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How to show that $G$ can be expressed as a semidirect product
The Next CEO of Stack OverflowProve that there is no element of order $8$ in $SL(2,3)$Can the semidirect product of two groups be abelian group?Show that every group of order 48 has a nontrivial normal subgroup.Semidirect Product of Two GroupsSemidirect product: general automorphism always results in a conjugationHow to find the images of the external semidirect product?Is semidirect product unique?External Semidirect product and isomorphismSemidirect product $mathbbF_q rtimes mathbbF_q'$The semidirect product $(C_7times C_13)rtimes C_3$Are groups constructed using semidirect product always non-abelian?
$begingroup$
Let $G$ be a group of order $42$. Prove that $G$ is a semidirect product of a normal subgroup of order $21$ and $mathbbZ_2$.
My attempt: $G$ has unique Sylow 7 subgroup and Sylow 3 subgroup is not unique (as $n_3 = 1$ (mod $3)$ and $n_3 | 14$, $implies n_3 = 7$?), and so the Sylow 3 subgroup is not normal (if it were, we could just take the product of Sylow 7 and Sylow 3), so I'm not sure how to construct a normal subgroup of order 21.
As far as construction of an explicit homomorphism goes $phi: mathbbZ_2rightarrow Aut(S)$, where $S$ is the normal subgroup of $G$ of order 21, $phi(0) = id_S$ and $phi (1)$ is the automorphism that sends each element to its conjugate as stated in the answer.
Also, how can we prove that elements of odd order form a subgroup of index 2 as suggested in the answer. Thanks.
abstract-algebra group-theory finite-groups sylow-theory semidirect-product
asked Mar 18 at 22:42
manifoldedmanifolded
49519
$endgroup$
add a comment |
$begingroup$
Let $G$ be a group of order $42$. Prove that $G$ is a semidirect product of a normal subgroup of order $21$ and $mathbbZ_2$.
My attempt: $G$ has unique Sylow 7 subgroup and Sylow 3 subgroup is not unique (as $n_3 = 1$ (mod $3)$ and $n_3 | 14$, $implies n_3 = 7$?), and so the Sylow 3 subgroup is not normal (if it were, we could just take the product of Sylow 7 and Sylow 3), so I'm not sure how to construct a normal subgroup of order 21.
As far as construction of an explicit homomorphism goes $phi: mathbbZ_2rightarrow Aut(S)$, where $S$ is the normal subgroup of $G$ of order 21, $phi(0) = id_S$ and $phi (1)$ is the automorphism that sends each element to its conjugate as stated in the answer.
Also, how can we prove that elements of odd order form a subgroup of index 2 as suggested in the answer. Thanks.
abstract-algebra group-theory finite-groups sylow-theory semidirect-product
asked Mar 18 at 22:42
manifoldedmanifolded
49519
$endgroup$
add a comment |
$begingroup$
Let $G$ be a group of order $42$. Prove that $G$ is a semidirect product of a normal subgroup of order $21$ and $mathbbZ_2$.
My attempt: $G$ has unique Sylow 7 subgroup and Sylow 3 subgroup is not unique (as $n_3 = 1$ (mod $3)$ and $n_3 | 14$, $implies n_3 = 7$?), and so the Sylow 3 subgroup is not normal (if it were, we could just take the product of Sylow 7 and Sylow 3), so I'm not sure how to construct a normal subgroup of order 21.
As far as construction of an explicit homomorphism goes $phi: mathbbZ_2rightarrow Aut(S)$, where $S$ is the normal subgroup of $G$ of order 21, $phi(0) = id_S$ and $phi (1)$ is the automorphism that sends each element to its conjugate as stated in the answer.
Also, how can we prove that elements of odd order form a subgroup of index 2 as suggested in the answer. Thanks.
abstract-algebra group-theory finite-groups sylow-theory semidirect-product
asked Mar 18 at 22:42
manifoldedmanifolded
49519
$endgroup$
Let $G$ be a group of order $42$. Prove that $G$ is a semidirect product of a normal subgroup of order $21$ and $mathbbZ_2$.
My attempt: $G$ has unique Sylow 7 subgroup and Sylow 3 subgroup is not unique (as $n_3 = 1$ (mod $3)$ and $n_3 | 14$, $implies n_3 = 7$?), and so the Sylow 3 subgroup is not normal (if it were, we could just take the product of Sylow 7 and Sylow 3), so I'm not sure how to construct a normal subgroup of order 21.
