Manipulating equations in groups

This is a survey article related to:group
View other survey articles about group

This article is about the basic rules of manipulating equations involving groups. It is written mainly for beginners to the subject of group theory.

A quick review of expressions in groups

Because of associativity of the group multiplication, we often omit the multiplication symbol. Thus, if $G$ is a group with multiplication operation $*$, we often write $abc$ for the product $a*(b*c)$. For expressions of the form $aaa\ldots a$, we simply write $a^{n}$. The identity element is denoted here as $e$ (it is also sometimes denoted as $1$).

The rules for manipulating equations in groups

A typical equation in groups has two expressions, with an equality sign between them. We can do the following:

• left multiply by an element: Both sides of the equation are multiplied on the left by the same element. For instance, the equation $B=C$ (for expressions $B,C$ involving elements of the group) can be multiplied on the left by an expression $A$ to get $AB=AC$.
• right multiply by an element: Both sides of the equation are multiplied on the right by the same element. For instance, the equation $B=C$ (for expressions $B,C$ involving elements of the group) can be multiplied on the right by an expression $A$ to get $BA=CA$.
• multiply two equations: Two equations are multiplied with each other. We need to decide in advance which of the equations is the left equation and which equation is the right equation. For instance, for expressions $A,B,C,D$, the equations $A=B$ and $C=D$ can be multiplied to give any of the equations $AC=BD,AD=BC,CA=DB,CB=DA$ but cannot be used to give $AC=DB$.

These are the basic operations.

Cancellation of elements

If we have an equation of the form:

$ab=ac$

we can cancel $a$ from the left, to conclude that $b=c$. The formal justification of this is via left multiplication by $a^{-1}$. Similarly, if $ba=ca$, we can cancel $a$ from the right, to conclude that $b=c$.

However, if we have an equation like:

$ab=ca$

we cannot cancel $a$. This is because $a$ appears on different sides.

Transposing terms to the other side

We can use multiplication by the inverse to transpose a term from one side to another. However, only those terms that are at the left or right end can be transposed. For instance:

$a_{1}a_{2}\ldots a_{n}=b_{1}b_{2}\ldots b_{m}\implies b_{1}^{-1}a_{1}a_{2}\ldots a_{n}=b_{2}b_{3}\ldots b_{m}$

Note that these transposing operations are reversible, because we can multiply back by the element to transpose back. Thus, we in fact have:

$a_{1}a_{2}\ldots a_{n}=b_{1}b_{2}\ldots b_{m}\iff b_{1}^{-1}a_{1}a_{2}\ldots a_{n}=b_{2}b_{3}\ldots b_{m}$

Transposing the term on the right end looks like this:

$a_{1}a_{2}\ldots a_{n}=b_{1}b_{2}\ldots b_{m}\iff a_{1}a_{2}\ldots a_{n}b_{m}^{-1}=b_{1}b_{2}\ldots b_{m-1}$

However, middle terms cannot be directly transposed; we first need to transpose enough terms to bring them to the edge, before transposing them.

Solving for an element

If an element occurs only once in the equation, then all the remaining elements can be gathered to the other side, so we can solve for that element in terms of the other elements. For instance, if we have the equation:

$a_{1}a_{2}a_{3}=b_{1}b_{2}b_{3}$

Then solving it for $b_{2}$ yields:

$b_{2}=b_{1}^{-1}a_{1}a_{2}a_{3}b_{3}^{-1}$

Not every equation can be solved for an element

An equation can be solved for an element only if that element occurs only once, on exactly one side of the equation. Given an equation of the form:

$x^{2}=g$

we cannot solve $x$ in terms of $g$ (there's no generic way of extracting a square root). Even worse, given an equation like:

$x^{2}gxh^{3}=g^{3}$

we cannot isolate the powers of $x$ on one side.

Note that in case the group is an abelian group, we can bring all the powers/occurrences of an element together. For more, see manipulating equations in abelian groups.

