First Proofs: Uniqueness, Cancellation and Inverses

Now we prove things from the group axioms alone. Every fact in this lesson then holds in every group at once: for integers, matrices, permutations and symmetries alike. The proofs are short, and they show the habits of a good proof: name your objects, justify each equals sign by an axiom or an earlier result, and end exactly at the claim.

The identity and inverses are unique

Why it works

Theorem. A group has exactly one identity, and each element has exactly one inverse.
Proof. Suppose $e$ and $e'$ are both identities. Then $e = ee'$ (because $e'$ is an identity) and $ee' = e'$ (because $e$ is an identity), so $e = e'$.
Suppose $b$ and $c$ are both inverses of $a$. Then $$b = be = b(ac) = (ba)c = ec = c,$$ using in turn: $e$ is the identity, $ac = e$, associativity, $ba = e$, and the identity again. $\square$

Because inverses are unique, to show that some $y$ is the inverse of $x$, it is enough to check $xy = e$. (From $xy = e$, multiply on the left by $x^{-1}$ to get $y = x^{-1}$.)

Cancellation and solving equations

Theorem. If $ab = ac$ then $b = c$, and if $ba = ca$ then $b = c$. The equations $ax = b$ and $xa = b$ have the unique solutions $x = a^{-1}b$ and $x = ba^{-1}$.

Proof of left cancellation. Multiply $ab = ac$ on the left by $a^{-1}$: $a^{-1}(ab) = a^{-1}(ac)$. By associativity $(a^{-1}a)b = (a^{-1}a)c$, so $eb = ec$ and $b = c$. $\square$ A consequence you can see: in a Cayley table, each row and each column contains every element exactly once.

The inverse of a product

Theorem. $(ab)^{-1} = b^{-1}a^{-1}$, and $(a^{-1})^{-1} = a$.
Proof. By associativity, $(ab)(b^{-1}a^{-1}) = a(bb^{-1})a^{-1} = aea^{-1} = aa^{-1} = e$, so $b^{-1}a^{-1}$ is the inverse of $ab$. And $a^{-1}a = e$ says $a$ is the inverse of $a^{-1}$. $\square$ Think of socks and shoes: you put on socks, then shoes, and take them off in the reverse order.

Worked example: Socks and shoes with matrices

In the group of $2 \times 2$ integer matrices with determinant $1$, let $A = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$ and $B = \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}$. Find $(AB)^{-1}$.

$A^{-1} = \begin{pmatrix} 1 & -1 \\ 0 & 1 \end{pmatrix}$ and $B^{-1} = \begin{pmatrix} 1 & 0 \\ -1 & 1 \end{pmatrix}$.
$(AB)^{-1} = B^{-1}A^{-1} = \begin{pmatrix} 1 & -1 \\ -1 & 2 \end{pmatrix}$.
Check: $AB = \begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix}$, and $\begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix}\begin{pmatrix} 1 & -1 \\ -1 & 2 \end{pmatrix} = I$.
The other order, $A^{-1}B^{-1} = \begin{pmatrix} 2 & -1 \\ -1 & 1 \end{pmatrix}$, is the inverse of $BA$, not of $AB$.

Answer: $\begin{pmatrix} 1 & -1 \\ -1 & 2 \end{pmatrix}$.

Worked example: A short proof

Prove: if $x^{2} = e$ for every $x$ in a group $G$, then $G$ is abelian.

Let $a, b \in G$. Since $x^{2} = e$ means $x = x^{-1}$, every element is its own inverse.
So $ab = (ab)^{-1}$, because $ab$ is an element too.
$(ab)^{-1} = b^{-1}a^{-1} = ba$.
Therefore $ab = ba$. $\square$

Answer: $ab = (ab)^{-1} = b^{-1}a^{-1} = ba$.

The Cayley table of $S_{3}$, the six rearrangements of $\{1, 2, 3\}$ (a later lesson explains the notation). Cancellation is why every row and column contains each element once. The two products you tapped differ: this group is not abelian, so the order in $(ab)^{-1} = b^{-1}a^{-1}$ matters.

Common mistake

Writing $(ab)^{-1} = a^{-1}b^{-1}$. That would need $ba^{-1} = a^{-1}b$, which is only true when $a$ and $b$ commute; the matrix example shows it failing. Another slip: cancelling on opposite sides. From $ab = ca$ you can only conclude $c = aba^{-1}$, not $b = c$. To check a proof, ask of each equals sign: which axiom, applied on which side?

Key idea

From the axioms alone: the identity and inverses are unique, you may cancel on the same side ($ab = ac \Rightarrow b = c$), $ax = b$ has the unique solution $x = a^{-1}b$, and $(ab)^{-1} = b^{-1}a^{-1}$.

First Proofs: Uniqueness, Cancellation and Inverses

The identity and inverses are unique

Cancellation and solving equations

The inverse of a product

Practice

Builds on

Leads to

Sources and further reading