Symmetry structure and homomorphisms

A structure-preserving map is more informative than a bare correspondence. For example, two finite sets can have the same number of elements while supporting incompatible group laws. We use a simple rectangular symmetry group to see how group axioms, element orders and homomorphisms distinguish structures.

1 Reversible transformations form a group

A group consists of a set with an associative binary operation, an identity, and inverses for every element. For bijections of a set, composition is associative, the identity function is the identity, and each inverse bijection exists. Any nonempty collection closed under composition and inversion is therefore a transformation group.

For a nonsquare rectangle centered at zero, let $h$ reflect across the horizontal axis and $v$ across the vertical axis. Then $h^2=v^2=e$ and $hv=vh=r$, the half-turn. The group is $G=\{e,h,v,r\}$. Every nonidentity element has order 2. It is abelian.

The cyclic additive group $\mathbb Z/4\mathbb Z$ also has four elements, but $[1]$ has order 4. These groups cannot be isomorphic because an isomorphism preserves element orders: $\phi(g)^k=e$ exactly when $\phi(g^k)=e$, and injectivity converts that to $g^k=e$.

A subgroup is a nonempty subset using the same operation and closed under products and inverses; these conditions also imply that it contains the identity. In $G$, the subgroups are $\{e\}$, the three two-element groups generated by $h,v,r$, and all of $G$. There is no three-element subgroup: equal-sized cosets of a subgroup partition a finite group, so subgroup size must divide 4. This counting statement is Lagrange's theorem.

2 Homomorphisms record structure rather than labels

A homomorphism $\phi:G\to K$ satisfies $\phi(ab)=\phi(a)\phi(b)$. It preserves identity and inverses. Its kernel is the set sent to identity; its image is a subgroup of the target.

Kernel criterion. A homomorphism is injective if and only if its kernel is $\{e\}$.

Proof. If it is injective, only $e$ can share the identity's image. Conversely, if $\phi(a)=\phi(b)$, then $\phi(a^{-1}b)=e$, so trivial kernel gives $a^{-1}b=e$ and $a=b$. $\square$

A homomorphism $\phi:G\to\mathbb Z/2\mathbb Z$ can record whether a symmetry flips the horizontal coordinate: set $\phi(e)=\phi(h)=[0]$ and $\phi(v)=\phi(r)=[1]$. Composition adds the number of horizontal-coordinate flips modulo 2. Its kernel is $H=\{e,h\}$; the map forgets whether a vertical-coordinate flip also occurred.

3 Quotients and the information that remains

A subgroup $N$ is normal when $gNg^{-1}=N$ for all $g$. Kernels are normal because conjugation maps the identity image back to identity. In an abelian group all subgroups are normal.

When $N$ is normal, cosets multiply by $(aN)(bN)=abN$. To check representatives, replace $a,b$ by $an,bm$: the product is $ab(b^{-1}nb)m$, belonging to the same coset $abN$. Normality makes the formula well defined.

For the rectangular group, $G/H$ has two elements, $H$ and $vH=\{v,r\}$, and its operation matches $\mathbb Z/2\mathbb Z$. In general,

$$G/\ker\phi\cong\operatorname{im}\phi.$$

The map $g\ker\phi\mapsto\phi(g)$ is well defined and injective because equal images are exactly equal kernel cosets; it is onto the image and preserves products. This proof shows why the quotient identifies precisely what the original map forgets.

Checking the map laws and the rectangle model

The identity and inverse claims used in the kernel argument have short proofs. From $\phi(e)=\phi(e)^2$, multiply by $\phi(e)^{-1}$ in the target group to obtain $\phi(e)=e$. Applying $\phi$ to $gg^{-1}=g^{-1}g=e$ shows that $\phi(g^{-1})$ is the two-sided inverse of $\phi(g)$. The kernel contains $e$ and is closed under $ab^{-1}$, while $\phi(a)\phi(b)^{-1}=\phi(ab^{-1})$ proves the image is a subgroup. These are consequences of the group laws, not extra assumptions in the definition of homomorphism.

For the rectangle, write $h(x,y)=(x,-y)$ and $v(x,y)=(-x,y)$. Each composition is obtained by multiplying the two coordinate signs, so the sign of the horizontal coordinate is a homomorphism to $\{1,-1\}$, or equivalently to the additive group modulo 2. This directly verifies the stated kernel. A geometric symmetry must send a chosen vertex to one of four vertices. Since the two side lengths are unequal, the image of its neighbor along the longer side is then forced; distances determine the remaining vertices. Hence there are at most four symmetries, and the four displayed coordinate-sign maps realize all of them. The nonsquare hypothesis is the step excluding the additional quarter-turn symmetries of a square.

4 Graded practice

E1 Foundation. In the rectangular group, compute $hr$, $vr$ and $r^2$.

E2 Homomorphism. List all homomorphisms from $\mathbb Z/4\mathbb Z$ to $\mathbb Z/2\mathbb Z$ and their kernels.

E3 Nonisomorphism. Show $\mathbb Z/6\mathbb Z$ and the permutation group $S_3$ are not isomorphic even though both have six elements.

E4 Proof. Show the image of an abelian group under any homomorphism is abelian. Must the entire target be abelian?

5 Full solutions

S1. Since $r=hv$ and the generators commute, $hr=h^2v=v$, $vr=vhv=h$, and $r^2=h^2v^2=e$.

S2. A homomorphism is determined by the image of $[1]$, which may be $[0]$ or $[1]$. These give the zero map with whole-group kernel and reduction modulo 2 with kernel $\{[0],[2]\}$. Both choices satisfy the relation $4[1]=[0]$ in the target.

S3. $\mathbb Z/6\mathbb Z$ is abelian, whereas $(1\ 2)(2\ 3)\ne(2\ 3)(1\ 2)$ in $S_3$. Isomorphisms preserve commutativity, so no isomorphism exists. Alternatively, $S_3$ has no element of order 6 but the cyclic group does.

S4. If $u=\phi(a)$ and $v=\phi(b)$, then $uv=\phi(ab)=\phi(ba)=vu$. The target need not be abelian: the trivial map from any abelian group to $S_3$ has the abelian one-element image inside a nonabelian target.

Summary

Transformation groups connect abstract axioms to visible symmetries. Homomorphisms preserve the operation, while kernels identify lost distinctions. A quotient recovers exactly the map's image. Equal size alone never establishes structural equivalence.