Groups and the Group Axioms

Let $a*b=a-b$. On $\mathbb Z$ this is a binary operation: the difference of two integers is always an integer, e.g. $3-5=-2\in\mathbb Z$. On $\mathbb N=\{0,1,2,\ldots\}$ the SAME formula is not a binary operation, because $3-5=-2\notin\mathbb N$: the rule sends a pair of inputs from $\mathbb N$ outside $\mathbb N$. Nothing about the formula changed — only the set it is asked to act on.

Write $a\bmod n$ for the remainder when $a$ is divided by $n$ — the hour a clock with $n$ marks shows. Let $a*b=(a+b)\bmod5$ on $S=\{0,1,2,3,4\}$. Take $a=3,b=4$: the ordinary sum is $3+4=7$, and $7\bmod5=2$, so $3*4=2\in S$. To confirm closure for every pair, not just this one, note that a remainder on division by $5$ is always one of $0,1,2,3,4$, so no sum can escape $S$.

Operations do not have to be familiar ones. On $\mathbb Z$ define $a*b=a+b+ab$. Then $2*3=2+3+2\cdot3=11$ and $3*2=3+2+3\cdot2=11$, while $(-1)*5=-1+5+(-1)\cdot5=-1$. It is a binary operation on $\mathbb Z$: sums and products of integers are integers. Working with rules like this one is the best practice there is for the group axioms, because nothing can be checked by habit — every property has to be checked from the formula.

$(\mathbb Z,+)$ is a group. Closure and associativity are familiar; the identity is $e=0$ since $0+a=a+0=a$ for every $a$; and every $a$ has the inverse $-a$, since $a+(-a)=0$, e.g. $7+(-7)=0$. Note that for addition the "inverse" is the negative — the word inverse always means "the element that gets you back to $e$", whatever $e$ is.

$(\mathbb Z,\times)$ is NOT a group. Closure, associativity and the identity $e=1$ all hold, since $1\times a=a$ for every $a$, but the inverse axiom fails: $2$ would need an inverse $x\in\mathbb Z$ with $2x=1$, and no integer satisfies this, since $x=\tfrac12\notin\mathbb Z$. One missing inverse is enough to disqualify the whole structure from being a group. (Enlarge the set to the nonzero rationals and the problem disappears: every $a\ne0$ has the inverse $\tfrac1a$.)

Does $\{0,1,2,3,4,5\}$ under multiplication mod $6$ (multiply, then keep the remainder on division by $6$) form a group? Closure holds, since any remainder lies in the set, $1$ is the identity, and multiplication mod $6$ is associative. But checking inverses for $2$: we need $x$ with $2x\equiv1\pmod6$. Checking all six values, $2\cdot0,2\cdot1,\ldots,2\cdot5\equiv0,2,4,0,2,4\pmod6$ — the value $1$ never appears. So $2$ has no inverse, and $(\mathbb Z_6,\times)$ is not a group.

Take $2\times2$ matrices with determinant $1$, multiplied the usual way (row of the left matrix times column of the right). Let $A=\begin{pmatrix}1&1\\0&1\end{pmatrix}$ and $B=\begin{pmatrix}1&0\\1&1\end{pmatrix}$, both of determinant $1$. Then $AB=\begin{pmatrix}2&1\\1&1\end{pmatrix}$ while $BA=\begin{pmatrix}1&1\\1&2\end{pmatrix}$ — different matrices from the same two inputs, multiplied in the opposite order. Determinant-$1$ matrices form a group under multiplication (identity $I=\begin{pmatrix}1&0\\0&1\end{pmatrix}$, and every one has an inverse matrix), so this single pair proves the group is non-abelian.

On $\mathbb Z$ let $a*b=a+b-1$. Identity: $a*e=a$ means $a+e-1=a$, so $e=1$, and indeed $1*a=1+a-1=a$ for every $a$. Inverse of $a$: $a*x=1$ means $a+x-1=1$, so $x=2-a$, which is an integer. For example $5^{-1}=-3$: check $5*(-3)=5-3-1=1=e$. Every integer has an inverse, and the operation is associative (both groupings give $a+b+c-2$), so $(\mathbb Z,*)$ is a group whose identity is $1$, not $0$.

On the rationals let $a*b=a+b+3ab$. Identity: $a+e+3ae=a$ gives $e(1+3a)=0$ for every $a$, so $e=0$. Inverse of $a$: $a+x+3ax=0$ gives $x(1+3a)=-a$, so $a^{-1}=\dfrac{-a}{1+3a}$ — which is undefined exactly when $a=-\tfrac13$. And $-\tfrac13$ really has no inverse: $-\tfrac13*b=-\tfrac13+b-b=-\tfrac13$ for every $b$, so combining with it can never reach $0$. Removing it fixes everything. The key is the identity $1+3(a*b)=(1+3a)(1+3b)$: if neither $1+3a$ nor $1+3b$ is $0$, neither is $1+3(a*b)$, so $\mathbb Q\setminus\{-\tfrac13\}$ is closed. For instance $2^{-1}=\dfrac{-2}{7}$, and the check $2*\left(-\tfrac27\right)=2-\tfrac27-\tfrac{12}{7}=0$ confirms it.

