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$\DeclareMathOperator{\Ima}{Im}$ By the first isomorphism theorem of groups, we have that $$G/\ker g \cong g(G) \leq S_4 $$ Since the group $S_4$ is solvable, we know that every subroup of $S_4$ is solvable. Hence $g(G)$ is solvable, and so is $G/\ker g$.

So, in order to show that $G$ is solvable, it suffices to show that $\ker g$ is solvable.

We have $|\ker g|=24=2^33$. So:

$$n_3 \equiv 1 \pmod 3 \ \ \ \ \ \ \text{and} \ \ \ \ n_3 \ | \ 8,$$ $$n_2 \equiv 1 \pmod 2 \ \ \ \ \ \ \text{and} \ \ \ \ n_2 \ | \ 3,$$ where $n_3$ represents the number of Sylow $3$-subgroups of $\ker g$, and $n_2$ represents the number of Sylow $2$-subgroups of $\ker g$. Hence $n_3 = 1$ or $n_3=4$ and $n_2 = 1$ or $n_2=3$.

If $n_3 = 1$, then there is a unique Sylow $3$-subgroup, $P_3$, which is a normal subroup of $\ker g$.

So we have the subnormal series

$$\{1\} \trianglelefteq P_3 \trianglelefteq \ker g $$

from which we can conclude that the group $\ker g$ is solvable, since $|\ker g / P_3|=8=2^3$ is a prime power, hence the group $\ker g / P_3$ is abelian.

Similarly, one could prove that $\ker g$ is solvable if $n_2 = 1$.

The problem is, I don't know how to exclude the case $n_3 = 4$ and $n_2 = 3$.

Any help would be greatly appreciated.

Shaun
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J.Spi
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    It is a duplicate here, that a group of order $24$ is solvable - see for example this post. – Dietrich Burde Jan 12 '22 at 20:46
  • If $n_2=3$, then the action on the Sylow subgroups by conjugation gives you a homomorphism to $S_3$. – Arturo Magidin Jan 12 '22 at 20:47
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    @DietrichBurde: Unfortunately, the (eventual) duplicate target simply asserts that groups of order less than $60$ are solvable; then the OP would not need the considerations in the first place. – Arturo Magidin Jan 12 '22 at 20:49
  • I think that this site has several posts showing that all groups of order $n<60$ are solvable. I haven't checked all of them (they are too many), how detailed or useful they are. For example, they should not use Burnside's $p^aq^b$- result of course. – Dietrich Burde Jan 13 '22 at 09:43

1 Answers1

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Let $N=\mathrm{ker}(G)$. Assume $n_2=3$. Then $N$ acts by conjugation on the set of $2$-Sylow subgroups, yielding a homomorphism $N\to S_3$. Because the action is transitive, the image cannot be trivial; the image must be either $A_3$ (cyclic of order $3$) or all of $S_3$. In both cases we have a solvable image, so again we can just go down to the kernel. If the image is cyclic of order $3$, then the kernel is of order $8$, and being a $2$-group, it is solvable (nontrivial center). If the image has order $6$, then the kernel has order $4$ and so must be abelian, hence solvable.

Arturo Magidin
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  • Hi! I have two questions: First, why can't the image of the homomorphism be of order 2? And second, what do you mean by "going down to the kernel?" Thanks! – J.Spi Sep 13 '22 at 19:19
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    @J.Spi: Image can't be of order $2$ because the subgroups of order $2$ are not transitive. (But even if the image were of order $2$, it would be solvable and the argument could proceed; but it can't be).Given a group $G$ and normal subgroup $N$ of $G$, it is well-known that $G$ is solvable if and only if both $N$ and $G/N$ are solvable. Here, we know that $G/N$ is solvable, so the solvability of $G$ depends only on establishing the solvability of $N$; that is, "going down" to just looking at $N$. – Arturo Magidin Sep 13 '22 at 19:33
  • Thank you for your time. – J.Spi Sep 14 '22 at 10:32