= Solution
A <group isomorphism> $f:G\to H$ is a bijection satisfying $f(xy)=f(x)f(y)$ for all $x,y\in G$. It sends the identity to the identity and satisfies $f(x^k)=f(x)^k$. Applying its inverse shows that it preserves the <order of a group element> exactly.
The <cyclic group> $C_8$ has an element of order eight. Every element of the <direct product of groups> $C_4\times C_2$ has fourth power equal to the identity, and a generator of the first factor paired with the identity of the second has order four. Every nonidentity element of $C_2\times C_2\times C_2$ has order two. Thus their maximum element orders are respectively $8,4,2$, which an isomorphism would preserve. \b[The three groups are pairwise nonisomorphic.]
For a fourth example take the <dihedral group> of symmetries of a square. It consists of four rotations and four reflections, so has order eight. With $r$ a quarter-turn and $s$ a reflection, $srs=r^{-1}\ne r$, showing $sr\ne rs$. This <group> is nonabelian, whereas all three displayed <direct products of groups> are <abelian groups>. Since a <group isomorphism> preserves commutativity, \b[the square's dihedral group is isomorphic to none of them]. Its order is eight regardless of whether a notation convention calls it $D_4$ or $D_8$.
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