Subgroup of index $2$ is Normal

Question

Please excuse the selfishness of the following question:

Let $G$ be a group and $H \le G$ such that $|G:H|=2$. Show that $H$ is normal.

Proof:

Because $|G:H|=2$, $G = H \cup aH$ for some $a \in G \setminus H$.

Let $x\in G$. Then $x \in H$ or $x \in aH$.

Suppose $x \in H$. Then $xhx^{-1} \in H$ because $H$ is a group.

Suppose $x \in aH$. Then $ahH(ah)^{-1} = ahHh^{-1}a^{-1} = a(hHh^{-1})a^{-1}$ ***1

***1: Can I say now that $x \in H$, based on the that $hHh^{-1} \in H$ and that $aa^{-1} = e$ ?

Thank you for the time you spent reading my doubts.

Answer 1

To answer your question, "No." It is true that $ahH(ah)^{-1}=ahHh^{-1}a^{-1}=aHa^{-1}$ (since $h,h^{-1} \in H$ we have $hHh^{-1}=H$). But it is not generally true that $aHa^{-1}=H$. In fact, if you make this jump, you are essentially assuming what you're trying to prove!

Let's work with an element instead to better see what needs to be done. Suppose $x\in aH$. Then there exists some $h\in H$ such that $x=ah$. Now suppose $k \in H$ and consider $xkx^{-1}=ahkh^{-1}a^{-1}$ [We need to show $xkx^{-1}\in H$].

Notice that $a hkh^{-1} \in aH$ and thus $ahkh^{-1} \not\in H$. Also, $H \not= Ha$ since $a \not\in H$. Therefore, $ahkh^{-1}$ does not belong to $H$ and thus must belong to the only other right coset: $Ha$. Thus there exists some $h' \in H$ such that $ahkh^{-1}=h'a$. So $xkx^{-1}=ahkh^{-1}a^{-1}=h'aa^{-1}=h' \in H$ and so $xHx^{-1} \subseteq H$.

And thus we arrive at the most convoluted proof of the index 2 theorem I've ever seen. :)

It's better to work with the condition (usually used as a definition of normality): "$H$ is normal in $G$ if and only if its left and right cosets are equal."

First, recall that cosets partition the group. So if there are only two cosets (one of which is the subgroup itself), then the second coset must be the stuff that's left over. So the cosets of $H$ in $G$ are $H$ and $G-H = \{x\in G\;|\; x\not\in H\}$ (the complement of $H$ in $G$).

Let $x\in G$. Case 1: $x\in H$. Then $xH=H=Hx$. Case 2: $x\not\in H$. Then $xH \not= H$ and so $xH=G-H$. Likewise $Hx \not= H$ so $Hx=G-H$. Therefore, $xH=G-H=Hx$. Thus the left and right of cosets match so $H$ is normal in $G$.

The idea behind the theorem is that the subgroup itself is always both a left and right coset. So if there are only two cosets, there's not enough room for the non-subgroup coset to be mismatched. Now if the index is three more can happen. In this case there is room for cosets to be mismatched and thus $H$ wouldn't necessarily normal. For example: $H = \{ (1),(12) \}$ is a subgroup of index 3 in $S_3$ (permutations of ${1,2,3}$). But $H$ is not normal.

Answer 2

Let $H$ have index $2$ in a group $G.$ Let $g$ be any element of $G$. If $g ∈ H,$ then $gH = H = Hg.$ If $g$ is not in $H$, then, since there are exactly two left cosets of $H$ in $G$ and $g$ is not in $H,$ they must be $H$ and $gH$. Since left cosets are disjoint, we know $gH = G − H$. But right cosets are also disjoint, so $Hg = G − H$. Hence $Hg = G − H = gH$. Thus $ gH = Hg$ for all $g ∈ G$, so $H$ is normal.

