# Examples of properties not preserved under homomorphism

An isomorphism indicates that two structures are the same, using different names for the elements. Therefore it's obvious that every (algebraic) property of the first structure must be present in the second.

However, homomorphisms only indicate that the two structures are "similar", so it's not quite as obvious that every property will be preserved. Yet all the properties I've ever seen are preserved under homomorphism: commutativity, cyclicality, solvability...

What are some examples of properties of algebraic structures not preserved under homomorphism? Feel free to use any algebraic structures you like, but I'm particularly interested in your garden variety structures: group and rings, say.

• (Should this be community-wiki?) – Jack M Apr 23 '14 at 21:27
• The homomorphic image of an integral domain can have zero divisors. – Daniel Fischer Apr 23 '14 at 21:29
• In the case of group the order of element isn't necessarily preserved. But I would like to say I don't really see existence of homomorphism between two groups has any deep meaning. Every group has homomorphism into trivial group and it doesn't make sense to say any group is 'similar' to trivial group. There is a reason why the word 'homomorphic' doesn't exist. – Jack Yoon Apr 23 '14 at 21:42
• By "preserved under homomorphism" you refer to properties which apply to the image of a homomorphism, rather than its codomain, yes? Otherwise commutativity and cyclicity aren't preserved. – Kevin Arlin Apr 23 '14 at 21:45
• @KevinCarlson Yes. Preserved in the image. – Jack M Apr 23 '14 at 21:46

A very simple example is cardinality.

• Not really an algebraic property. I would have included that requirement in the question, but I wasn't sure how to formalize it... – Jack M Apr 23 '14 at 21:29

An image of an algebraic object is equivalently a quotient in the most elementary cases. Taking a quotient is an identification process, so a general class of properties not preserved under images are those relating to uniqueness of solutions of equations.

For instance, in any free abelian group a linear equation with a solution has only one solution-but in abelian groups with torsion there may be many. Similarly, rings of polynomials over a field admit factorization theorems to the effect that a polynomial of degree $n$ has no more than $n$ roots, whereas there are nonzero polynomials over finite rings that annihilate the entire ring.

One way to get some properties not preserved under general homomorphisms is to take quantitative versions of some qualitative properties that are so preserved. For instance, in the case of groups, while solubility is preserved, the derived length is not. Likewise, nilpotence is preserved but the class of nilpotence is not. The exponent of a group is not preserved (but being of finite exponent is).

A few others that come to mind (for groups): residual finiteness (and lots of other residual properties), being centre-less, being free and being torsion-free.

Cancellation law in a commutative monoid. An example is given in an answer to this (kind of related) question. More natural example of the same phenomenon is: homomorphic image of an integral domain need not to be a domain.

For a group homomorphism $\varphi:G\to H$ you have $G\big/\ker\varphi\cong \varphi(G)$, so the properties of $\varphi(G)$ will always be those of a quotient of $G$. Similarly, any quotient is of course a homomorphic image under the quotient map $G\to G\big/ N$. Since the properties you named are preserved by quotients, they are also preserved in homomorphic images. Similar theorems exist for monoids, rings, modules, vector spaces, ... as well.

The example of homomorphic images of integral domains not being integral domains is explained this way as well, since quotients of integral domains are not integral domains in general, just look at $\mathbb Z\big/n\mathbb Z$ for $n$ not prime.

Commutativity isn't a property that gets preserved for a homomorphism. Consider the algebra A with a binary operation A such that it has the following table:

A  0  1
0  1  0
1  0  1


Now let's define a homomorphism H: H[A(x, y)]=B[H(x), H(y)] from {0, 1} to {3, 4, 5} where H(0)=3 and H(1)=5, where algebra B has the following table:

B  3  4  5
3  5  4  3
4  5  3  3
5  3  4  5


B and A thus are not similar structures. A is similar to a sub-structure of B.

It seems reasonable to conclude that no algebraic structure gets preserved under a homomorphism, only that an algebraic structure gets respected for a sub-structure of the target algebra.

• The OP was, as it turned out in comments, looking for properties preserved under image, since as you say nothing is preserved in codomains. Incidentally, for commutativity one could just map the trivial group to any noncommutative group. – Kevin Arlin Apr 23 '14 at 21:54
• I'm confused. If x and m commute, i.e., $x ⊕ m = m ⊕ x$ then so do their images under homomorphisms: f x ⊕ f m = f (x ⊕ m) = f (m ⊕ x) = f m ⊕ f x ...so homomorphisms preserve commutativity?? – Musa Al-hassy Feb 15 '16 at 18:53
• @MusaAl-hassy Homorphisms don't preserve commutavity for the entire structure. – Doug Spoonwood Feb 15 '16 at 19:12
• alrighty, but they take commutative pairs of elements to commutative pairs of elements yeah? And so homomorphisms induce operations between centers? – Musa Al-hassy Feb 15 '16 at 19:25

Just because you asked for an example, a loop (= quasigrup with identity element) with homomorphic image which is not a quasigroup is mentioned in line 4 of this article: A note on homomorphic mappings of quasigroups into multiplicative systems by G.E. Bates and F. Kiokemeister.

• Interestingly, this depends on the definition of a quasigroup / loop. If defined with a single operation (as in the article linked by MattAllegro), the homomorphic image of a quasigroup/loop need not preserve the uniqueness of $x$ in $x*y=z$ when given $y$ and $z$. If defined with 3 operations $*$, $/$, $\backslash$ (multiplication plus left- and right-division), the homomorphic image of a quasigroup is a quasigroup since the uniqueness of solutions is encoded in identities for the 3 operations. See Wikipedia. – Frentos Sep 6 '17 at 3:24

Almost by definition, the property of "being a free object" is not preserved under homomorphisms*. Moreover, being a free group, free abelian group, free nilpotent group, free monoid etc., of a fixed rank $$n$$ is not preserved under homomorphisms with non-trivial kernel (this is not always obvious, and is a theorem in the case of free groups).

*although will be preserved under some homomorphisms, such as isomorphisms or for example $$\mathbb{Z^3\rightarrow Z^2}$$.