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.
 A: A very simple example is cardinality.
A: 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. 
A: 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.
A: 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.
A: 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.
A: Roger Lyndon's "Properties Preserved Under Homomorphism" (1959) gives a complete classification of such properties that can be expressed in first order logic. A first-order property is preserved under homomorphism iff if can be expressed using just:

*

*Atomic Formulas

*And, Or

*$\forall, \exists$
Specifically, these formulas cannot involve negation or implication.
As a consequence of this, for instance, the cancellation laws cannot be expressed without negation or implication.
A: 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.
A: The order of the image of an element under a group homomorphism is a divisor of the order of the element. Therefore, elements' orders are not preserved under a homomorphism, in general. Moreover, the trivial homomorphism always exists.
A: 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.  
A: 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 $\mathbb{Z^3\rightarrow Z^2}$.
A: The homomorphic image of an indecomposable module need not be indecomposable. Consider $\Bbb Z \to \Bbb Z/6\Bbb Z$.
