A question on the proof "Normal subgroup of prime index"

Question

On Arturo's Magidin answer, it states that

Then $G/K$ is isomorphic to a subgroup of $S_p$, and so has order dividing $p!$"

Why it that true? I thought in order to have an isomorphism between $2$ groups, they must have the same order, since it's a $1-1$ and onto function.

How do you prove this?

If $H$ is a subgroup of $G$ then the order of $H$ divides the order of $G$. — lulu, Nov 20 '21 at 19:22
@lulu ohh I see its to A subgroup of $S_p$ not the group $S_p$ — 領域展開, Nov 20 '21 at 19:38
@lulu i just used the 1 isomorphism theorem and make the assuption it was the hole $S_p$ — 領域展開, Nov 20 '21 at 19:39

score 2 · Accepted Answer · answered Nov 20 '21 at 19:31

2

It's Lagrange's Theorem. For any subgroup $H$ of a finite group $G$, we have $$|H|\mid |G|.$$

A proof can be found in any textbook on group theory or, indeed, abstract algebra, worth its salt.

answered Nov 20 '21 at 19:31

Shaun

I have dyslexia, so I read "to the " subgroup etc, silly question, thanks for clarifying – 領域展開 Nov 20 '21 at 19:42
1

You're welcome, @lupusnox. Some of the best insights come from supposedly "silly" questions, so don't worry. – Shaun Nov 20 '21 at 19:44

1 Answers1