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

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

• Please let me know if the title of the question (#4, 2004) corresponds to the problem about which you are asking? – Namaste Oct 11 '16 at 21:33
• Yes, that is the problem I'm asking about thanks – eeser boy Oct 11 '16 at 21:36
• Possible duplicate of math.stackexchange.com/questions/164244/… – Starfall Oct 11 '16 at 21:37
• 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 – eeser boy Oct 11 '16 at 21:39
• im not necessarily asking about proving the statement as a whole, moreso about that step – eeser boy Oct 11 '16 at 21:40

## 1 Answer

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 $S_p$.