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Suppose $(G,\cdot)$ is a group and $H$ is a subgroup of $G$, with $|H|=|G|/2$. Is $H$ a normal subgroup of $G$?

It's a theorem that's asserted without proof in a book for my Abstract Algebra class, so it's probably true, I'm just having trouble convincing myself. Lagrange's theorem tells us that there are two unique (left) cosets of $H$, and that they form a partition for $G$. The same can be said for the right cosets, so we essentially have two partitions for $G$, that each split the number of elements evenly between the two cosets. In order for $H$ to be normal we need the left and right cosets to be equal, but:

(1) I fail to see how $xH \cup yH$ and $Hx \cup Hy$, with $x,y\in G$, must necessarily be the partitions of $G$. Why not $xH \cup yH$ and $Hw \cup Hz$, for $w,z\in G$?

(2) Even if $xH \cup yH$ and $Hx \cup Hy$ are the two partitions, what is preventing $xH=Hy$ and $yH=Hx$, for $x\neq y$?

kieransquared
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    Worth being careful: this only works (or makes sense even) for finite groups - for infinite groups you need more specifically that the index of $H$ in $G$ is $2$. (i.e. there are two cosets) – Milo Brandt Nov 29 '20 at 00:29
  • More generally: let $G$ be a finite group and let $p$ be the smallest prime that divides $|G|$. Then any subgroup of index $p$ is normal. – Pranav Chinmay Nov 29 '20 at 01:41

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We have $xH=H$ or $xH=G\setminus H$, where the former holds iff $x\in H$. Likewise $Hx=H$ or $Hx=G\setminus H$, where the former holds iff $x\in H$. At any rate $xH=Hx$.

Hagen von Eitzen
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  • I guess I'm still unsure why $xH$ must equal $H$ or $G\backslash H$. Is there some unstated result I'm missing? – kieransquared Nov 29 '20 at 00:34
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    Cosets are either equal or disjoint. If $xH$ is not $H$, then, it can't meet $H$ anywhere. If $G$ is finite, then it must consist of $|H|$ elements which do lot lie in $H$, which is all of the elements which do not lie in $H$. – user3482749 Nov 29 '20 at 00:37
  • Got it, thanks. That clarifies things. – kieransquared Nov 29 '20 at 00:39