Homomorphism, difference between $G'$ and $f(G)$ Could you please help me to understand.
Let $(G, \cdot )$ and $(G', \star )$ be groups. Let $f:G\to G'$ be a function with the property of $f(a \cdot b)=f(a)\star f(b)$. (That is, it construct homomorphism)  
There is such statement:
"If G is finite, then the order of $f(G)$ divides the order of $G$; if $G'$ is finite, then the order of $f(G)$ also divides the order of $G'$."  
I don't understand the difference between $G'$ and $f(G)$ and why they can have different orders.
P.S.
I'm the begginer in the abstract algebra, please answer as simple as possible. Examples are welcomed!
 A: Let $f:G\to G'$ be a finite groups homomorphism. It's a general fact (fundamental theorem actually) that
$$f(G)\cong\frac{G}{\mbox{ker}f}$$
and from that you get that $|G|=|\mbox{ker}f|\cdot |f(G)|$. So, if $\mbox{ker}f\not=\{e_G\}$, you get that the order of $f(G)$ is strictly smaller than the order of $G$ (a proper divisor, indeed).


*

*For example, take $f:(\mathbb{Z}_4,+)\to(\mathbb{Z}_8,+)$ defined as
$$f:\begin{cases}
\overline{0}\mapsto \overline{0}\\
\overline{1}\mapsto \overline{0}\\
\overline{2}\mapsto \overline{0}\\
\overline{3}\mapsto \overline{0}\\
\end{cases}$$
Notice that $f(\mathbb{Z}_4)=\{\overline{0}\}\subset\mathbb{Z}_8$, of order 1, while the whole group $\mathbb{Z}_8$ has 8 elements. Notice also that $f(\mathbb{Z}_4)$ is not the codomain $\mathbb{Z}_8$, it's just a subset. Those would be equal if and only if $f$ was surjective (onto).

*As another example, consider $g:(\mathbb{Z}_4,+)\to(\mathbb{Z}_2,+)$ as
$$g:\begin{cases}
\overline{0}\mapsto \overline{0}\\
\overline{1}\mapsto \overline{1}\\
\overline{2}\mapsto \overline{0}\\
\overline{3}\mapsto \overline{1}\\
\end{cases}$$
which is similar to the previous one, but pay attention to the change in the codomain. Now $g(\mathbb{Z}_4)=\{\overline{0},\overline{1}\}=\mathbb{Z}_2$ so in this case $g$ is surjective. Again, its order is 2, a proper divisor of $|\mathbb{Z}_4|=4$.

*Finally, take $h:(\mathbb{Z_4},+)\to(\mathbb{C}-\{0\},\cdot)$ as
$$h:\begin{cases}
\overline{0}\mapsto 1\\
\overline{1}\mapsto i\\
\overline{2}\mapsto -1\\
\overline{3}\mapsto -i\\
\end{cases}$$
which you can explicitly write as $h(\overline{n})=i^{n}$. This time $h(\mathbb{Z}_4)=\{1,i,-1,-i\}$ and it has the same order as the domain (h is injective!). Once again, the map is not surjective as the second group is actually infinite, so $f(\mathbb{Z}_4)\subset\mathbb{C}-\{0\}$.
