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The (old) exam I'm looking at has the following problem:

Suppose the order of $G$ is even, but is not divisible by $4$. Prove that $G$ is not simple.

A group with $2$ elements is clearly a counter-example to that.

Are those the only counterexamples?
(If yes, how would one prove that?)

Karl Kroningfeld
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  • FYI, sometimes prime order groups are not regarded as simple by convention (although they certainly satisfy the definition). The idea is roughly that every other simple group is interesting, and it allows statements like "A finite group is not solvable iff it has a simple composition factor". I wrote an algebra qual question with this convention once, without realizing it. Luckily the students just acknowledged the exception and moved on. – Pete L. Clark Sep 09 '13 at 22:52

3 Answers3

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It is indeed the only counterexample. Here's how you go about showing that.

Let $|G|=2n$ where $n$ is odd, and let $x\in G$ have order $2$. The permutation $G\to G$ given by multiplication (say on the left) by $x$ is an involution, and yet it fixes nothing. Thus its cycle decomposition is comprised of $n$ $2$-cycles, whence $x$ induces an odd permutation.

The set of elements of $G$ which induce even permutations forms a subgroup $H$. For any $g\in G\setminus H$, we note that $xg$ is an element of $H$, so $g\in x^{-1}H$. This means that the index of $H$ in $G$ is $2$, and and as result $H\lhd G$.

If $|G|>2$, then $H$ is a nontrivial normal subgroup of $G$.

I learned this argument in the text Finite Group Theory by Martin Isaacs, pg. 35, and I have hardly changed it here.

Karl Kroningfeld
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This is a standard result, and the proof -- show that the hypotheses imply that the left action of $G$ on itself does not land in the alternating subgroup $A_{|G|}$ and thus the kernel of the sign homomorphism gives an index $2$ normal subgroup -- is also standard. But I had a hard time remembering it: I would look it up in my undergraduate algebra text every few years. $\newcommand{\ra}{\rightarrow}$

A few months ago though I found a construction which is a bit more general but also more conceptual, and thus I think I will remember it from now on. The idea is that whenever a group $G$ acts on a finite set $X$ we get a homomorphism $G \ra \operatorname{Sym} X$; by composing with the natural sign map $\operatorname{Sym} X \ra \{\pm 1\}$ we get a signature homomorphism $\epsilon_X: G \ra \{ \pm 1\}$. It is interesting to try to determine $\epsilon_X$ for various natural representations: in fact, this is related to Zolotarev's proof of quadratic reciprocity and came up in some work of mine on "abstract" generalizations thereof.

If we start with the group $G$, what is the most natural $G$-set? Certainly it is $G$ acting on itself, say from the left (doing it on the right doesn't change what happens next). I call the associated homomorphism $\epsilon_G: G \ra \{ \pm 1\}$ the Cayley signature map of $G$. It turns out that there is a surprising clean characterization of $\epsilon_G$:

Lemma: Let $G$ be a finite group.
a) The following are equivalent:
(i) The Cayley signature homomorphism $\epsilon_G$ is nontrivial.
(ii) The Sylow $2$-subgroups of $G$ are cyclic and nontrivial.
b) If $\epsilon_G$ is nontrivial, its kernel is the unique index $2$ subgroup of $G$.

I discovered this result for myself and was pleased with it. I found though that part a) previously appeared in a 1979 MONTHLY note of Patrick Morton. For a proof and for further related results, see Lemma 3 here. Part a) is very elementary and comes from the easy identification of the cycle type of $g \cdot$ for any $g \in G$. Part b) uses what I think must be the very first "normal complement theorem", due in this case -- appropriately enough! -- to Cayley. (A reference to Cayley's normal complement theorem seems to be missing currently; maybe someone can help me out here...)

Now let $G$ be finite of order $n \equiv 2 \pmod{4}$. Then its Sylow $2$-subgroups have order $2$, so must be cyclic and nontrivial. It follows that $\epsilon_G$ is nontrivial, and its kernel is a normal, index $2$ subgroup of $G$ (of course index $2$ subgroups of any group are always normal). So $G$ cannot be simple unless it itself has order $2$. Moreover $\operatorname{Ker} \epsilon$ is the unique index $2$ subgroup of order $G$.

As the OP says, this problem is often asked on qualifying exams. In fact algebra quals often ask other problems which can be solved by this technique of extracting an index $2$ subgroup from the Cayley action of $G$ on itself. I think that the above Lemma will help with these as well...

Community
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Pete L. Clark
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  • "index subgroup" $:\mapsto:$ "index 2 subgroup" $;;;$ –  Sep 09 '13 at 05:42
  • Hmm. Honestly I thought this was really cool, and it's disappointing to only get one upvote. Oh, well. – Pete L. Clark Sep 09 '13 at 15:48
  • PeteL.Clark: I upvoted this shortly after you posted it. IIRC, my answer was accepted just before you put your answer here, and I honestly expected the checkmark to find a new home. The lemma not only solves handles many of the problems of this type, it helps conceptualize the idea. For instance, it tells us exactly why we cannot apply the same argument to an arbitrary group of order $60$ or in general a group of order $2^kn$ for $k>1$. – Karl Kroningfeld Sep 09 '13 at 21:44
  • @Karl: Thanks. I certainly don't have any problem with your answer being the accepted one: it was first, and it is a simpler take on the problem. But I'm glad you liked my answer! – Pete L. Clark Sep 09 '13 at 22:03
  • @ Pete L. Clark: Great answer! – leshik Sep 10 '13 at 15:56
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Suppose G is a nonabelian group of order 2m where m is odd, and it has a subgroup A = {1,a} of order 2 which is not normal. Note that $a^2$ = 1. Then $\exists$ y $\in$ G such that

(1) $y^{-1}ay$ =$b \neq$ a. Then

$b^2$ = $y^{-1}ayy^{-1}ay$ = 1
$a^2$ = $b^2$ so that

aa = bb $\Rightarrow$ $ab^{-1} =ba^{-1}$ $\Rightarrow$ ab= ba

Then the set {1,a,b,ab} is a group. You can see that by inspection (multiplying together any two elements). For example bab = bba = a.

But 4 does not divide 2m, so there cannot be a subgroup of order 4 in G. So there is no y as in (1).

Betty Mock
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    -1: Please check your new argument against the group that was given to you as a counterexample to your last argument. – Pete L. Clark Sep 12 '13 at 23:29
  • @PeteL.Clark $S_3$ has only normal subgroups of order 2, which can be checked by inspection. So my subgroup of order 4 would not appear there. My argument here says that if a subgroup of order 2 is not normal there must be a subgroup of order 4, and I show how to construct it. – Betty Mock Sep 13 '13 at 00:30
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    Note that $(123)^{-1}(12)(123)=(132)(12)(123)=(13)$. – Karl Kroningfeld Sep 13 '13 at 01:09
  • @Karl Thank you Karl. – Betty Mock Sep 13 '13 at 07:24