Some groups of order $40$ Is there some table on the web giving information about particular small groups, that would go up to order $40$ and that would give enough information so that one could be sure whether groups matching certain descriptions exist or don't?
Two I'm thinking of would be


*

*A group with a cyclic normal subgroup of order $10$, a generator of which let us call $g$, and there is an element $a$ of the group of order $40$ such that $a^{-1}ga=g^3$.

*A group with a cyclic normal subgroup of order $10$, such that the quotient of the whole group by that subgroup would be the Klein four-group.
The first I feel fairly sure ought to exist; the second I'm not so sure.
OK, slightly revised question: How big must a group be in order to have a cyclic subgroup of order $10$ generated by an element $g$ and also have an element $a$ (not, of course, in that cyclic subgroup) such that $a^{-1}ga=g^3$?  (This would imply that $a^4$ is in the cyclic subgroup generated by $g$.)
Still later edit: I see my second example is trivial.  But somehow I had in mind a non-abelian group, but by hindsight I don't know what sort of non-abelian relation might have been required.
 A: groupprops is your friend, remember it: http://groupprops.subwiki.org/wiki/Groups_of_order_40
A: If $H$ and $K$ are groups, then there always exists a group $G$ where $H$ is a normal subgroup and $G/H \cong K$. You can take $G$ to be the direct product $ H \times K$. So for the second one, the group $V \times C_{10}$ is an easy example. 
For the first one the following semidirect product $G = C_5 \rtimes_\phi C_8$ should work. Let $C_8 = \langle x \rangle$ and $C_5 = \langle y \rangle$. Define the action for the semidirect product by $y^x = y^3$. That is, $\phi$ maps the generator $x$ to the automorphism of $C_5$ which sends $y$ to $y^3$. Then $xyx^{-1} = y^3$ in $G$. Now $G$ contains a central element $z = x^4$ of order $2$. Thus $g = zy$ has order $10$, generates a normal subgroup and $xgx^{-1} = g^3$ since $z = z^3$.
A: For your revised question, $\langle b \rangle = \operatorname{Aut}(C_{10}) \cong \mathbf{Z}_{10}^{\star} \cong C_{4}$ (here $C_{n}$ is a multiplicatively written group of order $n$) is generated indeed by the map $\alpha : x \mapsto x^{3}$, which has order $4$.
So you can build a group of order $40$ as a semidirect product
$$
\langle b \rangle = C_{10} \rtimes C_{4}, \text{with $C_{4} = \langle a \rangle$,} 
$$
where $a^{-1} x a = x^{3}$ for $x \in C_{10}$. (There is another possibility to build such a group, namely as a semidirect product of a cyclic group of order $5$ by a cyclic group of order $8$, in case I can elaborate.)
For a non-abelian example in your second question (see m.k.'s answer), note that $\alpha^{2} : x \mapsto x^{-1}$ is an automorphism of order  $2$ of $C_{10}$.
So you can build another group of order $40$ as a semidirect product
$$
C_{10} \rtimes V, \text{with  $V = \langle b, c \rangle$,}
$$
where $b^{-1} x b = x^{-1}$ and $c^{-1} x c = x$  for $x \in C_{10}$.