As far as construction of an explicit homomorphism goes $phi: mathbbZ_2rightarrow Aut(S)$, where $S$ is the normal subgroup of $G$ of order 21, $phi(0) = id_S$ and $phi (1)$ is the automorphism that sends each element to its conjugate as stated in the answer.
Also, how can we prove that elements of odd order form a subgroup of index 2 as suggested in the answer. Thanks.
abstract-algebra group-theory finite-groups sylow-theory semidirect-product
abstract-algebra group-theory finite-groups sylow-theory semidirect-product
asked Mar 18 at 22:42
manifoldedmanifolded
49519
asked Mar 18 at 22:42
manifoldedmanifolded
49519
edited Mar 19 at 1:45
manifolded
asked Mar 18 at 22:42
manifoldedmanifolded
49519
asked Mar 18 at 22:42
manifoldedmanifolded
49519
asked Mar 18 at 22:42
manifoldedmanifolded
49519
49519
add a comment |
add a comment |
1 Answer
1
active
oldest
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$begingroup$
My attempt: $G$ has unique Sylow 7 and Sylow 3 subgroups which are normal
No, the 3-subgroup does not have to be normal. There is a non-abelian group of order $21$, in which $14$ elements have order $3$.
Once you can construct a subgroup $H$ of order $21$, the rest is easy. There is an element $x$ of order $2$, which is not in that subgroup. Look at how $x$ acts on $H$ by conjugation. That's the explicit homomorphism.
That leaves the unresolved part - finding an index-2 subgroup.
Consider the action of $G$ on itself by (left) multiplication as a permutation - the Cayley homomorphism $pi$. For an element $x$ of order $k$, the cycle representation of $pi(x)$ consists entirely of $k$-cycles. If the number of elements in $G$ is twice an odd number and $k$ is even, then that will be an odd number of even cycles - which makes it an odd permutation. The intersection of $pi(G)$ with the alternating group form a subgroup of index $1$ or $2$ in $pi(G)$ - and, since we've found an odd permutation $pi(x)$ for some $x$ of order $2$, said intersection can't be all of $pi(G)$.
The homomorphism $pi$ is injective, so after pulling back $pi^-1(A_Gcap pi(G))$ is a subgroup of $G$ of index $2$. It consists exactly of the elements of odd order in $G$. We proved this for any order that's twice an odd number, and $42-2cdot 21$ certainly qualifies, so we're done.
Thanks to Jyrki Lahtonen for this argument constructing the subgroup of index 2.
answered Mar 18 at 23:39
jmerryjmerry
16.8k11633
$endgroup$
1
$begingroup$
When a group has order $2cdottextodd$, it always has a normal subgroup of index 2 (you can see this by seeing how an element of order 2 acts in the regular representation).
$endgroup$
– Hempelicious
Mar 19 at 1:50
2
$begingroup$
Can't you do the following. Let $P$ be the normal Sylow $7$-subgroup. Then $G/P$ is a group of order six. It has a subgroup $H$ of order three. The preimage $pi^-1(H)$, where $pi:Gto G/P$ is the natural projection, is what you are looking for.
$endgroup$
– Jyrki Lahtonen
Mar 19 at 5:04
2
$begingroup$
Alternatively, let $rho: Gto operatornameSym(G)$ be the Cayley homomorphism coming from $G$ acting on itself by left multiplication. The image of every element of an even order is a disjoint product of an odd number of cycles of an even length, i.e. an odd permutation, On the other hand, the image of an element of an odd order is a disjoint product of cycles of an odd length, i.e. an even permutation. Therefore the elements of an odd order form a subgroup (=the preimage of the alternating group).
$endgroup$
– Jyrki Lahtonen
Mar 19 at 5:09
1
$begingroup$
(cont'd) This argument shows that elements of odd order form a subgroup whenever the group has order $equiv2pmod4$.
$endgroup$
– Jyrki Lahtonen
Mar 19 at 5:09
1
$begingroup$
@JyrkiLahtonen: yes this is the proof I had in mind. An element of order 2 acts via right multiplication as an odd product of transpositions (here the regular representation is the one given by Cayley's theorem, no character theory needed). Thus the regular representation contains an odd permutation, so the preimage of the alternating subgroup is of index 2. This same argument generalizes via induction to show groups with cyclic Sylow 2-subgroups have normal subgroups consisting of all odd-order elements.
