I have been going through some material on elementary group theory (Group Theory for Physicists, Christoph Ludeling) and in one place it states:

A subgroup which contains half of all elements, i.e. for which $|G|=2|H|$, is a normal subgroup.

Where a subgroup $H\subset G$ is called a normal subgroup if $gHg^{-1}=H~\forall g\in G$. How would I go about proving this? Sorry if the answer exists somewhere, but I tried searching and could not find it.

  • 4
    $\begingroup$ You will find it if you search for «subgroups of index 2» $\endgroup$ May 12, 2013 at 2:04
  • $\begingroup$ Thanks for that, that was what exactly I was looking for. $\endgroup$ May 12, 2013 at 2:12

2 Answers 2


The two left cosets of $H$ are $H$ and $G\setminus H$, and the two right cosets of $H$ are also $H$ and $G\setminus H$. Clearly $xH=Hx$ for every $x\in H$, and if $x\in G\setminus H$, then $xH=G\setminus H=Hx$ as well. Thus, $xH=Hx$ for each $x\in G$, and $H$ is normal in $G$.

  • $\begingroup$ Stylistic complaint: the $x$ you fixed in the the first part of the first sentence is not used until the second sentence :-) $\endgroup$ May 12, 2013 at 2:08
  • $\begingroup$ @Mariano: Yes, that’s pretty horrible. It was left over from an earlier version; I just forgot that it was there when I changed the first sentence. $\endgroup$ May 12, 2013 at 2:10
  • $\begingroup$ Thank-you for that, much appreciated. I will accept your answer in four minutes when it lets me. $\endgroup$ May 12, 2013 at 2:13
  • $\begingroup$ @Andrew: You’re very welcome. $\endgroup$ May 12, 2013 at 2:15

Not as slick but perhaps more direct (as given in Rotman):

It suffices to prove that if $h \in H$, then the conjugate $ghg^{-1}\in H$ for every $g \in G$. If $g \in H$, then $ghg^{-1}\in H$, because $H$ is a subgroup. If $g \notin H$, then $g = ah_0$, where $h_0\in H$ (for $G = H \cup aH$). If $ghg^{-1} \in H$, we are done. Otherwise, $ghg^{-1} = ah_1$ for some $h_1\in H$. But $ah_1 = ghg^{-1} = ah_0hh_0^{-1}a^{-1}$. Cancel $a$ to obtain $h_1 = h_0hh_0^{-1}a^{-1}$, contradicting $a \notin H$.

This was the proof I came up with before reading the slick proof in the text I was reading, seeing it appear in Rotman later made me feel less silly.

  • $\begingroup$ Thanks for that, it is always useful to see different ways of looking at things. $\endgroup$ May 16, 2013 at 1:40

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