Opposite ring

The smallest such ring ${\displaystyle R}$ has eight elements and it is the only noncommutative ring among 11 rings with unity of order 8, up to isomorphism.[5] It has the additive group ${\displaystyle C_{2}\times C_{2}\times C_{2}=C_{2}^{3}}$.[3]^: 76 Obviously ${\displaystyle R^{\text{op}}}$ is antiisomorphic to ${\displaystyle R}$, as is always the case, but it is also isomorphic to ${\displaystyle R}$. Below are the tables of addition and multiplication in ${\displaystyle R}$,[lower-alpha 4] and multiplication in the opposite ring, which is a transposed table.

Addition + 0 1 2 3 4 5 6 7 0 0 1 2 3 4 5 6 7 1 1 0 6 7 5 4 2 3 2 2 6 0 4 3 7 1 5 3 3 7 4 0 2 6 5 1 4 4 5 3 2 0 1 7 6 5 5 4 7 6 1 0 3 2 6 6 2 1 5 7 3 0 4 7 7 3 5 1 6 2 4 0

Multiplication ${\displaystyle \cdot }$ 0 1 2 3 4 5 6 7 0 0 0 0 0 0 0 0 0 1 0 1 2 3 4 5 6 7 2 0 2 1 3 7 5 6 4 3 0 3 5 3 6 5 6 0 4 0 4 4 0 4 0 0 4 5 0 5 3 3 0 5 6 6 6 0 6 6 0 6 0 0 6 7 0 7 7 0 7 0 0 7

Opposite multiplication ${\displaystyle \diamond }$ 0 1 2 3 4 5 6 7 0 0 0 0 0 0 0 0 0 1 0 1 2 3 4 5 6 7 2 0 2 1 5 4 3 6 7 3 0 3 3 3 0 3 0 0 4 0 4 7 6 4 0 6 7 5 0 5 5 5 0 5 0 0 6 0 6 6 6 0 6 0 0 7 0 7 4 0 4 6 6 7

To prove that the two rings are isomorphic, take a map ${\displaystyle f:R^{\text{op}}\to R}$ given by the table

Isomorphism between ${\displaystyle R}$ and ${\displaystyle R^{\text{op}}}$

${\displaystyle x}$	0	1	2	3	4	5	6	7
${\displaystyle f(x)}$	0	1	2	4	3	7	6	5

The map swaps elements in only two pairs: ${\displaystyle 3\leftrightarrow 4}$ and ${\displaystyle 5\leftrightarrow 7}$. Rename accordingly the elements in the multiplication table for ${\displaystyle \diamond }$ (arguments and values). Next, rearrange rows and columns to bring the arguments back to ascending order. The table becomes exactly the multiplication table of ${\displaystyle R}$. Similar changes in the table of additive group yield the same table, so ${\displaystyle f}$ is an automorphism of this group, and since ${\displaystyle f(1)=1}$, it is indeed a ring isomorphism.
The map is involutory, i.e. ${\displaystyle f\circ f={\text{id}}}$, so ${\displaystyle f^{-1}}$=${\displaystyle f}$ and it is an isomorphism from ${\displaystyle R}$ to ${\displaystyle R^{\text{op}}}$ equally well.
So, the permutation ${\displaystyle f}$ can be reinterpreted to define isomorphism ${\displaystyle f\colon (R,\cdot )\rightarrow (R,\diamond )}$ and then ${\displaystyle q=\iota \circ f}$ is an antiautomorphism of ${\displaystyle (R,\cdot )}$ given by the same permutation ${\displaystyle q=(3,4)(5,7)}$.
The ring ${\displaystyle R}$ has exactly two automorphisms: identity ${\displaystyle \operatorname {id} _{R}}$ and ${\displaystyle p=(3,5)(4,7)}$, that is ${\displaystyle \operatorname {Aut} (R)=\{\operatorname {id} _{R},p\}}$. So its full group ${\displaystyle G}$ has four elements with two of them antiautomorphisms. One is ${\displaystyle q}$ and the second, denote it by ${\displaystyle r}$, can be calculated

${\displaystyle r=q\circ p=(3,4)(5,7)(3,5)(4,7)=(3,7)(4,5)}$

${\displaystyle G=\{\operatorname {id} _{R},p,q,r\}=\{\operatorname {id} _{R},(3,5)(4,7),(3,4)(5,7),(3,7)(4,5)\}}$

There is no element of order 4, so the group is not cyclic and must be the group ${\displaystyle D_{2}}$ (or the Klein group ${\displaystyle K_{4}}$), which can be confirmed by calculation. The "symmetry group" of this ring is isomorphic to the symmetry group of rectangle.

