Back to the on-screen lesson ·
Uniqueness of the identity and of inverses, cancellation, the reversing rule for the inverse of a product, and why the left-hand and right-hand versions of each are different statements.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to prove that a group's identity and each element's inverse are unique, cancel correctly and say why cancelling across sides is not allowed, invert a product by reversing its factors, solve $ax = b$ and $xa = b$ and explain why the answers differ, and recognise where an argument has quietly assumed that the group commutes.
The definition promises an identity and an inverse for each element. It does not promise that there is only one of either, it says nothing about cancelling, and it says nothing about how to undo a product of two things. All of that is true in every group, and all of it has to be proved — from the four conditions and nothing else.
Left cancellation is the step from $ab = ac$ to $b = c$; right cancellation is the step from $ba = ca$ to $b = c$. The reversing rule (sometimes called socks and shoes) is $(ab)^{-1} = b^{-1}a^{-1}$. An element with $x^{-1} = x$ is self-inverse, equivalently $x^{2} = e$.
1. The identity is unique. If $e$ and $f$ are both identities, then $ef = f$ (because $e$ is one) and $ef = e$ (because $f$ is one), so $e = f$.
2. Inverses are unique. If $ab = ba = e$ and $ac = ca = e$ then
$$b = be = b(ac) = (ba)c = ec = c.$$
So each element has exactly one inverse, and writing $a^{-1}$ is legitimate.
3. Cancellation. From $ab = ac$, multiply on the left by $a^{-1}$: $b = c$. From $ba = ca$, multiply on the right. Note what is not claimed: $ab = ca$ gives nothing, because the repeated element is on different sides.
4. The reversing rule. $(ab)^{-1} = b^{-1}a^{-1}$, because $(ab)(b^{-1}a^{-1}) = a(bb^{-1})a^{-1} = e$. Also $(a^{-1})^{-1} = a$, since $a$ inverts $a^{-1}$ and inverses are unique.
5. Equations have unique solutions. $ax = b$ has the single solution $x = a^{-1}b$, and $xa = b$ has the single solution $x = ba^{-1}$. The two are different elements unless the group is abelian.
An algebraic fact is about the operation, not about what the elements happen to be. The same group turns up as rotations of a square, as residues under addition and as matrices, and a theorem proved once from the axioms holds in all of them at once.
Another way: picture
Socks and shoes. To get dressed you put on socks and then shoes; to undo that you take off the shoes and then the socks. The inverse of socks then shoes is shoes off then socks off — the same two operations, in the opposite order. That is the whole content of $(ab)^{-1} = b^{-1}a^{-1}$, and it is why the rule is remembered by that name.
Another way: steps
To manipulate an equation in a group: 1. Decide which side you need to multiply on, and multiply both sides on that same side. 2. Use associativity to regroup, and only then cancel a pair $aa^{-1}$. 3. When a product is inverted, reverse the order of the factors. 4. Never move two elements past each other unless the group is known to be abelian.
Almost every mistake in this part of the subject is the same mistake: treating $ab$ and $ba$ as the same element. The axioms do not allow it, and the consequences are concrete.
$ax = b$ and $xa = b$ have different solutions, $a^{-1}b$ and $ba^{-1}$. In the symmetries of a square, with $r$ the quarter turn and $s$ a reflection, $rs$ and $sr$ are different reflections, so $rx = s$ and $xr = s$ genuinely name different symmetries.
Cancellation has the same one-sided character. From $ab = ac$ you may cancel, because $a$ is on the left in both. From $ab = ca$ you may not: multiplying on the left by $a^{-1}$ gives $b = a^{-1}ca$, which is a different element from $c$ in general. That expression, $a^{-1}ca$, is called a conjugate of $c$, and it will be the central object of the lesson on normal subgroups. For now it is enough to notice that it appears exactly where an illegal cancellation was attempted.
One consequence of cancellation is worth stating on its own, because the Cayley table lesson used it: the map $x \mapsto ax$ is a bijection of the group onto itself. It is injective by cancellation and surjective because $a^{-1}b$ maps to $b$. That is why every row of a Cayley table is a rearrangement of the elements, and it is the single most reused counting fact in this course.
Writing $(ab)^{-1} = a^{-1}b^{-1}$. True in an abelian group, false in general, and it looks exactly like ordinary algebra, which is why it is rarely noticed.
Cancelling across sides. $ab = ca$ does not give $b = c$. What it gives is $b = a^{-1}ca$.
Expanding $(ab)^{n}$ as $a^{n}b^{n}$. $(ab)^{2} = abab$, and turning that into $a^{2}b^{2}$ requires swapping $b$ and $a$.
Assuming an element that squares to the identity must be the identity. Any reflection does. In a group of even order there is always at least one such element.
Thinking uniqueness of inverses is an axiom. It is a theorem, and it is why the notation $a^{-1}$ is allowed to name a specific element at all.
In the symmetries of a square, let $r$ be the quarter turn and $s$ a reflection. Solve $rx = s$.
One equation, one unknown.
Multiply on the left by $r^{-1}$, which is the three-quarter turn: $x = r^{-1}s$.
Multiply on the left, because $r$ is on the left.
The other equation $xr = s$ gives $x = sr^{-1}$, a different reflection. The two answers coincide only when the group is abelian, and this one is not.
Side matters.
In $U(9) = \{1, 2, 4, 5, 7, 8\}$, suppose $2x = 2y$. Multiplying by $2^{-1} = 5$ gives $x = y$.
Cancellation, with a concrete inverse.
The same statement in the ring of all residues modulo $9$ is false: $3 \times 2 = 3 \times 5$ but $2 \ne 5$.
Cancellation needs an inverse, not just a product.
That is the difference between a group and a mere set with multiplication, and in unit 5 it becomes the difference between a field and a ring with zero divisors.
The same fact, met twice.
$a^{2} = e$ says $aa = e$, which is exactly the statement that $a$ is an inverse of $a$.
Read the equation as a sentence about inverses.
Inverses are unique, so the inverse of $a$ is $a$ itself.
Uniqueness does the rest.
The converse holds too: multiply $a = a^{-1}$ on the right by $a$. So self-inverse and squares to the identity are the same condition, and either form may be used.
In the integers modulo $8$ under addition, write down the inverse of each element.
| Its inverse | |
|---|---|
| 1 | |
| 2 | |
| 3 |
Put the proof that an element of a group has only one inverse into order.
Number the steps in order (write the number in the box):
Select every step that is valid in every group.
This task has no paper form; do it on a device.
Match each expression in a group to what it is equal to.
| $b^{-1}a^{-1}$ | $a$ | $a^{-1}b$ | $c^{-1}b^{-1}a^{-1}$ | $ba^{-1}$ | |
|---|---|---|---|---|---|
| $(a^{-1})^{-1}$ | |||||
| $(ab)^{-1}$ | |||||
| $(abc)^{-1}$ | |||||
| the solution of $ax = b$ |
In a group that is not assumed abelian, what is $(ab)^{-1}$?
Build the proof that a group in which every element satisfies $x^{2} = e$ is abelian.
This task has no paper form; do it on a device.
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
In the integers modulo $12$ under addition, write down the inverse of each element.
| Its inverse | |
|---|---|
| 1 | |
| 2 | |
| 3 |
You can cancel, invert a product and solve an equation in a group, without assuming it is abelian. Say in your own words why the inverse of $ab$ is $b^{-1}a^{-1}$ rather than $a^{-1}b^{-1}$. Next: the order of an element, which is the first number a group hands you.
9. Your turn: in a group, show that if $a^{2} = e$ then $a = a^{-1}$., step 3