Prove that there doesn't exists a simple group of order 24 and order 48 .

Question

Edit : SOme people are linking my question to solvability of group of order 8p but I don't want to use the concept of solvability of groups. So, Please don't link it to that.

I have been following Joseph Gallian Abstract algabra ( Chapter - Sylow Theorems , Simple Groups)

I have following question: Prove that there doesn't exists a non abelian simple group of order 24.

I tried by finding possibility of sylow 2- subgroups , sylow 3-subgroups using 3 sylow theorems.

Let $n_i$ denote the number of sylow - i subgroups. | x means divides x.

then $n_2 =2k+1$ and $| 3 => k=0,1$ which implies $n_2 = 1,3$ and $n_3 =3k+1$ $| 8 => k=0,1$ so $n=1,4$ are the required possibilities.

Even If i prove that $n_2 \neq 1 $ and $n_3 \neq 1$ even then I am unable to reason how to deduce that there is no unique normal subgroup.

Can you please help me proving the result that there are no non abelian simple groups of order 24 ( using sylow theorems or theorems covered in 1st course of group theory)?

Edit : a very similar question I am struck at and looked for here along with the above asked question ( group of order 24):$G$ is non abelian simple group of order $<100$ then $G\cong A_5$

The argument given by the user 87543 are incomplete and not correct for these groups.

I used the sylow theorems for group of order 48. $n_i$ denote the number of sylow - i subgroups. | x means divides x.

Then using sylow theorems: $n_2 = 2k+1$ and $n_2 |3$ which implies k=0,1 and $n_3 = 3k+1 $ and divides 16 which implies k=0,1,5. are the required possiblity.

But I am unable to move foreward from this point and need help.

The question can be generalized as what statergies / propositions to use in the cases in which sylow 3rd theorem is unable to help proving that $n_1$ =1 to find the exact number of sylow subgroups?

I know that generalized question might be too broad to cover , so in that case please just answer the two parts asked originally. But i had the following generalizing in mind while answering it.

Answer 1

There's a useful trick when trying to show that a group is not simple:

Get an action of $G$ on some set of size $n$, then argue that the associated map $G \to \mathfrak{S}_n$ cannot be injective. Often this goes by cardinality, since $|\mathfrak{S}_n| = n!$.

Since groups are often best understood in terms of their actions, this gives us a very concrete way of attacking these kinds of problems. In your case:

Let $|G| = 24 = 2^3 \cdot 3$. Let's look at the $2$-sylow subgroups. As you've recognized, a good first step is to figure out how many $2$-sylow subgroups there are. The number divides $3$, and has to be odd, so the options are $1$ and $3$.

If there's only one $2$-sylow subgroup, then we're done: it must be normal.

If there's three $2$-sylow subgroups, then we have some work to do. But remember the trick! You should try and cook up a group action that works, but I'll leave the answer under the fold:

Remember that $G$ acts (transitively) on the set of $2$-sylow subgroups by conjugation. Since there's $3$ of them, have have a group action of $G$ on a set of size $3$. But now we use the fact that $3$ is small! The action gives us a homomorphism $G \to \mathfrak{S}_3$, but since $|\mathfrak{S}_3| = 6$, this cannot be injective. So the kernel of this homomorphism is our desired normal subgroup. Do you see why the kernel isn't everything?

What about for $|G| = 48 = 2^4 \cdot 3$? The same argument is going to work! Again, I'll put it under the fold for posterity:

We know the number of $2$-sylow subgroups is odd and divides $3$. If there's only one, then it's normal and we're done. If there's $3$, then we have a nontrivial map $G \to \mathfrak{S}_3$ which cannot be injective, since $|G| = 48 > 6 = |\mathfrak{S}_3|$.

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I hope this helps ^_^