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.


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 $xHx^{-1} \subseteq H$. $\endgroup$ – Matthias Klupsch Nov 22 '11 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. $\endgroup$ – Martin Sleziak Nov 22 '11 at 17:57
  • $\begingroup$ Possible duplicate math.stackexchange.com/questions/420583/… $\endgroup$ – Group Theory Feb 4 '14 at 13:44
  • $\begingroup$ Also a consequence of "strong Cayley's theorem" $\endgroup$ – qwr Nov 27 '18 at 6:37

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.

  • $\begingroup$ 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 Jan 27 '13 at 13:28
  • $\begingroup$ 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 Feb 4 '13 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. $\endgroup$ – Cristian Baeza Aug 9 '16 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 Aug 9 '16 at 22:03
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    $\begingroup$ Helps a lot thanks! $\endgroup$ – Cloud JR Sep 2 '18 at 14:33

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.


$G$ 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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