Let G be a finite group and let p be the smallest prime divisor of $| G|.$ Then every subgroup H of index p in G is normal in G .

(In https://people.math.osu.edu/all.3/algebra/group%20theory/group1.pdf Autumn 2004, problem 4)

They say that if the kernel of $\varphi$ is trivial then $H$ is normal in $G$ but i can't figure out why that is true

  • $\begingroup$ Please let me know if the title of the question (#4, 2004) corresponds to the problem about which you are asking? $\endgroup$ – Namaste Oct 11 '16 at 21:33
  • $\begingroup$ Yes, that is the problem I'm asking about thanks $\endgroup$ – eeser boy Oct 11 '16 at 21:36
  • $\begingroup$ Possible duplicate of math.stackexchange.com/questions/164244/… $\endgroup$ – Starfall Oct 11 '16 at 21:37
  • 2
    $\begingroup$ no, I'm interested in following this line of reasoning by conjugation and am just confused by that particular line of the proof. My question is essentially asking why that line is true $\endgroup$ – eeser boy Oct 11 '16 at 21:39
  • $\begingroup$ im not necessarily asking about proving the statement as a whole, moreso about that step $\endgroup$ – eeser boy Oct 11 '16 at 21:40

Note that $\ker \phi \not = G$, as otherwise we know that $G = \ker \phi \le H$, which means that $G = H$.

I guess that's what the author thinks when he says that the kernel is non-trvial, although to be fair in my opinion when somebody say non-trivial I understand that $\ker \phi \not = \{e\}$. If you want to say that $\ker \phi \not = G$, you usually say that the kernel is proper (a proper subgroup of $G$). My best guess is that this is the author's thought and he just used a "different" language.

The fact $\ker \phi \not = G$ is important because it means that $[G:\ker \phi] \not = 1$. This is used to conlcude that $[G:\ker \phi] = p$, as otherwise note that $1$ is a number that divides the order of both $G$ and $S_p$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.