Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
Your $a^{-1} g a = g^3$ is $a^{-1} g a = a^{-1} g^3 a$, i.e., $g = g^3$, so $g^2 = e$. This isn't a generator of a cyclic group of order 10. –  vonbrand Mar 24 '13 at 2:46
    
I think you wrote in fact " $\,a^{-1}ga=g^3\Longrightarrow a^{-1}ga=a^{-1}g^3a\,$ "...why? –  DonAntonio Mar 24 '13 at 3:41
1  
@vonbrand : I don't see how you're making the least bit of sense. You need to read carefully. I said $g$ is a generator of a cyclic group of order $10$, and there is an element $a$ of the group of order $40$ such that $a^{-1}ga=g^3$. That certainly does not imply that $g=g^3$. You can't cancel the $a^{-1}$ with the $a$ when there's something between them; we don't have $a^{-1}a$ anywhere in that expression. –  Michael Hardy Mar 24 '13 at 4:55
    
I think that $g^{10}=1$, $a^{-1}ga=a^3$ only imply that $a^4$ commutes with $g$. It does not have to be in the group generated by $g$ unless by other means you have shown that the group $\langle g\rangle$ is its own centralizer. –  Jyrki Lahtonen Mar 24 '13 at 20:26
add comment

3 Answers 3

groupprops is your friend, remember it: http://groupprops.subwiki.org/wiki/Groups_of_order_40

share|improve this answer
    
Does this page answer the question? Wehter the "semidirect product of Z5 and Z8 via square map" or the "semidirect product of Z5 and Z8 via inverse map" does it, doesn't seem immediately clear. –  Michael Hardy Mar 24 '13 at 13:36
    
.....ok, it's not that one. –  Michael Hardy Mar 24 '13 at 18:48
add comment

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 $z = z^3$.

share|improve this answer
add comment

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 $x \in C_{10}$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.