The ring of the upper triangular 2 x 2 matrices over the field with 3 elements ${\displaystyle {\text{F}}_{3}}$ has 27 elements and is a noncommutative ring. It is unique up to isomorphism, that is, all noncommutative rings with unity and 27 elements are isomorphic to it.[5][6] The largest noncommutative ring ${\displaystyle S}$ listed in the "Book of the Rings" has 27 elements, and is also isomorphic. In this section the notation from "The Book" for the elements of ${\displaystyle S}$ is used. Two things should be kept in mind: that the element denoted by ${\displaystyle 18}$ is the unity of ${\displaystyle S}$ and that ${\displaystyle 1}$ is not the unity.[4]^: 369 The additive group of ${\displaystyle S}$ is ${\displaystyle C_{3}^{3}}$.[4]^: 330 The group of all automorphisms ${\displaystyle \operatorname {Aut} (S)}$ has 6 elements:

${\displaystyle {\begin{aligned}h_{1}&=\operatorname {id} _{S}\\h_{2}&=(1,13,25)(2,26,14)(4,16,19)(5,20,17)(7,10,22)(8,23,11)\\h_{3}&=(1,25,13)(2,14,26)(4,19,16)(5,17,20)(7,22,10)(8,11,23)=h_{2}^{-1}=h_{2}^{2}\\h_{4}&=(4,16)(5,17)(7,22)(8,23)(13,25)(14,26)(3,15)(6,21)(12,24)\\h_{5}&=(1,13)(2,26)(4,19)(8,11)(10,22)(17,20)(3,15)(6,21)(12,24)\\h_{6}&=(1,25)(2,14)(5,20)(7,10)(11,23)(16,19)(3,15)(6,21)(12,24).\end{aligned}}}$

Since ${\displaystyle S}$ is self-opposite, it has also 6 antiautomorphisms. One isomorphism ${\displaystyle f\colon (S,\cdot )\to (S,\diamond )}$ is ${\displaystyle f=(1,14,13,2,25,26)(4,20,16,17,19,5)(7,8,10,23,22,11)(3,15)(6,21)(12,24),}$
which can be verified using the tables of operations in "The Book" like in the first example by renaming and rearranging. This time the changes should be made in the original tables of operations of ${\displaystyle S=(S,\cdot )}$. The result is the multiplication table of ${\displaystyle S^{\operatorname {op} }=(S,\diamond )}$ and the addition table remains unchanged. Thus, one antiautomorphism

${\displaystyle q_{1}=\iota \circ f=(1,14,13,2,25,26)(4,20,16,17,19,5)(7,8,10,23,22,11)(3,15)(6,21)(12,24)}$

is given by the same permutation. The other five can be calculated (in the multiplicative notation the composition symbol ${\displaystyle \circ }$ can be dropped):

${\displaystyle {\begin{aligned}q_{2}&=q_{1}h_{2}=(1,14,13,2,25,26)(4,20,16,17,19,5)(7,8,10,23,22,11)(3,15)(6,21)(12,24)\cdot \\&\quad \cdot (1,13,25)(2,26,14)(4,16,19)(5,20,17)(7,10,22)(8,23,11)\\&=[(1,14,13,2,25,26)(1,13,25)(2,26,14)][(4,20,16,17,19,5)(4,16,19)(5,20,17)]\cdot \\&\quad \cdot [(7,8,10,23,22,11)(7,10,22)(8,23,11)](3,15)(6,21)(12,24)\\&=(1,2)(13,26)(14,25)(4,17)(5,16)(19,20)(7,23)(8,22)(10,11)(3,15)(6,21)(12,24)\\&=(1,2)(4,17)(5,16)(7,23)(8,22)(10,11)(13,26)(14,25)(19,20)(3,15)(6,21)(12,24)\\q_{3}&=q_{1}h_{3}=(1,26,25,2,13,14)(4,5,19,17,16,20)(7,11,22,23,10,8)(3,15)(6,21)(12,24)=q_{1}^{-1}\\q_{4}&=q_{1}h_{4}=(1,14)(2,25)(4,17)(5,19)(7,11)(8,22)(10,23)(13,26)(16,20)\\q_{5}&=q_{1}h_{5}=(1,2)(4,5)(7,8)(10,11)(13,14)(16,17)(19,20)(22,23)(25,26)\\q_{6}&=q_{1}h_{6}=(1,26)(2,13)(4,20)(5,16)(7,23)(8,10)(11,22)(14,25)(17,19)\end{aligned}}}$

${\displaystyle G=\{h_{1},h_{2},h_{3},h_{4},h_{5},h_{6},q_{1},q_{2},q_{3},q_{4},q_{5},q_{6}\}.}$
The group ${\displaystyle G}$ has 7 elements of order 2 (3 automorphisms and 4 antiautomorphisms) and can be identified as the dihedral group ${\displaystyle D_{6}}$[lower-alpha 5] (see List of small groups). In geometric analogy the ring ${\displaystyle S}$ has the "symmetry group" ${\displaystyle G}$ isomorphic to the symmetry group of 3-antiprism[lower-alpha 6], which is the point group ${\displaystyle D_{3d}}$ in Schoenflies notation or ${\displaystyle {\overline {3}}m}$ in short Hermann-Mauguin notation for 3-dimensional space.