$endgroup$
– Hempelicious
Mar 19 at 5:38
|
show 4 more comments
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$begingroup$
My attempt: $G$ has unique Sylow 7 and Sylow 3 subgroups which are normal
No, the 3-subgroup does not have to be normal. There is a non-abelian group of order $21$, in which $14$ elements have order $3$.
Once you can construct a subgroup $H$ of order $21$, the rest is easy. There is an element $x$ of order $2$, which is not in that subgroup. Look at how $x$ acts on $H$ by conjugation. That's the explicit homomorphism.
That leaves the unresolved part - finding an index-2 subgroup.
Consider the action of $G$ on itself by (left) multiplication as a permutation - the Cayley homomorphism $pi$. For an element $x$ of order $k$, the cycle representation of $pi(x)$ consists entirely of $k$-cycles. If the number of elements in $G$ is twice an odd number and $k$ is even, then that will be an odd number of even cycles - which makes it an odd permutation. The intersection of $pi(G)$ with the alternating group form a subgroup of index $1$ or $2$ in $pi(G)$ - and, since we've found an odd permutation $pi(x)$ for some $x$ of order $2$, said intersection can't be all of $pi(G)$.
The homomorphism $pi$ is injective, so after pulling back $pi^-1(A_Gcap pi(G))$ is a subgroup of $G$ of index $2$. It consists exactly of the elements of odd order in $G$. We proved this for any order that's twice an odd number, and $42-2cdot 21$ certainly qualifies, so we're done.
Thanks to Jyrki Lahtonen for this argument constructing the subgroup of index 2.
answered Mar 18 at 23:39
jmerryjmerry
16.8k11633
$endgroup$
1
$begingroup$
When a group has order $2cdottextodd$, it always has a normal subgroup of index 2 (you can see this by seeing how an element of order 2 acts in the regular representation).
$endgroup$
– Hempelicious
Mar 19 at 1:50
2
$begingroup$
Can't you do the following. Let $P$ be the normal Sylow $7$-subgroup. Then $G/P$ is a group of order six. It has a subgroup $H$ of order three. The preimage $pi^-1(H)$, where $pi:Gto G/P$ is the natural projection, is what you are looking for.
$endgroup$
– Jyrki Lahtonen
Mar 19 at 5:04
2
$begingroup$
Alternatively, let $rho: Gto operatornameSym(G)$ be the Cayley homomorphism coming from $G$ acting on itself by left multiplication. The image of every element of an even order is a disjoint product of an odd number of cycles of an even length, i.e. an odd permutation, On the other hand, the image of an element of an odd order is a disjoint product of cycles of an odd length, i.e. an even permutation. Therefore the elements of an odd order form a subgroup (=the preimage of the alternating group).
$endgroup$
– Jyrki Lahtonen
Mar 19 at 5:09
1
$begingroup$
(cont'd) This argument shows that elements of odd order form a subgroup whenever the group has order $equiv2pmod4$.
$endgroup$
– Jyrki Lahtonen
Mar 19 at 5:09
1
$begingroup$
@JyrkiLahtonen: yes this is the proof I had in mind. An element of order 2 acts via right multiplication as an odd product of transpositions (here the regular representation is the one given by Cayley's theorem, no character theory needed). Thus the regular representation contains an odd permutation, so the preimage of the alternating subgroup is of index 2. This same argument generalizes via induction to show groups with cyclic Sylow 2-subgroups have normal subgroups consisting of all odd-order elements.
$endgroup$
– Hempelicious
Mar 19 at 5:38
|
show 4 more comments
$begingroup$
My attempt: $G$ has unique Sylow 7 and Sylow 3 subgroups which are normal
No, the 3-subgroup does not have to be normal. There is a non-abelian group of order $21$, in which $14$ elements have order $3$.
Once you can construct a subgroup $H$ of order $21$, the rest is easy. There is an element $x$ of order $2$, which is not in that subgroup. Look at how $x$ acts on $H$ by conjugation. That's the explicit homomorphism.
That leaves the unresolved part - finding an index-2 subgroup.
Consider the action of $G$ on itself by (left) multiplication as a permutation - the Cayley homomorphism $pi$. For an element $x$ of order $k$, the cycle representation of $pi(x)$ consists entirely of $k$-cycles. If the number of elements in $G$ is twice an odd number and $k$ is even, then that will be an odd number of even cycles - which makes it an odd permutation. The intersection of $pi(G)$ with the alternating group form a subgroup of index $1$ or $2$ in $pi(G)$ - and, since we've found an odd permutation $pi(x)$ for some $x$ of order $2$, said intersection can't be all of $pi(G)$.