Answer 3

$G$ acts by multiplication from the left on $G/H$, giving a group morphism $f$ from $G$ to the permutations of this two-element set. The stabiliser in this action of the coset $H$, in other words the subgroup $\{\, g\in G\mid g\cdot H=H\,\}$, is equal to $H$ itself (this is true for any subgroup). But a permutation that fixes one of the two cosets also fixes the other (it has nowhere else to go), so this stabiliser$~H$ is also the kernel of$~f$, and therefore a normal subgroup.

Answer 4

If $[G:H]=2$, then necessarily $Ha=aH$ for every $a\in G\setminus H$.

Answer 5

Let $G$ be a group and $H$ be it's subgroup of index $2$. Since $H$ is of index $2$, the only right costs are $ H$ and $Ha$, where $a\in G$. Therefore,

(1) $G=H\cup Ha$.

Similarly,

(2) $G=H\cup aH$.

Hence, from (1) and (2), $H\cup Ha=H\cup aH\implies Ha=aH$. Therefore, $H$ is normal.

Answer 6

Here's an overkill using character theory, just for fun.
Since $2=|G:H|=(1_H)^G(1)$ and $1_G$ is an irreducible constituent of $(1_H)^G$ it follows that $$(1_H)^G=1_G+\chi,$$ for some linear character $\chi$. Now, since $0\not=[(1_H)^G,\chi]=[1_H,\chi_H]$ and both $1_H,\chi_H\in\text{Irr}(H)$ it follows that $1_H=\chi_H$ and hence $$H=\ker1_H=\ker\chi_H=\ker\chi\cap H.$$ Thus, $$H\subseteq\ker\chi=\ker(1_H)^G=\bigcap_{g\in G}\ker(1_H)^g=\bigcap_{g\in G}H^g=\text{Core}_G(H),$$ from where we conclude that $H=\text{Core}_G(H)$ and $H\unlhd G$.

Answer 7

One can also try proving that $H$ is the kernel of the homomorphism $\varphi:G\to \mathbb Z/2\mathbb Z$ which sends every element of $H$ to $[0]$ and every other element of $G$ to $[1]$. The hard part of this proof is not showing that $H$ is the kernel, but rather showing that $\varphi$ is a homomorphism!

Answer 8

Another approach:

Since $|G:H|=2$, $G = H \cup aH$ for some $a \in G \setminus H$.

We define a group homomorphism $\varphi: G\to S$ such that ker$\varphi=H$, so, corresponding to $H$ and $aH$ defined above, $G$ has two fibres under $\varphi$, and whatever element of G that has the image $e$ under $\varphi$ is in ker$\varphi=H$.

Obviously for any $g,h \in H$, $ghg^{-1} \in H$ holds. For any $h \in H$ and $g \notin H$, we examine the image of $ghg^{-1} $ under $\varphi$ and find:

$$\varphi(ghg^{-1} )=\varphi(g)\varphi(h)\varphi(g^{-1})=\varphi(g)e\varphi(g^{-1})=\varphi(g)\varphi(g^{-1})=e$$

Hence $ghg^{-1} $is in ker$\varphi=H$.

Answer 9

Here is my proof. It consists of smaller parts and I thought of sharing it.

Because of $(G:H)=2$, we have a decomposition $G= H\dot\cup gH$ for some $g\in G\backslash H$. We want to show that $gH=Hg$ for all $g\in G$. For $g\in H$ this is obvious, therefore assume $g\in G\backslash H$. Note, that the map $f: gH\rightarrow Hg$ given by $f(gh):=hg$ is bijective, hence $|gH|=|Hg|$. If we can show $Hg\subseteq gH$, we are finished. Let $hg\in Hg$. Assume $hg\notin gH$, i.e. $hg\in H$. Then there exist $h_1,\dots,h_r \in H$, s.t. $hg= h_1\cdot \cdots\cdot h_r$, but then $g=h^{-1}h_1\cdot\cdots\cdot h_r\in H$, which contradicts the assumption $g\in G\backslash H$. This concludes the proof.