Bringing everything to one side

Given any equation, we can transform it to the form where all the terms are on a single side. For instance, the equation:

$a_{1}a_{2}\ldots a_{n}=b_{1}b_{2}\ldots b_{m}$

solves to give:

$b_{m}^{-1}b_{m-1}^{-1}\ldots b_{1}^{-1}a_{1}a_{2}\ldots a_{n}=e$

Inverting a product

Further information: inverse map is involutive

This is a special case of the above, but may be useful to remember in isolation:

$(a_{1}a_{2}\ldots a_{n})^{-1}=a_{n}^{-1}a_{n-1}^{-1}\ldots a_{1}^{-1}$

Equations involving subsets

Arbitrary subsets

If $A,B$ are subsets of a group, then we define:

$AB:=\{ab\mid a\in A,b\in B\}$

If $A$ is a subset and $x$ is an element, we define:

$Ax:=\{ax\mid a\in A\}$

and similarly we define $xA$. Many of the results we have for manipulating elements, work with subsets. For instance:

We define:

$A^{-1}=\{a^{-1}\mid a\in A\}$

The rules for manipulating elements give rules for manipulating subsets. Below, $A,B$ are subsets of a group and $x$ is an element of the same group:

• $(A^{-1})^{-1}=A$
• $(AB)^{-1}=B^{-1}A^{-1}$
• $Ax=B\implies A=Bx^{-1}$
• $xA=B\implies A=x^{-1}B$
• Multiplication of subsets is associative

What's not true

It is not true that $AA^{-1}$ is the identity element. To understand this, note that $AA^{-1}$ is defined as:

$\{a_{1}a_{2}^{-1}\mid a_{1},a_{2}\in A\}$.

In other words, it may well happen that $a_{1}\neq a_{2}$. In this case, it is not true that $a_{1}a_{2}^{-1}$ is the identity element.

Similarly, we can define $A^{2}=AA$ as follows:

$A^{2}=\{a_{1}a_{2}\mid a_{1},a_{2}\in A\}$.

Then, every element of the form $a^{2},a\in A$, is in $A^{2}$, but $A^{2}$ may well be strictly bigger.

In particular, we cannot do manipulations of the form:

$AB=C\implies A=CB^{-1}$

All we can say is that if $B$ is non-empty:

$AB=C\implies A\subseteq CB^{-1}$.

Symmetric subsets

Symmetric subsets are subsets that contain the identity element and are closed under the inverse map. If $A$ and $B$ are symmetric subsets, then:

$(AB)^{-1}=BA$

Subgroups

Subgroups are special kinds of subsets. If $H$ is a subgroup of $G$, then the sets $Hx$ form a partition of $G$, called right cosets, and the sets $xH$ also form a partition of $G$, called left cosets.

Notions of conjugate elements

A useful notion when manipulating equations is that of conjugate elements. Given a group $G$ and elements $g,x\in G$, the conjugate of $x$ by $g$ is the element $gxg^{-1}$ The key point is that conjugation is an automorphism, in the sense that:

$(gxg^{-1})(gyg^{-1})=g(xy)g^{-1}$

and:

$gx^{-1}g^{-1}=(gxg^{-1})^{-1}$

The significance of this is that it allows us to do some computational manipulations.

Using equations to deduce abelianness

There are situations where we can use equations that are true for a group, to deduce that the group is Abelian. Here are some examples.

Inverse map is homomorphism implies group is abelian

Further information: Inverse map is automorphism iff abelian

For a group $G$, if the map $x\mapsto x^{-1}$ is an automorphism, then $G$ is abelian. This is a simple equation-manipulation. If we have:

$(xy)^{-1}=x^{-1}y^{-1}$

Then, multiplying by $xy$ yields:

$e=xyx^{-1}y^{-1}$

Now multiplying by $yx$ yields:

$yx=xy$

Thus showing that $x$ and $y$ commute.

Square map is endomorphism implies group is abelian

Further information: Square map is endomorphism iff abelian

If we have:

$(xy)^{2}=x^{2}y^{2}$

Then expanding this yields:

$xyxy=xxyy$

We can now cancel the leftmost $x$ and rightmost $y$, and get:

$yx=xy$

Cube map is endomorphism...

Further information: Cube map is endomorphism iff abelian (if order is not a multiple of 3)

The map $x\mapsto x^{3}$, the so-called cube map, is an endomorphism only if the group is Abelian, provided the group is finite and its order is not a multiple of 3. This involves a somewhat more tricky argument, that makes use of cancellation, and the fact that if the order is not a multiple of 3, then every element is a cube.

More on manipulating equations

The extent to which we can manipulate systems of equations involving groups is tremendous. In fact, given a group with a system of equations satisfied by some elements, it is in general impossible to determine whether that system of equations forces all elements to be the identity element.

This is related to the notion of a presentation of a group, the question of whether a group has a solvable word problem, and other notions. Usually, many of the apparently unmotivated formal manipulations involving groups actually come from deeper theoretical insights, that are beyond the scope of this survey article.