The homomorphism $pi$ is injective, so after pulling back $pi^-1(A_Gcap pi(G))$ is a subgroup of $G$ of index $2$. It consists exactly of the elements of odd order in $G$. We proved this for any order that's twice an odd number, and $42-2cdot 21$ certainly qualifies, so we're done.
Thanks to Jyrki Lahtonen for this argument constructing the subgroup of index 2.
answered Mar 18 at 23:39
jmerryjmerry
16.8k11633
$endgroup$
1
$begingroup$
When a group has order $2cdottextodd$, it always has a normal subgroup of index 2 (you can see this by seeing how an element of order 2 acts in the regular representation).
$endgroup$
– Hempelicious
Mar 19 at 1:50
2
$begingroup$
Can't you do the following. Let $P$ be the normal Sylow $7$-subgroup. Then $G/P$ is a group of order six. It has a subgroup $H$ of order three. The preimage $pi^-1(H)$, where $pi:Gto G/P$ is the natural projection, is what you are looking for.
$endgroup$
– Jyrki Lahtonen
Mar 19 at 5:04
2
$begingroup$
Alternatively, let $rho: Gto operatornameSym(G)$ be the Cayley homomorphism coming from $G$ acting on itself by left multiplication. The image of every element of an even order is a disjoint product of an odd number of cycles of an even length, i.e. an odd permutation, On the other hand, the image of an element of an odd order is a disjoint product of cycles of an odd length, i.e. an even permutation. Therefore the elements of an odd order form a subgroup (=the preimage of the alternating group).
$endgroup$
– Jyrki Lahtonen
Mar 19 at 5:09
1
$begingroup$
(cont'd) This argument shows that elements of odd order form a subgroup whenever the group has order $equiv2pmod4$.
$endgroup$
– Jyrki Lahtonen
Mar 19 at 5:09
1
$begingroup$
@JyrkiLahtonen: yes this is the proof I had in mind. An element of order 2 acts via right multiplication as an odd product of transpositions (here the regular representation is the one given by Cayley's theorem, no character theory needed). Thus the regular representation contains an odd permutation, so the preimage of the alternating subgroup is of index 2. This same argument generalizes via induction to show groups with cyclic Sylow 2-subgroups have normal subgroups consisting of all odd-order elements.
$endgroup$
– Hempelicious
Mar 19 at 5:38
|
show 4 more comments
$begingroup$
My attempt: $G$ has unique Sylow 7 and Sylow 3 subgroups which are normal
No, the 3-subgroup does not have to be normal. There is a non-abelian group of order $21$, in which $14$ elements have order $3$.
Once you can construct a subgroup $H$ of order $21$, the rest is easy. There is an element $x$ of order $2$, which is not in that subgroup. Look at how $x$ acts on $H$ by conjugation. That's the explicit homomorphism.
That leaves the unresolved part - finding an index-2 subgroup.
Consider the action of $G$ on itself by (left) multiplication as a permutation - the Cayley homomorphism $pi$. For an element $x$ of order $k$, the cycle representation of $pi(x)$ consists entirely of $k$-cycles. If the number of elements in $G$ is twice an odd number and $k$ is even, then that will be an odd number of even cycles - which makes it an odd permutation. The intersection of $pi(G)$ with the alternating group form a subgroup of index $1$ or $2$ in $pi(G)$ - and, since we've found an odd permutation $pi(x)$ for some $x$ of order $2$, said intersection can't be all of $pi(G)$.
The homomorphism $pi$ is injective, so after pulling back $pi^-1(A_Gcap pi(G))$ is a subgroup of $G$ of index $2$. It consists exactly of the elements of odd order in $G$. We proved this for any order that's twice an odd number, and $42-2cdot 21$ certainly qualifies, so we're done.
Thanks to Jyrki Lahtonen for this argument constructing the subgroup of index 2.
answered Mar 18 at 23:39
jmerryjmerry
16.8k11633
$endgroup$
My attempt: $G$ has unique Sylow 7 and Sylow 3 subgroups which are normal
No, the 3-subgroup does not have to be normal. There is a non-abelian group of order $21$, in which $14$ elements have order $3$.
Once you can construct a subgroup $H$ of order $21$, the rest is easy. There is an element $x$ of order $2$, which is not in that subgroup. Look at how $x$ acts on $H$ by conjugation. That's the explicit homomorphism.
