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If $G$ is a group and $H$ is a subgroup of index $2$ in $G$, prove that $H$ is a normal subgroup of $G$.

Proof: We know that $i_G(H)=2$ $\Rightarrow$ subgroup $H$ has two cosets in $G$. One of the cosets is $He=H.$ Since the cosets decompose the group $G$ $\Rightarrow$ another coset is $H^c$ , complement of $H$.

It's easy to prove that if $g\in H$ $\Rightarrow$ $Hg=gH=H$. And also if $g\in H^c$ $\Rightarrow$ $gH=Hg=H^c$. Let's consider two cases:

1) If $g\in H\subset G$ $\Rightarrow$ $gHg^{-1}=g(Hg^{-1})=gH=H$ $\Rightarrow$ $H$ is a normal subgroup of $G$.

2) If $g\in H^c\subset G$ $\Rightarrow$ $gHg^{-1}=g(Hg^{-1})=gH^c$ since $g^{-1}\in H^c$. Let's prove that $gH^c\subset H$. Taking any element $z\in gH^c$ $\Rightarrow$ $z=gh'$ for some $h'\in H^c$ and since $H^c=g^{-1}H$ $\Rightarrow$ $h'=g^{-1}h''$ for some $h''\in H$ $\Rightarrow$ $z=gh'=gg^{-1}h''=h''\in H$ $\Rightarrow$ $gHg^{-1}=gH^c\subset H$ so $H$ is a normal subgroup of $G$.

Remark: Please do not duplicate it since I would like to know the correctness of my solution. Is this proof correct? Please check it out and assess my solution. However, my proof is not short.

RFZ
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    It is correct.. – Yanko Dec 14 '17 at 10:56
  • @yanko, Thanks a lot! – RFZ Dec 14 '17 at 11:00
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    This is a duplicate of https://math.stackexchange.com/questions/84632 and of many other posts that have themselves been closed as duplicates. Unfortunately, although I can close questions as duplicates, I cannot vote to do so, otherwise I would! – Derek Holt Dec 14 '17 at 11:46
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    Why don't you first review the numerous excellent proofs given at all the duplicates, before posting again a question, which has appeared so often. This site is neither a homework service site, nor a "checking it out and assessing my solution"-site. – Dietrich Burde Dec 14 '17 at 20:22

3 Answers3

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The proof has the ingredients, but they're quite mixed up.

Especially in points 1 and 2, you should not end them with “$\Rightarrow$ $H$ is a normal subgroup of $G$”. It's the combination of the two parts that shows normality. A proof written like this would get no positive score, because it shows lack of understanding what you need to prove.

Anyway, besides this important point, the proof is unnecessarily long. After you prove that

  • for $g\in H$, we have $gH=Hg$
  • for $g\in H^c$, we have $gH=Hg$

the proof is complete, because you have $gH=Hg$ for all $g\in G$, so $gHg^{-1}=Hgg^{-1}=H$.

Why is $gH=Hg$ for $g\notin H$? There are just two right cosets and two left cosets. In both cases one coset is $H$, so the other one is $H^c$.

egreg
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    Dear egreg! Thanks a lot for remark. They are precious for me since I am in the begining of math road and it is important for me to see these corrections and try not to repeat them again. Thanks a lot! +1 – RFZ Dec 14 '17 at 12:39
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Yet another possible approach is the following.

Lemma. Let $f$ be the group operation, $H \lneq G$ and $H^c:=G\setminus H$. Then:

$$f(H^c \times H^c)=H\iff[G:H]=2 \tag 1$$

Proof.

Take $a \in H^c$ (such element exists, because $H^c\ne\emptyset$ by hypothesis); then $a^{-1}\in H^c$ and $h \in H \Rightarrow ah\in H^c$. Therefore:

$$h\in H \Rightarrow h= a^{-1}ah \in f(H^c\times H^c) \tag 2$$

whence:

$$H \subseteq f(H^c\times H^c) \tag 3$$

This was in general. Now:

$\Leftarrow$) If $[G:H]=2$, then $a\in H^c\Rightarrow H^c=aH$; therefore, $\forall b \in f(H^c,H^c) \Rightarrow \exists h_1,h_2 \in H\mid b=ah_1ah_2$. By contrapositive, suppose $ah_1a\in H^c$; then $H^c\ni(ah_1a)(ah_1a)^{-1}=ah_1aa^{-1}h_1^{-1}a^{-1}=e$: contradiction. Then, $ah_1a\in H$ and thence $b=(ah_1a)h_2\in H$. Therefore, $f(H^c,H^c)\subseteq H$, and by $(3)$:

$$f(H^c,H^c)= H\tag 4$$

$\Rightarrow$) Coversely, if $f(H^c\times H^c)=H$, then $a,b\in H^c \Rightarrow a^2\in H \wedge ab\in H \Rightarrow a^2H=abH \Rightarrow aH=bH$, i.e. there's one coset only, further than $H$ itself: $[G:H]=2$.

$\Box$

Corollary. If $[G:H]=2$, then $H\lhd G$.

Proof. If $g\in H$, then trivially $ghg^{-1} \in H, \forall h \in H$. If $g\in H^c$, then $gh\in H^c, \forall h \in H$, and -by the Lemma- $ghg^{-1}\in H, \forall h \in G$. $\space\Box$

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I am reading "Topics in Algebra 2nd Edition" by I. N. Herstein.
This problem is the same problem as Problem 2 on p.53 in Herstein's book.
I solved this problem as follows:

Let $g\in G$.
Let $h\in H$.
If $g\in H$, then $ghg^{-1}\in H$.
If $g\notin H$, then $(ghg^{-1})(g^{-1})^{-1}=gh\in gH$.
Since $gH\cap H=\emptyset$, $(ghg^{-1})(g^{-1})^{-1}=gh\notin H$.
So, $ghg^{-1}\notin Hg^{-1}$.
Since $i_G(H)=2$, $G=H\cup Hg^{-1}$.
So, $ghg^{-1}\in H$.
Therefore $H$ is a normal subgroup of $G$.

By the way, the index of $A_n$ in $S_n$ is $2$.
So, $A_n$ is a normal subgroup of $S_n$.

tchappy ha
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