Do "embedding" and "injective homomorphism" mean the same thing? On page 6 of A Shorter Model Theory, it says

For example if $G$ and $H$ are groups and $f : G \to H$ is a
  homomorphism, then (2.1) says that $f(1^G)=1^H$ and $f(a^{(-1)^{G}}) =
> f(a)^{(-1)^{H}}$. This is exactly the usual definition of homomorphism
  between groups. Clause (2.2) adds nothing in this case since there are
  no relation symbols in the signature. For the same reason (2.4) is
  vacuous for groups. So a homomorphism between groups is an embedding
  if and only if it is an injective homomorphism.

Emphasis added.
Hodges always bolds terms when providing a definition like this, which I think means that embedding and injective homomorphism refer to separately-defined notions that happen to line up in this specific case.
Intuitively I think of an embedding and an injective homomorphism as the same thing, not even differing intensionally, but I think this intuition is wrong.
How do embedding and injective homomorphism differ in meaning? Is there a good example that demonstrates this?
 A: For groups they are the same thing. For topological spaces, they are not. There can be an injective continuous map $f:X\to Y$ where the image $f(X) $ is not homeomorphic to $X$. A quick example is $X=[0, 2\pi)$, $Y=S^1$, $f(x) =e^{ix} $. 
A: Here is a finite example involving partial orders. Intuitively, if we take a partial order that is not a linear order and linearize it, the $\le$ relation now has extra pairs of elements in it that were not there originally.
Let $\left(*\right)^W$ denote the set of pairs inhabiting the $*$ relation in $W$. $*$ itself may be thought of as an abstract predicate or as a set of pairs, $\left(*\right)^W$ is always a set of pairs.
Let's define a single non-logical binary predicate $\le \mathop{:} \mathcal{D} \times \mathcal{D} \to \left\{ \top, \bot \right\} $ , with $\mathcal{D}$ being the domain of the model.
$\le$ can also be thought of as a set of pairs $P\left( \mathcal{D} \times \mathcal{D} \right)$ .
The theory of partial orders consists the following laws:
$$ a \le a \tag{1a}$$
$$ a \le b \land b \le c \implies a \le c \tag{1b} $$
Let's define two partial orders on 3 elements, $A$ and $B$ . Let the three elements be $\mathcal{D} = \left\{ g, h, k \right\}$ .
        A                   B


        g                   g
       / \                   \
      /   \                   h
     h     k                   \
                                k

Let $f : \left\{ g, h, k \right\} \to \left\{ g, h, k \right\} = \text{id} $ denote a function from the domain of $A$ to the domain of $B$. I will show that $f$ is an injective homomorphism and that $f$ is not an embedding.
Here (3a-3b) are the elements of $\left(\le\right)^A$ and $\left(\le\right)^B$ . 
Note that 
$$\left(=\right)^A = \left(=\right)^B \tag{2} $$
Examining $\left(\le\right)$ specifically.
$$ \left\{ \left(g,h\right), \left(g,k\right) \right\} \cup \left(=\right)^A = \left(\le\right)^A \tag{3a} $$
$$ \left\{ \left(g,h\right), \left(g,k\right), \left(h,k\right) \right\} \cup \left(=\right)^B = \left(\le\right)^B \tag{3b} $$
From (2) and (3a-3b) it is clear that $\left(\le\right)^A \subsetneq \left(\le\right)^B$, which means that $f$ is a homomorphism and a non-embedding. $f$ is injective because the graph of $f$ is the identity function, which is a bijection.
A: Embedding is a term used for the underlying set of space, for instance we may embed the set of integers $\mathbb{Z}$ by a list of distinct points $\{x_n\}\in\mathbb{R}$. Homomorphism is used for maps that preserve addition and multiplication. For instance we only have two homomorphism $f:\mathbb{Z}\rightarrow\mathbb{R}$, i.e. zero map and identity map. But we can construct many embeddings $g: \mathbb{Z}\mapsto\mathbb{R}$ by choosing different lists $\{x_n\}$ 
