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Let $k$ be a positive integer. Is it true that any finite group $H$ of cardinal $4k+2$ whose center contains an element $h$ of order $2$ is isomorphic to the direct product $H=(\mathbb{Z}/2\mathbb{Z})\times G$, where $G=H/\{1,h\}$?

An equivalent statement would be: Let $G$ be a finite group of odd cardinal. Is it true that the second cohomology group $H^2(G,\mathbb{Z}/2\mathbb{Z})$ with respect to the trivial action of $G$ on $\mathbb{Z}/2\mathbb{Z}$, vanishes?

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2 Answers 2

up vote 10 down vote accepted

There is a reasonably simple argument that any group $H$ of twice odd order has a normal subgroup $N$ of index 2. Given that, if you know also that $H$ has a central subgroup $Z$ of order 2, then it is straightforward to show that $H \cong N \times Z$.

The argument goes like this. By Cayley's Theorem, there is an isomorphism $\phi$ that maps $G$ to a permutation group of degree $|G|$ in which all non-identity elements of $G$ act without fixed points. So the elements of order 2 in $G$ are mapped by $\phi$ onto odd permutations, and hence the inverse image $N$ in $G$ of the intersection of the image of $\phi$ with the alternating group has index 2 in $G$.

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This is exactly the kind of argument I was looking for, thank you Derek! –  Andrei Moroianu Nov 14 '10 at 17:57
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Yes, because $G$ and $Z/2Z$ have coprime order, every extension of one by the other splits by the Schur-Zassenhaus theorem (http://en.wikipedia.org/wiki/Schur-Zassenhaus_theorem). If you prefer, $H^2(G,M)$ is killed both by $|G|$ and by $|M|$, so it must be trivial in your case.

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Thanks, Tim; I have found some more references about the S-Z theorem here: math.uconn.edu/~kconrad/blurbs/grouptheory/schurzass.pdf I wonder, however, if there is some simple argument in this particular case, since I have the impression that one burns the house to fright the mouse away (this is the english equivalent I have found for the french expression 'tuer une mouche avec un canon'). –  Andrei Moroianu Nov 14 '10 at 17:21
    
Actually, Tim's second argument about the vanishing of $H^2$ is independent of Schur-Zassenhaus and I don't think you can hope for anything simpler: the hypothesis that the order of your subgroup is co-prime to its index is crucial, because otherwise the result would be false in general. Since you asked whether $H^2$ vanished, it doesn't really get much simpler than just saying "yes, because it is killed by restriction-corestriction, which is multiplication by $2k+1$, and by 2". –  Alex B. Nov 14 '10 at 18:04
    
Well, Derek's argument appears simpler to me, though. –  Andrei Moroianu Nov 14 '10 at 18:08
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