Show isomorphy by mapping generators onto generators only If I want to show that two cyclic groups are isomorphic, is it enough to show that their cardinality is the same and that the generators of the groups are mapped onto each other?
To be precise: I am given a group G with 
   G = <a,b> , |G| = 2q

and I am supposed to show that G is isomorphic to Dq, the Dihedral Group with
    Dq = <s,o> , |Dq| = 2q

is it enough to show that 
    a --> s and b --> o

?
Thanks already for any help!
 A: If $G$ and $H$ are finite groups of the same cardinality, then any surjective homomorphism $\phi:G\to H$ is bijective, hence an isomorphism. If $H = \langle h_1,\ldots,h_n\rangle$ and each $h_i$ is in the image of $\phi$, then $\phi$ is surjective.
So in your situation, if you can prove that there is a homomorphism $\phi:G\to D_q$ such that $\phi(a)=s$ and $\phi(b)=o$, then indeed $\phi$ is an isomorphism. The question is whether there exists such a homomorphism. There does exist such a homomorphism if and only if $s$ and $o$ satisfy all the same relations together that $a$ and $b$ satisfy. This can be checked using a presentation of $G$.
A presentation of a group $G$ is a set $S$ of generators for $G$, together with a minimal set $R$ of relations on that set of generators. This means that every relation satisfied by the elements of $S$ can be derived from the relations in $R$. We usually write this as $G = \langle S |  R \rangle$. 
For example, one presentation of the dihedral group of order $2q$ is
$$D_q = \langle \rho,\tau | \rho^q=1, \tau^2=1, (\rho\tau)^2=1\rangle$$ where $\rho$ is a rotation and $\tau$ is a reflection. ($1$ is the identity element.)
So if $\phi:D_q\to H$ is a homomorphism, then we must have
$$\phi(\rho)^q=1, \phi(\tau)^2=1, (\phi(\rho)\phi(\tau))^2=1.$$ The usefulness of a presentation is that these relations are equivalent to $\phi$ defining a homomorphism. That is, if $x$ and $y$ are elements of $H$ such that 
$$x^q=1, y^2=1, (xy)^2=1$$ then there exists a unique homomorphism $\phi:D_q\to H$ such that $\phi(\rho)=x$ and $\phi(\tau)=y$. The analogous statement is true for an arbitrary presentation $\langle S|R\rangle$.
