Which group of order 24 is this group?

Question

I have a group $G$ which is presented as

$$\langle g, m \mid m^{12} = 1, g^2 = 1, gm = m^5g\rangle$$

That is, $G$ is a semidirect product $C_2 \rtimes C_{12}$, but it is not the usual semi-direct product.

The action $C_2 \to Aut(C_{12})$ sends $$g \mapsto (m \mapsto gmg^{-1} = m^5)$$

I am vexed! It is not obviously the usual dihedral or dicyclic group. If we try to put it in the form of the dicyclic group, with $\langle a, x \mid a^{12} = 1, x^2 = a^6, x^{-1}ax = a^{-1} \rangle$ we find that setting $a = g$, and $x = gm$ does not satisfy the last relation.

Which of the familiar groups of order 24 is this beastie?

Answer 1

Here is a proof using Sylow theory.

$C_2$ acts trivially on the copy of $C_4$ in $C_{12}$ so the Sylow $2$-subgroup of $G$ is isomorphic to $C_2\times C_4$. Also, $G$ has only one Sylow $3$-subgroup (a Sylow $3$-subgroup of $G$ is characteristic in $C_{12}$ which is normal in $G$). Then $G\cong C_3\rtimes(C_2\times C_4)$ where $C_4$ acts trivially on $C_3$ and where $C_2$ acts nontrivially on $C_3$. This is isomorphic to $(C_3\rtimes C_2)\times C_4$ where $C_2$ acts nontrivially on $C_3$. Thus, $G\cong S_3\times C_4$.

Answer 2

Notice that $G$ is generated by $m^4$, $m^3$ and $g$.

Also $gm^4g^{-1}=m^8$, so $\langle m^4,g\rangle\cong S_3$ (e.g. identify $m^4$ with $(1,2,3)$ and $g$ with $(2,3)$ ).

For $m^3$, $gm^3g^{-1}=m^3$, so $m^3\in Z(G)$.

Hence $G\cong S_3\times C_4$.

Answer 3

$\DeclareMathOperator{\Aut}{Aut} gm=m^5g\implies gmgm^7=1$.

In GAP:

gap> f:=FreeGroup("m","g");m:=f.1;g:=f.2;
<free group on the generators [ m, g ]>
m
g
gap> rel5:=[m^12,g*g,g*m*g*m^7];G:=f/rel5;
[ m^12, g^2, (g*m)^2*m^6 ]
<fp group on the generators [ m, g ]>
gap> IdGroup(G);
[ 24, 5 ]
gap> IdGroup(AutomorphismGroup(G));
[ 24, 14 ]
gap> S:=SylowSubgroup(G,2);
Group(<fp, no generators known>)
gap> NrConjugacyClasses(S);
8
gap> IdGroup(AutomorphismGroup(S));
[ 8, 3 ]
gap> IdGroup(Center(G));
[ 4, 1 ]
gap> ch:=CharacterTable(G);
CharacterTable( <fp group of size 24 on the generators [ m, g ]> )
gap> Display(ch);
CT2

      2  3  3  3  3  2  3  3  3  2  3   2   2
      3  1  .  .  1  1  1  .  .  1  1   1   1

        1a 2a 4a 2b 3a 4b 2c 4c 6a 4d 12a 12b
     2P 1a 1a 2b 1a 3a 2b 1a 2b 3a 2b  6a  6a
     3P 1a 2a 4c 2b 1a 4d 2c 4a 2b 4b  4d  4b
     5P 1a 2a 4a 2b 3a 4b 2c 4c 6a 4d 12a 12b
     7P 1a 2a 4c 2b 3a 4d 2c 4a 6a 4b 12b 12a
    11P 1a 2a 4c 2b 3a 4d 2c 4a 6a 4b 12b 12a

X.1      1  1  1  1  1  1  1  1  1  1   1   1
X.2      1 -1 -1  1  1  1 -1 -1  1  1   1   1
X.3      1 -1  1  1  1 -1 -1  1  1 -1  -1  -1
X.4      1  1 -1  1  1 -1  1 -1  1 -1  -1  -1
X.5      1 -1  A -1  1 -A  1 -A -1  A  -A   A
X.6      1 -1 -A -1  1  A  1  A -1 -A   A  -A
X.7      1  1  A -1  1  A -1 -A -1 -A   A  -A
X.8      1  1 -A -1  1 -A -1  A -1  A  -A   A
X.9      2  .  .  2 -1 -2  .  . -1 -2   1   1
X.10     2  .  .  2 -1  2  .  . -1  2  -1  -1
X.11     2  .  . -2 -1  B  .  .  1 -B  -A   A
X.12     2  .  . -2 -1 -B  .  .  1  B   A  -A

A = -E(4)
  = -Sqrt(-1) = -i
B = -2*E(4)
  = -2*Sqrt(-1) = -2i

Thus your group $G$'s Sylow 2-group $S$ is abelian with $|S|=|\Aut S|=8$, thus $C_2\times C_4$ as Robert Chamberlain said. Its centre is $C_4$. This identifies it as 24/9 in the notation of Thomas and Wood*, who also call this group $S_3\times C_4$, thus agreeing with Robert Chamberlain and Thomas Browning. GAP's character tables seems to match that in Thomas and Wood for their 24/9. Strange thing, though -- Thomas and Wood claims $|\Aut G|=12$ but GAP IDs Aut $G$ as [24,14].

*A D Thomas and G V Wood. Group Tables. Pub Shiva 1980