem 2: Let $V,W$ be vector spaces and $\mathrm{H}\mathrm{o}\mathrm{m}(V,W)$ be the set of linear maps from $V$ to $W$. Show that $\mathrm{H}\mathrm{o}\mathrm{m}(V,W)$ forms an abelian group by defining $(f+g)(v)=f(v)+g(v)$.

Problem 3: Let ${\mathbb{Q}}_8$ be generated by the elements $i,j,k,m$ such that $i^2=j^2=k^2=m$, $m^2=e$ and $ij=mji=k$. Show that ${\mathbb{Q}}_8$ forms a group. Are any of the conditions above redundant? When the identity e is written 1 and m = −1, then ${\mathbb{Q}}_8$ is called the quaternion group. The $i,j,k$ are imaginary units. Using 1 and one of these as a basis for a number plane results in the complex number plane.

Problem 4: Let $S$ be any nonempty set and consider the set $G^S$. Show that $G^S$ has a natural group structure.

Answer

$G^S$ is the set of functions $f:S\to G$. Let $f_1,f_2\in G^S$ and define the binary operation $(f_1\ast f_2)(x)=f_1(x)f_2(x)$ for all $x\in S$. Then $G^S$ is a group with identity $0$ such that $0(x)=e$ for all $x\in X$ and inverses $f^{-1}(x)=f(x{)}^{-1}$ for all $f\in G^S,x\in X$.∎

Problem 5: Let $G$ be a group with two distinct elements $a$ and $b$, both of order 2. Show that $G$ has a third element of order 2.

Answer

We consider first the case where $ab=ba$. Then $(ab{)}^2=abab=aabb=e$ and $ab$ is distinct from $e,a$ and $b$. If $ab\ne ba$, then $(ab{a}^{-1}{)}^2=aba{a}^{-1}b{a}^{-1}=abb{a}^{-1}=e$ and $ab{a}^{-1}$ is distinct from $e,a$ and $b$. ∎

Problem 6: Let $G$ be a group with one and only one element $f$ of order 2. Show that $\prod _{g\in G}g=f$.

Answer

Since the product of two elements generally depends on the order in which we multiply them, the stated product is not necessarily well-defined. However, it works out in this case.

Since every element of $G$ appears once in the product, for every element $g\in G$, the inverse of $g$ must appear somewhere in the product. That, is, unless $|g|=2$ in which case $g$ is its own inverse by Theorem 14. Now, applying Corollary 13 to the product shows that its order is that same as the order of the product of all elements of order 2 in $G$. But there is only one such element, $f$, so the order of the product is 2. Since the only element in $G$ having order 2 is $f$, the equality follows. ∎