Groups of order 56 with Sylow 2-subgroup isomorphic $Q_8$
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I try to classify non-abelian groups of order $56$ with sylow $2$-subgroup isomorphic to quaterion group $Q_8$. More accurately I want to construct $2$ non-isomorphic such groups. This is an excercise 5.3.7 from Dummit and Foote's book.
I can construct such groups as semidirect product as follows
$G_1 = C_7 rtimes_{phi_{1}} Q_8$, $G_2 = C_7 rtimes_{phi_{2}} Q_8$, $G_3 = C_7 rtimes_{phi_{3}} Q_8$, where $phi_n: Q_8 to operatorname{Aut}(C_7)$.
Let $Q_8 = <i,j>$ and $<sigma>$ be a unique subgroup of order $2$ of $operatorname{Aut}(C_7) cong C_6$, where $sigma$ inverts elements of $C_7$. We define
$phi_1$ as
$$phi_1(i) = sigma, phi_1(j) = operatorname{id}$$
(here $operatorname{id}$ is an identical automorphism),
$$phi_2(i) = operatorname{id}, phi_2(j) = sigma$$
and
$$phi_3(i) = sigma, phi_3(j) = sigma$$
My question. Am I right that all this groups are isomorphic? And if all this groups are isomorphic what is the second type of such gruops?
I use the following result about isomorphism of semidirect product. Let $tauin operatorname{Aut}(H)$ than $Nrtimes_{phi} H cong Nrtimes_{tauphi} H$.
group-theory finite-groups group-isomorphism semidirect-product
asked Dec 13 '18 at 17:53
Mikhail GoltvanitsaMikhail Goltvanitsa
623414
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add a comment |
$begingroup$
I try to classify non-abelian groups of order $56$ with sylow $2$-subgroup isomorphic to quaterion group $Q_8$. More accurately I want to construct $2$ non-isomorphic such groups. This is an excercise 5.3.7 from Dummit and Foote's book.
I can construct such groups as semidirect product as follows
$G_1 = C_7 rtimes_{phi_{1}} Q_8$, $G_2 = C_7 rtimes_{phi_{2}} Q_8$, $G_3 = C_7 rtimes_{phi_{3}} Q_8$, where $phi_n: Q_8 to operatorname{Aut}(C_7)$.
Let $Q_8 = <i,j>$ and $<sigma>$ be a unique subgroup of order $2$ of $operatorname{Aut}(C_7) cong C_6$, where $sigma$ inverts elements of $C_7$. We define
$phi_1$ as
$$phi_1(i) = sigma, phi_1(j) = operatorname{id}$$
(here $operatorname{id}$ is an identical automorphism),
$$phi_2(i) = operatorname{id}, phi_2(j) = sigma$$
and
$$phi_3(i) = sigma, phi_3(j) = sigma$$
My question. Am I right that all this groups are isomorphic? And if all this groups are isomorphic what is the second type of such gruops?
I use the following result about isomorphism of semidirect product. Let $tauin operatorname{Aut}(H)$ than $Nrtimes_{phi} H cong Nrtimes_{tauphi} H$.
group-theory finite-groups group-isomorphism semidirect-product
asked Dec 13 '18 at 17:53
Mikhail GoltvanitsaMikhail Goltvanitsa
623414
$endgroup$
add a comment |
$begingroup$
I try to classify non-abelian groups of order $56$ with sylow $2$-subgroup isomorphic to quaterion group $Q_8$. More accurately I want to construct $2$ non-isomorphic such groups. This is an excercise 5.3.7 from Dummit and Foote's book.
I can construct such groups as semidirect product as follows
$G_1 = C_7 rtimes_{phi_{1}} Q_8$, $G_2 = C_7 rtimes_{phi_{2}} Q_8$, $G_3 = C_7 rtimes_{phi_{3}} Q_8$, where $phi_n: Q_8 to operatorname{Aut}(C_7)$.
Let $Q_8 = <i,j>$ and $<sigma>$ be a unique subgroup of order $2$ of $operatorname{Aut}(C_7) cong C_6$, where $sigma$ inverts elements of $C_7$. We define
$phi_1$ as
$$phi_1(i) = sigma, phi_1(j) = operatorname{id}$$
(here $operatorname{id}$ is an identical automorphism),
$$phi_2(i) = operatorname{id}, phi_2(j) = sigma$$
and
$$phi_3(i) = sigma, phi_3(j) = sigma$$
My question. Am I right that all this groups are isomorphic? And if all this groups are isomorphic what is the second type of such gruops?
I use the following result about isomorphism of semidirect product. Let $tauin operatorname{Aut}(H)$ than $Nrtimes_{phi} H cong Nrtimes_{tauphi} H$.
group-theory finite-groups group-isomorphism semidirect-product
asked Dec 13 '18 at 17:53
Mikhail GoltvanitsaMikhail Goltvanitsa
623414
$endgroup$
I try to classify non-abelian groups of order $56$ with sylow $2$-subgroup isomorphic to quaterion group $Q_8$. More accurately I want to construct $2$ non-isomorphic such groups. This is an excercise 5.3.7 from Dummit and Foote's book.
