0
$\begingroup$

Consider the group $$ G=\langle a,b \mid a^5=b^4=e,b^{-1}ab=a^{-1}\rangle $$ It looks like a dihedral group but it is not isomorphic to a dihedral group. Is this a well-known group?

$\endgroup$
  • 3
    $\begingroup$ See the list here: groupprops.subwiki.org/wiki/Groups_of_order_20 $\endgroup$ – Hongyi Huang May 24 at 9:03
  • 1
    $\begingroup$ It has 20 elements, so it ought to be known. Classifying it (there are three non-abelian order-20 groups, by the looks of it) is a different proposition, though. $\endgroup$ – Arthur May 24 at 9:03
5
$\begingroup$

$\newcommand{\Span}[1]{\left\langle #1 \right\rangle}$It's called a dicyclic group. You can see it by noting that $b^{-2} a b^{2} = b^{-1} a^{-1} b = a$, so that $a b^{2} = b^{2} a$, and then taking $c = a b^{2}$, and rewriting the presentation as $$ \Span{ c, b : c^{10} = e, b^{2} = c^{5}, b^{-1} c b = c^{-1} } $$

$\endgroup$
3
$\begingroup$

The group is the semidirect product of a cyclic group $\langle a \rangle$ of order $5$ and a cyclic group $\langle b \rangle$ of order $4$, where $b$ acts on $\langle a \rangle$ by an automorphism of order $2$.

$\endgroup$
2
$\begingroup$

Here's a tool for you to use in future: GAP.

It describes the group as isomorphic to a semidirect product of $\Bbb Z_5$ and $\Bbb Z_4$ as delineated in @spin's answer.

Look:

gap> F:=FreeGroup(2);
<free group on the generators [ f1, f2 ]>
gap> rel:=[F.1^5, F.2^4, F.2^-1*F.1*F.2*F.1];
[ f1^5, f2^4, f2^-1*f1*f2*f1 ]
gap> G:=F/rel;
<fp group on the generators [ f1, f2 ]>
gap> StructureDescription(G);
"C5 : C4"
gap> IdGroup(G);
[ 20, 1 ]

The latter identification, via known libraries like this, gives the dicyclic group of order twenty.

$\endgroup$
  • 2
    $\begingroup$ GAP is an essential tool for my own research work. Still, a caveat in this particular case. The GAP command List(AllSmallGroups(20), StructureDescription); yields [ "C5 : C4", "C20", "C5 : C4", "D20", "C10 x C2" ] showing that the description does not distinguish the two semidirect, non-direct products of $C_{5}$ by $C_{4}$, the difference being of course whether an element of order $4$ induces on the subgroup of order $5$ an automorphism of order $2$ or $4$. $\endgroup$ – Andreas Caranti May 24 at 13:14
  • $\begingroup$ Thank you, @AndreasCaranti; that's why I said "a semidirect product". Nonetheless, I have augmented this answer using IdGroup(G);. This identifies the group specifically to be the one you describe. $\endgroup$ – Shaun May 24 at 13:23
  • 1
    $\begingroup$ @AndreasCaranti thank you for pointing that out! A good further reading is 7.12: Can non-isomorphic groups have equal structure descriptions? from the GAP F.A.Q. $\endgroup$ – Alexander Konovalov May 24 at 19:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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