Let $G$ be a finite group of order $2m$ where $m>1$ is an odd natural number.
Question. Is it true that any such $G$ has a subgroup $H$ of index 2?
If yes, I would be grateful for a reference or a proof.
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asked Oct 20, 2023 at 6:49
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5$\begingroup$ This is a classical one. See for instance this MathSE post and this one. I guess it also appeared on MO many times. For instance, in this MO answer, it is explained that the same signature argument shows that if the 2-Sylow of $G$ are cyclic and nontrivial then $G$ has a subgroup of index 2. $\endgroup$– YCorCommented Oct 20, 2023 at 9:40
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2$\begingroup$ @YCor: Thank you! I never know which question is suitable for MO and which for MSE. Feel free to vote to migrate to MSE or to close as duplicate. $\endgroup$– Mikhail BorovoiCommented Oct 20, 2023 at 10:57
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1$\begingroup$ This is Exercise 4.2.13 of 'Abstract Algebra, 3rd ed.' by David S. Dummit and Richard M. Foote. Many solutions for exercises in this text are online. For this exercise, see linearalgebras.com/… $\endgroup$– Keith KearnesCommented Oct 20, 2023 at 18:30
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1 Answer
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Yes, in fact in algebra classes in Germany, this is a well-known example or homework problem: Consider the regular action of $G$. Then an element of order $2$ in $G$ is a product of $m$ transpositions, so it is an odd permutation. Thus $G\cap\text{Alt}_{2m}$ is a subgroup of index $2$.
This generalization follows from Burnside's normal p-complement theorem: Let $p$ be the smallest prime divisor of $\lvert G\rvert$, and suppose that the Sylow $p$-subgroup $P$ is cyclic. Then $G$ has a normal subgroup of index $\lvert P\rvert$.
answered Oct 20, 2023 at 7:14
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$\begingroup$ Many-many thanks! I took the liberty to add details to your answer (for the other users of MO). Roll back if you don't like my editing. $\endgroup$ Commented Oct 20, 2023 at 7:26