A question in the answer of the following question : Prove that a group of order 72 can't be a simple group

Question

I had the following question: A group of order 72 can't be a simple group. As it was asked and user Dietrich Burde gave links to already asked questions.

So, I got to know about : How to prove "a group $G$ of order $72$ can't be a simple group"?

But I have a question in answer of Ben West and the last question he answered was about 2 months ago. So, I am asking it as a separate question.

Why the kernel of the map from $G \to S_4$ cannot be trivial?

If it is trivial then what will be the contradiction?

A group of order $72$ can't inject into a group of order $24$. — lulu, Sep 19 '21 at 13:01
@lulu Got it. I just got confused. Could have thought it myself. — , Sep 19 '21 at 13:01

score 1 · Accepted Answer · answered Sep 19 '21 at 12:57

1

If $G$ is a simple group then any homomorphism has kernel either $1$ or $G$. Normally, by construction the image of the map is non-trivial, so the kernel is $1$. Then $|G|=72$ and $|S_4|=24$, so the existence of an injective map is impossible.

answered Sep 19 '21 at 12:57

David A. Craven

11,561

A question in the answer of the following question : Prove that a group of order 72 can't be a simple group

1 Answers1