Suppose that $A \cong \oplus_{i = 1}^{n} Z_{p_{i}^{k_{i}}}$ is some finite abelian group, and $(a_1, a_2, \ldots a_n)$ generates a subgroup $N$. If $\langle (a_1, a_2, \ldots a_n) \rangle$ was a direct sum of subgroups in the decomposition, then the quotient $A / N$ would be easy to compute by everyone's favorite isomorphism theorem. But what about the case when it doesn't decompose nicely? Is it possible to redo the decomposition so that it does?
For instance, how would one "see at a glance" what $Z_8 \times Z_2 / \langle (2,1) \rangle$ is?
(These computations are coming up a lot in computing homology, hence the algebraic topology tag.)
$Z_8 \times Z_2$ is the abelian group generated by $a$ and $b$ subject to $8a=0$, $2b=0$.
$(Z_8 \times Z_2) / \langle (2,1) \rangle$ is the abelian group generated by $a$ and $b$ subject to $8a=0$, $2b=0$, $2a+b=0$. From these equations, it is clear that $4a=0$ and $b=2a$. Hence $(Z_8 \times Z_2) / \langle (2,1) \rangle$ is the cyclic group of order $4$.
The Smith normal form is the systematic way of doing these manipulations:
$$ \pmatrix { 8 & 0 \\ 0 & 2 \\ 2 & 1 } \space\to\space \pmatrix { 0 & 8 \\ 2 & 0 \\ 1 & 2 } \space\to\space \pmatrix { 1 & 2 \\ 2 & 0 \\ 0 & 8 } \space\to\space \pmatrix { 1 & 2 \\ 0 & -4 \\ 0 & 8 } \space\to\space \pmatrix { 1 & 0\\ 0 & -4 \\ 0 & 8 } \space\to\space \pmatrix { 1 & 0\\ 0 & 4 \\ 0 & 0 } $$