Homomorphisms vs Isomorphisms on preserving structure I am listening to an abstract algbra lecture, and something the professor said made very little sense to me. 
He brought up the determinant function from the set of invertible $n \times n$ matrices to the non-zero real numbers as an example of a homorphism that preserves the structure of the group, but isn't an isomorphism (because $\det$ isn't one-to-one, for example). 
However, he later said that though homomorphisms need not be bijective, they also do not need to preserve the order of elements. This made little sense to me, since it seems that the distinguishing feature of a homomorphism is preserving the product operation of a group, and hence its structure. The only additional feature of an isomorphism is bijectivity. 
In sacrificing bijectivity, do we lose the structure-preserving features of the homomorphism? Are only some preserved? Indeed, we may lose the cardinality of the sets, but the order of elements sounds somewhat fundamental. (Then again, his earlier example didn't make much sense because the set of invertible matrices is not abelian, but the non-zero reals under multiplication are, so I do not quite see how we are "preserving the group structure.") 
 A: A homeomorhism preserves group structure in that $h(a*b) = h(a)\circ h(b)$
$*$ and $\circ$ would be the group operations in their respective groups.
And this would imply that identities map to identities, and inverses map to inverses.
A map from the integers under addition to the integers $\mod n$ would be a homemorphism.
It is not 1-1 since $h(a) \equiv h(a+n)$
The map $f(z)= -z$ from the integers (under addition) to the integers is a group isomorphism.
Structure is preserved as $ -(x+z) = -x + -z.$ And the mapping is bijective.
A: For an example of a easy homomorphism that is not an isomorphism and doesn't preserve order recall that the group $\mathbb Z/n$ can be defined as the set of integers $\mathbb Z/n = \{0, 1, 2, \ldots, n - 1\}$ with group operation addition followed by taking the remainder when dividing by $n$.  So, for example, in $\mathbb Z/4$ we have $3 + 3 = 2$ because the usual value of $3 + 3$ is $6$ and $2$ is the remainder when we divide $6$ by $4$.
Now consider the map
$$\mathbb Z/4 \to \mathbb Z/2$$
$$0 \mapsto 0$$
$$1 \mapsto 1$$
$$2 \mapsto 0$$
$$3 \mapsto 1$$
Since there are so few elements you can check that this is a homomorphism by just brute force checking all the possible sums.  Note that in $\mathbb Z/4$ the element $1$ has order $4$, but in $\mathbb Z/2$ the element $1$ has order $2$.  So homomorphisms do not need to preserve order.
Think of this homomorphism as folding $\mathbb Z/4$ in half.  The elements $0, 2$ are pasted together and the elements $1, 3$ are pasted together.  So the structure of $\mathbb Z/4$ is preserved in the map to $\mathbb Z/2$ but not perfectly; some information is lost.
An example of structure being preserved is the following: If $f\colon G \to H$ is a homomorphism and $g \in G$ has order $t$ then $g^t = 1$.  Since $f$ is a homomorphism this means $f(g)^t = f(g^t) = f(1) = 1$ so the order of $f(g)$ must divide the order of $g$.  So there is a relationship between the orders of these elements, but the relationship is not as straightforward as "they're equal".
