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fixes a typo
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Peter Mueller
Peter Mueller
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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 $P$$\lvert P\rvert$.

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 $P$.

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 $\lvert P\rvert$.

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Mikhail Borovoi
Mikhail Borovoi
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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 $P$.

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 is odd. 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 $P$.

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 $P$.

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Peter Mueller
Peter Mueller
  • 22.5k
  • 1
  • 75
  • 107

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 is odd. 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 $P$.