That leaves the unresolved part - finding an index-2 subgroup.
Consider the action of $G$ on itself by (left) multiplication as a permutation - the Cayley homomorphism $pi$. For an element $x$ of order $k$, the cycle representation of $pi(x)$ consists entirely of $k$-cycles. If the number of elements in $G$ is twice an odd number and $k$ is even, then that will be an odd number of even cycles - which makes it an odd permutation. The intersection of $pi(G)$ with the alternating group form a subgroup of index $1$ or $2$ in $pi(G)$ - and, since we've found an odd permutation $pi(x)$ for some $x$ of order $2$, said intersection can't be all of $pi(G)$.
The homomorphism $pi$ is injective, so after pulling back $pi^-1(A_Gcap pi(G))$ is a subgroup of $G$ of index $2$. It consists exactly of the elements of odd order in $G$. We proved this for any order that's twice an odd number, and $42-2cdot 21$ certainly qualifies, so we're done.
Thanks to Jyrki Lahtonen for this argument constructing the subgroup of index 2.
answered Mar 18 at 23:39
jmerryjmerry
16.8k11633
edited Mar 19 at 5:25
answered Mar 18 at 23:39
jmerryjmerry
16.8k11633
answered Mar 18 at 23:39
jmerryjmerry
16.8k11633
answered Mar 18 at 23:39
jmerryjmerry
16.8k11633
16.8k11633
1
$begingroup$
When a group has order $2cdottextodd$, it always has a normal subgroup of index 2 (you can see this by seeing how an element of order 2 acts in the regular representation).
$endgroup$
– Hempelicious
Mar 19 at 1:50
2
$begingroup$
Can't you do the following. Let $P$ be the normal Sylow $7$-subgroup. Then $G/P$ is a group of order six. It has a subgroup $H$ of order three. The preimage $pi^-1(H)$, where $pi:Gto G/P$ is the natural projection, is what you are looking for.
$endgroup$
– Jyrki Lahtonen
Mar 19 at 5:04
2
$begingroup$
Alternatively, let $rho: Gto operatornameSym(G)$ be the Cayley homomorphism coming from $G$ acting on itself by left multiplication. The image of every element of an even order is a disjoint product of an odd number of cycles of an even length, i.e. an odd permutation, On the other hand, the image of an element of an odd order is a disjoint product of cycles of an odd length, i.e. an even permutation. Therefore the elements of an odd order form a subgroup (=the preimage of the alternating group).
$endgroup$
– Jyrki Lahtonen
Mar 19 at 5:09
1
$begingroup$
(cont'd) This argument shows that elements of odd order form a subgroup whenever the group has order $equiv2pmod4$.
$endgroup$
– Jyrki Lahtonen
Mar 19 at 5:09
1
$begingroup$
@JyrkiLahtonen: yes this is the proof I had in mind. An element of order 2 acts via right multiplication as an odd product of transpositions (here the regular representation is the one given by Cayley's theorem, no character theory needed). Thus the regular representation contains an odd permutation, so the preimage of the alternating subgroup is of index 2. This same argument generalizes via induction to show groups with cyclic Sylow 2-subgroups have normal subgroups consisting of all odd-order elements.
$endgroup$
– Hempelicious
Mar 19 at 5:38
|
show 4 more comments
1
$begingroup$
When a group has order $2cdottextodd$, it always has a normal subgroup of index 2 (you can see this by seeing how an element of order 2 acts in the regular representation).
$endgroup$
– Hempelicious
Mar 19 at 1:50
2
$begingroup$
Can't you do the following. Let $P$ be the normal Sylow $7$-subgroup. Then $G/P$ is a group of order six. It has a subgroup $H$ of order three. The preimage $pi^-1(H)$, where $pi:Gto G/P$ is the natural projection, is what you are looking for.
$endgroup$
– Jyrki Lahtonen
Mar 19 at 5:04
2
$begingroup$
Alternatively, let $rho: Gto operatornameSym(G)$ be the Cayley homomorphism coming from $G$ acting on itself by left multiplication. The image of every element of an even order is a disjoint product of an odd number of cycles of an even length, i.e. an odd permutation, On the other hand, the image of an element of an odd order is a disjoint product of cycles of an odd length, i.e. an even permutation. Therefore the elements of an odd order form a subgroup (=the preimage of the alternating group).
