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?

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    $\begingroup$ See the list here: groupprops.subwiki.org/wiki/Groups_of_order_20 $\endgroup$ – Hongyi Huang May 24 at 9:03
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    $\begingroup$ 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

$\newcommand{\Span}[1]{\left\langle #1 \right\rangle}$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.


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.

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    $\begingroup$ 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
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    $\begingroup$ @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

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