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Let $H$ a subgroup of $G$ and the index of $G\backslash H$ is $2$. Prove that $H$ is a normal subgroup of $G$.

I have already know that $1_G *H $ is an element of $G\backslash H$. I miss the little flash to continue the problem. Will someone be able to give me advice?

2 Answers2

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Suppose $H≤G$ such that $[G:H]=2$.

Thus $H$ has two left cosets (and two right cosets) in $G$.

If $g\in H$, then $gH=H=Hg$.

If $g∉H$, then $gH=G\backslash H$ as there are only two cosets and the cosets partition G.

For the same reason, $g∉H\implies Hg=G\backslash H$.

That is, $gH=Hg$.

The result follows from the definition of normal subgroup.

Relure
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  • no need to reproduce duplicates. – uniquesolution Oct 28 '15 at 18:05
  • "If $g∉H$, then $gH=G\backslash H$ " is not clear at all for me. It's clear that $gH \subset G\backslash H$, but $G\backslash H \subset gH $ it not because all elements of $1_GH$ are not in $gH$. Am I wrong? –  Oct 28 '15 at 19:02
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Hint: Show that $xH = Hx$, for every $x \in G$, by using that $\sim_E$ is an equivalence relation, therefore G is a disjoint union of its equivalent classes and $xH = G-H$, similarly $Hx = G - H$.

Aaron Maroja
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