$endgroup$
– Jyrki Lahtonen
Mar 19 at 5:09
1
$begingroup$
(cont'd) This argument shows that elements of odd order form a subgroup whenever the group has order $equiv2pmod4$.
$endgroup$
– Jyrki Lahtonen
Mar 19 at 5:09
1
$begingroup$
@JyrkiLahtonen: yes this is the proof I had in mind. An element of order 2 acts via right multiplication as an odd product of transpositions (here the regular representation is the one given by Cayley's theorem, no character theory needed). Thus the regular representation contains an odd permutation, so the preimage of the alternating subgroup is of index 2. This same argument generalizes via induction to show groups with cyclic Sylow 2-subgroups have normal subgroups consisting of all odd-order elements.
$endgroup$
– Hempelicious
Mar 19 at 5:38
1
1
$begingroup$
When a group has order $2cdottextodd$, it always has a normal subgroup of index 2 (you can see this by seeing how an element of order 2 acts in the regular representation).
$endgroup$
– Hempelicious
Mar 19 at 1:50
$begingroup$
When a group has order $2cdottextodd$, it always has a normal subgroup of index 2 (you can see this by seeing how an element of order 2 acts in the regular representation).
$endgroup$
– Hempelicious
Mar 19 at 1:50
2
2
$begingroup$
Can't you do the following. Let $P$ be the normal Sylow $7$-subgroup. Then $G/P$ is a group of order six. It has a subgroup $H$ of order three. The preimage $pi^-1(H)$, where $pi:Gto G/P$ is the natural projection, is what you are looking for.
$endgroup$
– Jyrki Lahtonen
Mar 19 at 5:04
$begingroup$
Can't you do the following. Let $P$ be the normal Sylow $7$-subgroup. Then $G/P$ is a group of order six. It has a subgroup $H$ of order three. The preimage $pi^-1(H)$, where $pi:Gto G/P$ is the natural projection, is what you are looking for.
$endgroup$
– Jyrki Lahtonen
Mar 19 at 5:04
2
2
$begingroup$
Alternatively, let $rho: Gto operatornameSym(G)$ be the Cayley homomorphism coming from $G$ acting on itself by left multiplication. The image of every element of an even order is a disjoint product of an odd number of cycles of an even length, i.e. an odd permutation, On the other hand, the image of an element of an odd order is a disjoint product of cycles of an odd length, i.e. an even permutation. Therefore the elements of an odd order form a subgroup (=the preimage of the alternating group).
$endgroup$
– Jyrki Lahtonen
Mar 19 at 5:09
$begingroup$
Alternatively, let $rho: Gto operatornameSym(G)$ be the Cayley homomorphism coming from $G$ acting on itself by left multiplication. The image of every element of an even order is a disjoint product of an odd number of cycles of an even length, i.e. an odd permutation, On the other hand, the image of an element of an odd order is a disjoint product of cycles of an odd length, i.e. an even permutation. Therefore the elements of an odd order form a subgroup (=the preimage of the alternating group).
$endgroup$
– Jyrki Lahtonen
Mar 19 at 5:09
1
1
$begingroup$
(cont'd) This argument shows that elements of odd order form a subgroup whenever the group has order $equiv2pmod4$.
$endgroup$
– Jyrki Lahtonen
Mar 19 at 5:09
$begingroup$
(cont'd) This argument shows that elements of odd order form a subgroup whenever the group has order $equiv2pmod4$.
$endgroup$
– Jyrki Lahtonen
Mar 19 at 5:09
1
1
$begingroup$
@JyrkiLahtonen: yes this is the proof I had in mind. An element of order 2 acts via right multiplication as an odd product of transpositions (here the regular representation is the one given by Cayley's theorem, no character theory needed). Thus the regular representation contains an odd permutation, so the preimage of the alternating subgroup is of index 2. This same argument generalizes via induction to show groups with cyclic Sylow 2-subgroups have normal subgroups consisting of all odd-order elements.
$endgroup$
– Hempelicious
Mar 19 at 5:38
$begingroup$
@JyrkiLahtonen: yes this is the proof I had in mind. An element of order 2 acts via right multiplication as an odd product of transpositions (here the regular representation is the one given by Cayley's theorem, no character theory needed). Thus the regular representation contains an odd permutation, so the preimage of the alternating subgroup is of index 2. This same argument generalizes via induction to show groups with cyclic Sylow 2-subgroups have normal subgroups consisting of all odd-order elements.
$endgroup$
– Hempelicious
Mar 19 at 5:38
|
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