Group of order $24$ with no subgroup of order $6$

Question

Here is a problem I am working on:

Let $G$ be a group with order $24$, which happens not to have any subgroup of order $6$. Show that the Sylow $3$-subgroup of $G$ is normal in $G$.

Below is my work so far:

Since $|G| = 24 = 2^3 \cdot 3$, denoting the number of Sylow 3-subgroups of $G$ by $n_3$, we have $n_3 | 8$ and $n_3 \equiv 1$ (mod $3$). Thus, we have two possibilities for $n_3$, $n_3 = 1$ or $n_3 = 4$.

Denoting the number of Sylow $2$-subgroups of $G$ by $n_2$, I thought to also compute the possibilities for $n_2$. We have $n_2 | 3$ and $n_2 \equiv 1$ (mod $2$). Thus, we have two possibilities for $n_2$, $n_2 = 1$ or $n_2 = 3$.

If we knew that $n_3 = 1$, we would have the desired result, that we have a Sylow $3$-subgroup of $G$ which is normal in $G$. I'm struggling with ruling out $n_3 = 4$ as a possibility.

If $n_3 = 4$ and $n_2 = 3$, I believe a simple counting argument shows that this is not possible : With each Sylow $2$-subgroup having order $8$ and each Sylow $3$-subgroup having order $3$, this would give $7(3) + 2(4) = 29$ nonidentity elements in $G$, which already contradicts the order of $G$. Thus, I just need help finding a contradiction to the case that $n_3 = 4$ and $n_2 = 1$. I know I have to use the fact given that our particular $G$ does not contain a subgroup of order $6$ for this, but I'm just not sure how to utilize this fact.

Thanks!

Answer 1

Recall that $G$ acts transitively by conjugation on $\operatorname{Syl}_3(G)$ (this is Sylow's second theorem). So, if $n_3=\left|\operatorname{Syl}_3(G)\right|=4$, then the pointwise stabilizers have order $6$ (orbit-stabilizer theorem), which is a contradiction, in the assumption of problem. Therefore, $n_3=1$ and the Sylow $3$-subgroup is normal. Btw, this all holds true irrespective of whether such a $G$ exists or not.

Answer 2

There is no such group.

A group G of order 24 has a subgroup of order 3, call it H

H has 8 left cosets, on which it acts by left multiplication.

Since it's an action, it must stabilize 2, 5, or 8 cosets (including itself).

Those cosets form the normalizer of H, which is a subgroup, of order 6, 15, or 24.

You can't have a subgroup of order 15, so we're left with a normalizer of order 6, or of order 24. Call that N.

(... You could actually just stop here, if there's no subgroup of order 6 then H's normalizer is the whole group, so it's normal, but actually, there's no way to avoid a subgroup of order 6, because ...)

H is normal in its normalizer, so we can quotient out by it, and that quotient group N/H has order 2 or 8.

That means that the quotient group has a subgroup of order 2.

That subgroup is H and one of its cosets, and together they form a subgroup of order 6 in the big group G.

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This is just the usual proof of the first Sylow theorem but adapted to extend the order three subgroup rather than the order two subgroup.

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Visually, with a group of order 24, there's going to be a Cayley diagram with eight triangles in it. The one that contains the identity is H.

Each of the other edge types either connects one triangle to three other triangles, or one triangle to one other triangle three different ways.

The action of H on its cosets is to whirl the sets of three triangles around H.

The triangles that are connected one-to-one don't whirl, and they're the normalizer.

If there are two of those you're done.

If there are eight of them then they're connected up like an order 8 group, and so you're going to find two somewhere that form a subgroup.