If $H\leq G$ And $|G:H|=2$ Then $H$ Is Normal

Question

I want over few proofs here, but I wanted to be sure I got it right.

We will look at $2$ cases: case 1: $a\in H$ case 2: $a\notin H$

Case 1: $a\in H$ therefore $aH=H=Ha$

a. We can say that $aH=H$ because $H$ is a subgroup and is closed, why can we conclude that $H=Ha$? again because $H$ is closed? So the trial coset is always normal?

b. We have shown that $H$ is normal, but did not use the fact that the index is $2$

case 2: if $a\notin H$ then the cosets of $H$ in $G$ are $\{H,aH\}$ or $\{H,Ha\}$ but because there are only $2$ cosets is must be that $aH=Ha$

a. Yes, because $H$ is a subgroup. b. clear, you need to use that the index is $2$. — Dietrich Burde, Feb 24 '20 at 19:04
So we showed that $H$ is abilean too? How do we use the fact that the index is 2 in the first case? Which is the other coset? It is $bH$ when $b /notin H$? — newhere, Feb 24 '20 at 19:13
No, $H$ is not abelian. This was not showed. May be you confuse it with $Ha=aH$? It only means $ab=b'a$, but not necessarily $b=b'$. — Dietrich Burde, Feb 24 '20 at 19:14
I know that normal does not imply abilean. Sec we say that just one element commute, which is $a$? — newhere, Feb 24 '20 at 19:21
Well, what are the left and right cosets of a subgroup of index 2? — The Thin Whistler, Feb 24 '20 at 19:32

score 1 · Answer 1 · 2020-02-24T20:47:51.390

1

More simply, there are just $2$ right/left cosets; one is $H$, the other is either $Ha$ and $aH$, which then coincide, and $H$ is normal in $G$.

edited Feb 24 '20 at 20:47

answered Feb 24 '20 at 19:40

score 0 · Answer 2 · 2020-02-24T20:16:55.380

0

Consider that, if we denote by $f$ group's operation, it is:

$$[G:H]=2 \Longrightarrow f(G\setminus H \times G\setminus H)=H \tag 1$$

Then, $[G:H]=2⇒H⊴G$; in fact, take $h∈H$; if $g∈H$, then trivially $g^{−1}hg∈H$; if $g∈G\setminus H$, then $g^{−1}h∈G\setminus H$ and finally, by $(1)$, $g^{-1}hg=(g^{−1}h)g∈H$.

edited Feb 24 '20 at 20:16

answered Feb 24 '20 at 19:30

If $H\leq G$ And $|G:H|=2$ Then $H$ Is Normal

2 Answers2