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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.

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    $\begingroup$ You should not try to prove $x \in H$, but $xHx^{-1} \subseteq H$. $\endgroup$ Commented Nov 22, 2011 at 17:42
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    $\begingroup$ One of equivalent definitions of normal subgroup is that The sets of left and right cosets of are the same. If you know this, you could use it to find a very simple proof. $\endgroup$ Commented Nov 22, 2011 at 17:57
  • $\begingroup$ Possible duplicate math.stackexchange.com/questions/420583/… $\endgroup$
    – user53259
    Commented Feb 4, 2014 at 13:44
  • $\begingroup$ Also a consequence of "strong Cayley's theorem" $\endgroup$
    – qwr
    Commented Nov 27, 2018 at 6:37

9 Answers 9

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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.

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  • $\begingroup$ Is there anything wrong in the following proof: As $|G:H|=2$, $H$ divides $G$ into $H$ and $G-H=aH$. As multiplication of two cosets $aH \cdot H=(a \cdot e)H=aH$, and also the other group axioms are satisfied. $G/H$ has a group structure, and hence $H$ is normal. I am not sure if the above statement can be used in its converse form, and if two cosets can be multplied together. $\endgroup$
    – user23238
    Commented Jan 27, 2013 at 13:28
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    $\begingroup$ A subgroup is normal if and only if coset multiplication is well defined. The other group axioms aren't really in question (if the operation is well defined, they'll follow for free). Your argument seems to be using the particular representative "$e$" for "$H=eH$". You would need to show that any other representative would do just as well (the same if "$a$" is replaced by some equivalent representative "$b$" where $bH=aH$). I'm sure your argument could be repaired but this doesn't seem to be a fruitful path -- it's much harder than it needs to be. $\endgroup$
    – Bill Cook
    Commented Feb 4, 2013 at 16:03
  • $\begingroup$ Just a little detail, what did you mean by $(1)$ at the end when defining $H$ in your example? i.e. when you wrote $H=\{(1),(12)\}$ I understand $(12)$ of course, but what did you mean by $(1)$? where does the $1$ goes? thanks a lot. $\endgroup$ Commented Aug 9, 2016 at 21:26
  • $\begingroup$ $(12) $ is the transposition that switches 1 and 2 (and leaves 3 fixed). $(1) $ is the identity element (everything is fixed). In other words, this is the cyclic subgroup generated by $(12) $. $\endgroup$
    – Bill Cook
    Commented Aug 9, 2016 at 22:03
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    $\begingroup$ Helps a lot thanks! $\endgroup$
    – Cloud JR K
    Commented Sep 2, 2018 at 14:33
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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.

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    $\begingroup$ There might be a little technical issue: I've seen many cases where the index $(G:H)$ is defined as the number of left cosets, and your argument here assumes that the number of right cosets is the same as the number of left cosets. This is clear if G is finite, and in general the bijection $G\to G, g\mapsto g^{-1}$ induces a bijection between left and right cosets through mapping $aH$ to $Ha^{-1}$. Just thought I'd write this down as a younger me would have gotten confused over this point :) $\endgroup$ Commented Dec 16, 2022 at 14:33
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$G$ 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.

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

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  • $\begingroup$ so $Ha = aH$, since they both equal $G\backslash H$, right? $\endgroup$
    – orangeskid
    Commented Nov 29, 2020 at 2:43
  • $\begingroup$ @orangeskid, Yes that's right. $\endgroup$
    – user810157
    Commented Nov 29, 2020 at 9:08
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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.

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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 $H\unlhd G$.

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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!

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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$.

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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.

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