Is this a well-known group? $\langle a,b \mid a^5=b^4=e,b^{-1}ab=a^{-1}\rangle$

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?

• See the list here: groupprops.subwiki.org/wiki/Groups_of_order_20 – Hongyi Huang May 24 at 9:03
• 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. – Arthur May 24 at 9:03

3 Answers

$$\newcommand{\Span}[1]{\left\langle #1 \right\rangle}$$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 $$\Span{ c, b : c^{10} = e, b^{2} = c^{5}, b^{-1} c b = c^{-1} }$$

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 $$2$$.

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.

• 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 $4$. – Andreas Caranti May 24 at 13:14
• 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. – Shaun May 24 at 13:23
• @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. – Alexander Konovalov May 24 at 19:36