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

Question

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?

Answer 1

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

Answer 2

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 $2$.

Answer 3

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.