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I have this problem to solve

Let $G$ be a group and $H$ a subgroup. Show that $H<G$ such that $[G:H]:=\dfrac{|G|}{|H|}=2$ -- where $|\cdot|$ denotes the order of a group --- is necessarily normal. But if $[G:H]=3$ the subgroup $H$ is not necessarily normal, then show an example.

I tried to solve like so:

Firstly, let $x\in G$ be an element of group $G$. Then $x\in H$ OR $x\in\mathbb{L}$, the coset of $H$. Being $[G:H]=2\implies G$ is abelian, then $xH=Hx\implies H\triangleleft G$.

Secondly, I considered as $G=S_3$ and $H=\langle(1,2)\rangle$.

If my solution is wrong, please correct me.

Thank you so much

Puck
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    Why would $G$ be Abelian? It doesn't follow. Also, which subgroup of $S_3$ have you picked? – Berci Oct 02 '21 at 21:53
  • To show $H$ is normal you can pick $g\in G$ and prove that $Hg=gH$. try thinking what happens for $g\in H$ and for $g\in G\setminus H$. – Oria Oct 02 '21 at 21:55

1 Answers1

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Your conclusion that $G$ is abelian is definitely wrong. There are lots of non-abelian groups which contain a subgroup of index $2$. For example, $S_n$ contains such a subgroup for every $n$.

Thing is, $H$ is a left coset. Since $G$ is the disjoint union of the left cosets of $H$, the other coset must be the complement $G\setminus H$. Same thing about the right cosets. So now, if $x\in H$ then both $xH$ and $Hx$ are equal to $H$. If $x\notin H$ then $xH, Hx$ are both not equal to $H$, and so both must be equal to the other left/right coset $G\setminus H$. So in either case, we have $xH=Hx$ for all $x\in G$.

As for index $3$, taking $G=S_3$ is a good idea. However, it is unclear what subgroup did you take. You need to take some subgroup of order $2$. Take any such subgroup and show it isn't normal.

Mark
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