In $\mathbb Z_n=\{0,1,\ldots,n-1\}$ under addition mod $n$, the identity is $0$ and the inverse of $a$ is $n-a$ (and $0$ is its own inverse). In $\mathbb Z_{12}$, $5^{-1}=7$ because $5+7=12\equiv0\pmod{12}$. Under multiplication the identity is $1$, and the inverse of $a$ is the $x$ with $ax\equiv1$: mod $11$, $7^{-1}=8$ because $7\cdot8=56=5\cdot11+1$. The nonzero remainders mod a prime $p$ all have such inverses, so they form a group under multiplication, written $U(p)$; mod $6$ they do not, as $2$ showed above.

In $(\mathbb Z_{12},+)$, the inverse of $5$ is $7$, as found in the last section, since $5+7=12\equiv0\pmod{12}$. By the same rule, the inverse of $7$ is whatever added to $7$ gives $0\pmod{12}$, namely $5$, since $7+5=12\equiv0\pmod{12}$. So $(5^{-1})^{-1}=7^{-1}=5$ — undoing the undo returns the original element, exactly as $(a^{-1})^{-1}=a$ predicts.

Reuse the determinant-$1$ matrices $A=\begin{pmatrix}1&1\\0&1\end{pmatrix}$, $B=\begin{pmatrix}1&0\\1&1\end{pmatrix}$ from the non-abelian example. Their inverses are $A^{-1}=\begin{pmatrix}1&-1\\0&1\end{pmatrix}$ and $B^{-1}=\begin{pmatrix}1&0\\-1&1\end{pmatrix}$. Since $AB=\begin{pmatrix}2&1\\1&1\end{pmatrix}$, direct inversion gives $(AB)^{-1}=\begin{pmatrix}1&-1\\-1&2\end{pmatrix}$. And $B^{-1}A^{-1}=\begin{pmatrix}1&0\\-1&1\end{pmatrix}\begin{pmatrix}1&-1\\0&1\end{pmatrix}=\begin{pmatrix}1&-1\\-1&2\end{pmatrix}$ — the same matrix, confirming $(AB)^{-1}=B^{-1}A^{-1}$.

In the group $U(11)=\{1,\ldots,10\}$ under multiplication mod $11$, solve $7x\equiv3\pmod{11}$. Cancellation says we may multiply both sides by $7^{-1}$. We found $7^{-1}=8$ in the last section, since $7\cdot8=56\equiv1\pmod{11}$. So $x\equiv8\cdot3=24\equiv2\pmod{11}$. Check: $7\cdot2=14\equiv3\pmod{11}$.

Build the table for $\{0,1,2,3\}$ under addition mod $4$. For instance row $2$, column $3$: $2+3=5\equiv1\pmod4$. The table is symmetric about its diagonal, confirming the group is abelian; the row and column of $0$ reproduce $0,1,2,3$ unchanged, confirming $0$ is the identity; and reading along row $3$ to find the entry $0$ shows it sits in column $1$, so $3^{-1}=1$, matching $3+1=4\equiv0\pmod4$.

Label the six permutations of $\{1,2,3\}$ by one-line notation: $e=123$, $r=231$, $r^2=312$ (the rotations), and $s=213$, $t=321$, $u=132$ (the three transpositions), composed as functions where $x*y$ means "apply $y$, then $x$". Computing $r*s$: $1\mapsto r(s(1))=r(2)=3$, $2\mapsto r(s(2))=r(1)=2$, $3\mapsto r(s(3))=r(3)=1$, giving $r*s=321=t$. The other order, $s*r$: $1\mapsto s(r(1))=s(2)=1$, $2\mapsto s(r(2))=s(3)=3$, $3\mapsto s(r(3))=s(1)=2$, giving $s*r=132=u$. Since $t\ne u$, $r*s\ne s*r$: the table is NOT symmetric, so $S_3$ is not abelian.

From the $S_3$ table, find every element $x$ with $x*x=e$. Testing the transpositions, each undoes itself: $s*s=e$, $t*t=e$, $u*u=e$. Testing a rotation: $r*r$ sends $1\mapsto r(r(1))=r(2)=3$, $2\mapsto r(r(2))=r(3)=1$, $3\mapsto r(r(3))=r(1)=2$, giving $r*r=312=r^2\ne e$; so $r$ is not its own inverse (its inverse is $r^2$, since $r*r^2$ cycles back to $e$). The self-inverse elements of $S_3$ are exactly $\{e,s,t,u\}$ — four of the six elements.