Mystified by Rotman's proof of normality of index-2 subgroups

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On page 43 of his Advanced Modern Algebra (2d ed., 2010), Rotman gives a mystifyingly elaborate proof (by contradiction no less) of the normality of any subgroup of index 2 (Proposition 1.86 (ii)).

This proof is not incorrect, AFAICT, but it makes me wonder what's wrong with this naive 1-line proof: If $H < G$ are groups with $[G\;:\:H\;] = 2$, then (regarding $G, H, aH$, and $Ha$ as sets):

$$aH = G - H = Ha \;,\;\;\;\forall \; a \in G - H$$

(I.e. $G = H \; \cup \; aH = H \; \cup \; Ha$, and both unions are disjoint.)

Am I missing something?

Thanks!

Edit: OK, when transcribing Rotman's proof (in response to Arturo's request), I noticed a footnote that explains the mystery, or most of it (details below). My apologies for my careless reading.

(Perhaps I should point out that, I am not reading this 1000-page behemoth cover-to-cover, but rather I consult it occasionally as a reference.)

Anyway, FWIW, here's Rotman's proof:

(ii) 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 $a \notin H$.

The prospect of making sense of this oddly worded, belabored proof was what prodded my lazy brain into casting about for a simpler one...

Anyway, as I said earlier, upon transcribing the above I noticed for the first time an unobtrusive footnote marker leading to the remark: "[a]nother proof of this is given in Exercise 1.57 on page 45." Sure enough, exercise 1.57 says: "(i) Show that if $H$ is a subgroup $bH = Hb = \{hb\;:\;h\in H\;\}$ for every $b\in G$, then $H$ must be a normal subgroup. (ii) Use part (i) to give a second proof of...: if $H\subseteq G$ has index $2$, then $H \triangleleft G$."

It is surprising to me that Rotman gives such vanishingly discreet billing to the equivalence $$H \triangleleft G \;\;\; \Leftrightarrow \;\;\; \forall\,g\in G, \;\; gH = Hg$$

I remember the RHS of this equivalence as the definition of normality (rather than Rotman's $h\in H, g\in G \Rightarrow ghg^{-1}\in H$). If it is not made the definition of normality, then its equivalence with normality is certainly a more general, and IMHO, important, theorem than the one about the normality of an index-$2$ subgroup. But, be that as it may, the fact that this result was not available at this point in the text explains, perhaps, the choice of the more complex proof.

Once again, my apologies for my careless reading.

Solution 1:

You're not missing anything. Your proof is fine.