Can anyone explain decomposable and indecomposable groups? in my book,
Groups are decomposable if it is direct product of two proper-nontrivial groups.
Otherwise, they are indecomposable.
What I don't understand is that what is " product of two proper-nontrivial groups"?
$\mathbb{Z} _{21}$ is isomorphic to $\mathbb{Z} _{3} \times \mathbb{Z} _{7}$
Is it indecomposable? $\mathbb{Z} _{3}$ and $\mathbb{Z} _{7}$ are cycles groups
Can anyone give me an example decomposable groups?
 A: The word "decomposable" is used in connection with various algebraic objects: permutations, modules, and (as here) groups.  Often (but not always) the sense is consistent between settings, that an object can be expressed in a well-defined way in terms of two or more smaller objects.
For groups the definition is given that a group $G$ is decomposable iff it is the direct product of two proper, nontrivial subgroups, say $G = H_1 \times H_2$ where $H_1,H_2$ are subgroups of $G$, but neither is all of $G$ and neither is the trivial subgroup (consisting only of the identity element).
Now it happens that there is some redundancy in this definition, because if $H_1$ were the trivial subgroup, that would force $H_2$ to be all of $G$, and conversely.  So, as long as we are considering internal direct products for $G$, it would have been enough to require $H_1,H_2$ both nontrivial, or to require both to be proper subgroups.
This makes for a nice analogy with composite integers, which may be defined as nonzero numbers that can be expressed as the product of two "smaller" integers (or as the product of two nonunit integers).  Indeed in a finite group $G$ necessarily $|G|$ would be a composite number because $|G|=|H_1|\cdot |H_2|$.
One thing to keep your eye on is the requirements of a direct product, that $H_1 \cap H_2 = \{e\}$ is the trivial subgroup, and that every element of $G = H_1 H_2$ can be expressed as an element of $H_1$ times an element of $H_2$.
Finally a comment about the relationship between being indecomposable (not a decomposable group) and being a simple group.  A simple group has no nontrivial, proper normal subgroups (by definition).  But a decomposable group $G = H_1 \times H_2$ has at least two nontrivial, proper normal subgroups, namely $H_1$ and $H_2$.
Thus decomposable implies not simple, and taking the contrapositive, a simple group must be an indecomposable group.  However the converse is not true; some indecomposable groups are not simple.  For more discussion of this, see the older Question Are finite indecomposable groups necessarily simple?.
