The "structure" preserved is how the elements relate to each other under the operations of the algebraic structure. For example, for commutative rings, if $\ a = b^2 - c^2 = (b-c)(b+c)\ $ is a difference of squares then applying a ring hom $\,h\, :\, r\mapsto \bar r\,$ shows that it remains a difference of squares in the image ring, viz. $\ \bar a = \bar b^2-\bar c^2 = (\bar b - \bar c)(\bar b +\bar c),\,$ so this particular ring-theoretic structure is preserved. Similarly preserved are polynomial expressions, i.e. expressions composed of basic ring operations (addition, multiplication) and constants $\,0,1.\,$
For example, a root of a polynomial remains a root of the image of the polynomial. So we can prove that a polynomial has no roots by showing it has no roots in a simpler homomorphic image, e.g. a ring with size so small that we can easily test all of its elements to see if they are roots, e.g. common parity arguments: $\,f(x) = x(x\!+\!1)+2n\!+\!1\,$ has no integer roots since $\,x(x\!+\!1)\,$ is even, so adding $2n\!+\!1$ yields an odd hence $\ne 0;\,$ equivalently $\!\bmod 2\!:\ f(0)\equiv 1\equiv f(1),\,$ i.e. $\,f\,$ has no roots mod $\,2,\,$ hence it has no integer roots. This idea generalizes as follows
Parity Root Test $\ $ A polynomial $\,f(x)\,$ with integer coefficients
has no integer roots when its $\rm\,\color{#0a0}{constant\,\ coefficient}\,$ and $\,\rm\color{#c00}{coefficient\,\ sum}\,$ are both odd.
Proof $\ $ If so then $\ \color{#0a0}{f(0)} \equiv 1\equiv \color{#c00}{f(1)}\,\pmod 2,\ $ i.e. $\:f\:$ has no roots in $\,\Bbb Z/2 = $ integers mod $\,2,\,$ therefore $\,f\,$ has no integer roots. $\ $ QED
In the same way, we can often reduce problems to "smaller", simpler problems in modular images. Because homs preserve the ambient algebraic structure, as above, we can often deduce information about the original problems from information gleaned in the simpler modular images. Such problem solving by modular reduction is an algebraic way of "dividing and conquering".
Best Answer
Let $G$ and $H$ be cyclic groups of order $n$. Then there are $g\in G$ such that each element in $G$ can be written as $g^k$ for some $k$ and $h\in H$ such that each element in $H$ can be written as $h^k$ for some $k$. Can you find an isomorphism $f:G\to H$?