All the rings with unity of orders ranging from 9 up to 15 are commutative,[5] so they are self-opposite. The rings, that are not self-opposite, appear for the first time among the rings of order 16. There are 4 different non-self-opposite rings out of the total number of 50 rings with unity[7] having 16 elements (37[8] commutative and 13[5] noncommutative).[6] They can be coupled in two pairs of rings opposite to each other in a pair, and necessarily with the same additive group, since an antiisomorphism of rings is an isomorphism of their additive groups.
One pair of rings ${\displaystyle R_{1}}$[3]^: 330 and ${\displaystyle R_{2}=R_{1}^{\text{op}}}$ has the additive group ${\displaystyle C_{4}\times C_{2}\times C_{2}}$[3]^: 262 and the other pair ${\displaystyle R_{3}}$[3]^: 535 and ${\displaystyle R_{4}=R_{3}^{\text{op}}}$,[3]^: 541 the group ${\displaystyle C_{2}\times C_{2}\times C_{2}\times C_{2}=C_{2}^{4}}$.[3]^: 433 Their tables of operations are not presented in this article, as they can be found in the source cited, and it can be verified that ${\displaystyle R_{3}^{\text{op}}=R_{4}}$, they are opposite, but not isomorphic. The same is true for the pair ${\displaystyle R_{1}}$ and ${\displaystyle R_{2}}$, however, the ring ${\displaystyle {\widetilde {R}}_{2}}$[3]^: 335 listed in "The Book of the Rings" is not equal but only isomorphic to ${\displaystyle R_{2}}$.
The remaining 13−4=9 noncommutative rings are self-opposite.

The free algebra ${\displaystyle k\langle x,y\rangle }$ over a field ${\displaystyle k}$ with generators ${\displaystyle x,y}$ has multiplication from the multiplication of words. For example,

${\displaystyle {\begin{aligned}(2x^{2}yx+3yxy)\cdot (xyxy+1)=&{\text{ }}2x^{2}yx^{2}yxy+2x^{2}yx\\&+3yxyxyxy+3yxy\end{aligned}}}$

Then the opposite algebra has multiplication given by

${\displaystyle {\begin{aligned}(2x^{2}yx+3yxy)*(xyxy+1)&=(xyxy+1)\cdot (2x^{2}yx+3yxy)\\&=2xyxyx^{2}yx+3xyxy^{2}xy+2x^{2}yx+3yxy\end{aligned}}}$

which are not equal elements.

The quaternion algebra ${\displaystyle H(a,b)}$[9] over a field ${\displaystyle F}$ with ${\displaystyle a,b\in F^{\times }}$ is a division algebra defined by three generators ${\displaystyle i,j,k}$ with the relations

${\displaystyle i^{2}=a,\ j^{2}=b,\ k=ij=-ji}$

All elements ${\displaystyle x\in H(a,b)}$ are of the form

${\displaystyle x=x_{0}+x_{i}i+x_{j}j+x_{k}k}$, where ${\displaystyle x_{0},x_{i},x_{j},x_{k}\in F}$

For example, if ${\displaystyle F=\mathbb {R} }$, then ${\displaystyle H(-1,-1)}$ is the usual quaternion algebra.

If the multiplication of ${\displaystyle H(a,b)}$ is denoted ${\displaystyle \cdot }$, it has the multiplication table

${\displaystyle \cdot }$	${\displaystyle i}$	${\displaystyle j}$	${\displaystyle k}$
${\displaystyle i}$	${\displaystyle a}$	${\displaystyle k}$	${\displaystyle aj}$
${\displaystyle j}$	${\displaystyle -k}$	${\displaystyle b}$	${\displaystyle -bi}$
${\displaystyle k}$	${\displaystyle -aj}$	${\displaystyle bi}$	${\displaystyle -ab}$

Then the opposite algebra ${\displaystyle H(a,b)^{\text{op}}}$ with multiplication denoted ${\displaystyle *}$ has the table

${\displaystyle *}$	${\displaystyle i}$	${\displaystyle j}$	${\displaystyle k}$
${\displaystyle i}$	${\displaystyle a}$	${\displaystyle -k}$	${\displaystyle -aj}$
${\displaystyle j}$	${\displaystyle k}$	${\displaystyle b}$	${\displaystyle bi}$
${\displaystyle k}$	${\displaystyle aj}$	${\displaystyle -bi}$	${\displaystyle -ab}$

A commutative ring ${\displaystyle (R,\cdot )}$ is isomorphic to its opposite ring ${\displaystyle (R,*)=R^{\text{op}}}$ since ${\displaystyle x\cdot y=y\cdot x=x*y}$ for all ${\displaystyle x}$ and ${\displaystyle y}$ in ${\displaystyle R}$. They are even equal ${\displaystyle (R,*)=(R,\cdot )}$, since their operations are equal, i.e. ${\displaystyle *=\cdot }$.