Htykuut

I'm engaged in group theory (at least I am trying to get better) and so I found a problem dealing with the symmetric group $S_3$.

The first question is to find an (equational) law $gamma$, which holds in $S_3$, but doesn't hold in some other groups. Furthermore I should find a group $G$ with $ |G| > |S_3| $, which satisfies all laws from $S_3$.

Sadly I'm even struggling at the first point.
I started listing all members of $ S_3 = {e, (12), (13), (23), (123), (132) } $ and tried to find out which of the well known laws hold:

Due to $(12)circ(13) neq (13)circ(12)$ it can be seen that $S_3$ isn't an abelian group.

$S_3$ is a group, so associativity must hold. Furthermore (same reason) there has to exist a neutral element $e$.

Does anyone have an idea which law is meant and how I can find out how many laws hold in $S_3$ overall?

Thanks a lot!

Since the orders of the elements of $S_3$ are 1, 2 or 3, any element raised to the sixth power is equal to the identity element. This does not hold in most other groups (but it does in some, such as the cyclic group of 6 elements).

Besides the law
$$x^6=etag1$$
another equational law obeyed by $S_3$ is
$$x^2y^2=y^2x^2tag2$$
since the set of squares in $S_3$ is the Abelian subgroup $A_3$.

The alternating group $A_4$ obeys $(1)$ but does not obey $(2)$ since
$$(1 2 3)^2(1 2 4)^2ne(1 2 4)^2(1 2 3)^2.$$

For the second part, for any $I$, $S_3^I$ satisfies the equational laws that $S_3$ does. For $|I|geq 2$, $|S_3^I|>|S_3|$.

More generally, any homomorphic image of a subgroup of a product of $S_3$'s satisfies the same equational laws as $S_3$. Moreover the class of these groups is defined by a set of equations by a general theorem (Birkhoff's HSP theorem), this set of equations contains the equations of $S_3$, and is included in the equations of $S_3$. Thus the class of groups satisfying the equations satisfied by $S_3$ is precisely the class of homomorphic images of subgroups of products of $S_3$.

The symmetric group $S_{3}$ is isomorphic to the dihedral group $D_{6}$ of order 6 (the group of symmetries of the equilateral triangle) and has, therefore, the same "equational laws" as that group. Therefore, it has presentation (here, $e$ denotes the identity element, and $a$ and $x$ group generators):

$langle x,a mid a^3 = x^2 = e, xax^{-1} = a^{-1} rangle$.

As group generators, take $a=(123)$ and $x=(12)$. Note that $a^3=x^2=e$ (first law). Of course, by this first law, one also has:

$a^6=(a^3)^2=e^2=e$

$x^6=(x^2)^3=e^3=e$

where we have used the laws of exponents (since these apply to elements of any group under any group operations, that one usually writes in multiplicative notation, in the case of $S_{3}$ multiplication amounts to composition of permutations). Note that $a^6=x^6=e$ is not a proper law of this group, because it is derived from a more fundamental law (the first law above). The law $x^6=e$ (just only one generator) corresponds to the cyclic group $C_{6}$, which is isomorphic to, for example, the integers modulo 6 under addition.

With the laws in the presentation of the group you are able to construct the Cayley table of the whole group (generating all its elements consistently). As your intuition rightly says, the group is non-abelian since the law $xax^{-1} = a^{-1}$ which is equivalent to $xa=a^{-1}x$, says precisely this. This is not obeyed by the cyclic group of order 6 with only just one generator and which is abelian (as all cyclic groups are).

The presentation of the group contains the laws. If you consider direct products of this group, they have higher orders and all trivially obey the same laws.

Note also that any generic symmetric group $S_{n}$ with $n>3$ is not isomorphic to any dihedral group (which is always the semidirect product of two cyclic groups) and, therefore, has not the same laws. That $S_{3}$ is isomorphic to the dihedral group $D_{6}=D_{2cdot 3}$ is an exception between symmetric groups. Of course $S_{3}$ is always a subgroup of $S_{n}$, but $S_{n}$ has additional and/or different laws.