I can construct such groups as semidirect product as follows
$G_1 = C_7 rtimes_{phi_{1}} Q_8$, $G_2 = C_7 rtimes_{phi_{2}} Q_8$, $G_3 = C_7 rtimes_{phi_{3}} Q_8$, where $phi_n: Q_8 to operatorname{Aut}(C_7)$.
Let $Q_8 = <i,j>$ and $<sigma>$ be a unique subgroup of order $2$ of $operatorname{Aut}(C_7) cong C_6$, where $sigma$ inverts elements of $C_7$. We define
$phi_1$ as
$$phi_1(i) = sigma, phi_1(j) = operatorname{id}$$
(here $operatorname{id}$ is an identical automorphism),
$$phi_2(i) = operatorname{id}, phi_2(j) = sigma$$
and
$$phi_3(i) = sigma, phi_3(j) = sigma$$
My question. Am I right that all this groups are isomorphic? And if all this groups are isomorphic what is the second type of such gruops?
I use the following result about isomorphism of semidirect product. Let $tauin operatorname{Aut}(H)$ than $Nrtimes_{phi} H cong Nrtimes_{tauphi} H$.
group-theory finite-groups group-isomorphism semidirect-product
group-theory finite-groups group-isomorphism semidirect-product
asked Dec 13 '18 at 17:53
Mikhail GoltvanitsaMikhail Goltvanitsa
623414
asked Dec 13 '18 at 17:53
Mikhail GoltvanitsaMikhail Goltvanitsa
623414
asked Dec 13 '18 at 17:53
Mikhail GoltvanitsaMikhail Goltvanitsa
623414
asked Dec 13 '18 at 17:53
Mikhail GoltvanitsaMikhail Goltvanitsa
623414
asked Dec 13 '18 at 17:53
Mikhail GoltvanitsaMikhail Goltvanitsa
623414
623414
add a comment |
add a comment |
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You are correct: those are all isomorphic. The second type is of course the direct product. That is different as $C_7$ is central in the direct product of $C_7$ with $Q_8$.
answered Dec 13 '18 at 18:17
C MonsourC Monsour
6,2291325
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$begingroup$
Thank you very much! For some reason I do not know why I thought that we should construct precisely non-trivial homomorphisms =)
$endgroup$
– Mikhail Goltvanitsa
Dec 13 '18 at 18:24
add a comment |
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1 Answer
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$begingroup$
You are correct: those are all isomorphic. The second type is of course the direct product. That is different as $C_7$ is central in the direct product of $C_7$ with $Q_8$.
answered Dec 13 '18 at 18:17
C MonsourC Monsour
6,2291325
$endgroup$
$begingroup$
Thank you very much! For some reason I do not know why I thought that we should construct precisely non-trivial homomorphisms =)
$endgroup$
– Mikhail Goltvanitsa
Dec 13 '18 at 18:24
add a comment |
$begingroup$
You are correct: those are all isomorphic. The second type is of course the direct product. That is different as $C_7$ is central in the direct product of $C_7$ with $Q_8$.
answered Dec 13 '18 at 18:17
C MonsourC Monsour
6,2291325
$endgroup$
$begingroup$
Thank you very much! For some reason I do not know why I thought that we should construct precisely non-trivial homomorphisms =)
$endgroup$
– Mikhail Goltvanitsa
Dec 13 '18 at 18:24
add a comment |
$begingroup$
You are correct: those are all isomorphic. The second type is of course the direct product. That is different as $C_7$ is central in the direct product of $C_7$ with $Q_8$.
answered Dec 13 '18 at 18:17
C MonsourC Monsour
6,2291325
$endgroup$
You are correct: those are all isomorphic. The second type is of course the direct product. That is different as $C_7$ is central in the direct product of $C_7$ with $Q_8$.
answered Dec 13 '18 at 18:17
C MonsourC Monsour
6,2291325
answered Dec 13 '18 at 18:17
C MonsourC Monsour
6,2291325
answered Dec 13 '18 at 18:17
C MonsourC Monsour
6,2291325
answered Dec 13 '18 at 18:17
C MonsourC Monsour
6,2291325
6,2291325
$begingroup$
Thank you very much! For some reason I do not know why I thought that we should construct precisely non-trivial homomorphisms =)
$endgroup$
– Mikhail Goltvanitsa
Dec 13 '18 at 18:24
add a comment |
$begingroup$
Thank you very much! For some reason I do not know why I thought that we should construct precisely non-trivial homomorphisms =)
$endgroup$
– Mikhail Goltvanitsa
Dec 13 '18 at 18:24
$begingroup$
Thank you very much! For some reason I do not know why I thought that we should construct precisely non-trivial homomorphisms =)
$endgroup$
– Mikhail Goltvanitsa
Dec 13 '18 at 18:24
$begingroup$
Thank you very much! For some reason I do not know why I thought that we should construct precisely non-trivial homomorphisms =)
$endgroup$
– Mikhail Goltvanitsa
Dec 13 '18 at 18:24
add a comment |
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