Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

(I'm not sure if this question is suitable for Mathematics SE. Please comment if not.)

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.

share|improve this question
3  
But, let's be honest. Your proof is 2 lines (3 lines if you count the extra explanation at the end). You just highlighted one line. –  Graphth Dec 8 '11 at 13:57
    
When I tried to prove the statement, the image that came instantly to my mind was what I highlighted. The part "Let $H<G$ be groups with $[G:H]=2$" just recaps part of the statement of the theorem; it's not part of the proof. The stuff in parentheses (before and after) reflects the fact, unlike Rotman, I cannot assume the reader is familiar with my notation (e.g. that $G - H$ means set difference) and with some earlier results (like $G = H \cup aH$). All that's left is the $a\in G-H\;\;$ bit. I've incorporated it into the "one-line", which is now, literally, the proof. –  kjo Dec 8 '11 at 15:26
1  
Well, in the Fourth edition, the problem has been relegated to an exercise (2.30). Could you give a sketch of the argument given? It's possible that he has an eye on the standard generalization ("if $[G:H]$ is the smallest prime that divides $|G|$, then $H$ is normal"). –  Arturo Magidin Dec 8 '11 at 16:19
1  
"(I'm not sure if this question is suitable for Mathematics SE....)" I am quite sure that it is suitable. Also, as Qiaochu writes, your proof is dead-on correct, and anything much more elaborate than this is probably, at best, an overly careful explanation of what you wrote. –  Pete L. Clark Dec 8 '11 at 16:51
    
@ArturoMagidin: you must be thinking of his An introduction to the theory of groups. The book I'm referring to is this one. –  kjo Dec 9 '11 at 2:46

1 Answer 1

up vote 10 down vote accepted

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

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.