How do i show that finite abelian group is solvable? Let $G$ be a finite abelian group.
How do i show that $G$ is solvable using Fundamental theorem of finite abelian groups?
 A: Terminology can change from one book to another.
From J. S. Rose, A course in the theory of groups, Cambridge University Press

If $G$ is a group, a series in $G$ is a sequence of subgroups
  $$
\{1\}=H_0\unlhd H_1\unlhd H_2\unlhd\dots\unlhd H_{n-1}\unlhd H_n=G
$$
  where $H_{k-1}$ is normal in $H_k$ ($k=1,2,\dots,n$). […] Some authors call this a normal series. […] A series is abelian if every factor $H_k/H_{k-1}$ is abelian.
A group is called soluble (solvable by American authors) if it has an abelian series.

From S. Lang, Algebra, Springer-Verlag

Let $G$ be a group. A sequence of subgroups
  $$
G=G_0\supset G_1\supset G_2\supset \dots\supset G_m
$$
  is called a tower of subgroups. A tower of subgroups is said to be normal if each $G_{i+1}$ is normal in $G_i$ ($i=0,1,\dots,m-1$). It is said to be abelian (resp. cyclic) if it is normal and each factor group $G_i/G_{i+1}$ is abelian (resp. cyclic).
[…] 
A group is said to be solvable if it has an abelian tower whose last element is the trivial subgroup (i.e. $G_m=\{e\}$ in the above notation).


I'll follow the first cited book. If we denote by $G'$ the subgroup of $G$ generated by the commutators $[g,h]=ghg^{-1}h^{-1}$, it's easy to see that $G'$ is normal in $G$ and that $G/G'$ is abelian. Now we can continue and define, by recursion,
$$
G^{(0)}=G,\qquad G^{(k+1)}=(G^{(k)})'
$$
It's also easy to show that a group is solvable if and only there exists $n$ such that $G^{(n)}=\{1\}$. In particular every abelian group is solvable, because $G'=\{1\}$ for an abelian group $G$.
Note: in the article on Wikipedia the author uses composition series where normal series (or simply series) should be used. A composition series is a series in which every factor is simple. An abelian group may fail to have composition series, the simplest example is the group of integers, but it is solvable nonetheless. So the equivalence stated at the beginning of the article is false, if composition series are used instead of normal series.
In the case of finite groups, it is indeed true that a group is solvable if and only if it has an abelian composition series. One direction is trivial.
For the other direction, consider an abelian series
$$
\{1\}=H_0\unlhd H_1\unlhd H_2\unlhd\dots\unlhd H_{n-1}\unlhd H_n=G.
$$
Now we can consider a factor $H_{i}/H_{i-1}$. Call it $K$. Then $K$ is finite abelian, so it has a simple subgroup $K_1$. The group $K/K_1$ is again abelian, so it has a simple subgroup $K_2/K_1$; we can continue until we reach $K$. Thus we get a set of subgroups of $H_i$ containing $H_{i-1}$ that we can insert in the original series; repeat for each $i=1,2,\dots,n$ and finally you get a composition series with abelian factors.
