Structure of semi direct product.
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I want to verify that structure of group $Q_8 rtimes C_2$, i.e. semi direct product of Quaternion group of order $8$ and $C_2 = {1, a}$ (cyclic group of order $2$) can be defined like:
If we write $$Q_8 = {pm 1, pm x, pm y, pm xy}$$ Define $$h:C_2 to Aut(Q_8): a mapsto g(a) = begin{cases}x to y\ y to x\ -1to -1end{cases} $$
The define product in above group as $$(d_1, b_1)*(d_2, b_2) = (d_1d_2, b_1(g(d_1)(b_2))$$ Is it the correct representation or am I missing something?
group-theory semidirect-product
asked Dec 7 '18 at 11:44
Mittal GMittal G
1,193516
$endgroup$
add a comment |
$begingroup$
I want to verify that structure of group $Q_8 rtimes C_2$, i.e. semi direct product of Quaternion group of order $8$ and $C_2 = {1, a}$ (cyclic group of order $2$) can be defined like:
If we write $$Q_8 = {pm 1, pm x, pm y, pm xy}$$ Define $$h:C_2 to Aut(Q_8): a mapsto g(a) = begin{cases}x to y\ y to x\ -1to -1end{cases} $$
The define product in above group as $$(d_1, b_1)*(d_2, b_2) = (d_1d_2, b_1(g(d_1)(b_2))$$ Is it the correct representation or am I missing something?
group-theory semidirect-product
asked Dec 7 '18 at 11:44
Mittal GMittal G
1,193516
$endgroup$
$begingroup$
It's think it might be a little unclear what your question is. Did you have some particular semi-direct product of $Q_8$ and $C_2$ in mind and you are wondering you have the right one?
$endgroup$
– Ben
Dec 7 '18 at 12:42
$begingroup$
@Ben yes exactly.
$endgroup$
– Mittal G
Dec 7 '18 at 13:08
add a comment |
$begingroup$
I want to verify that structure of group $Q_8 rtimes C_2$, i.e. semi direct product of Quaternion group of order $8$ and $C_2 = {1, a}$ (cyclic group of order $2$) can be defined like:
If we write $$Q_8 = {pm 1, pm x, pm y, pm xy}$$ Define $$h:C_2 to Aut(Q_8): a mapsto g(a) = begin{cases}x to y\ y to x\ -1to -1end{cases} $$
The define product in above group as $$(d_1, b_1)*(d_2, b_2) = (d_1d_2, b_1(g(d_1)(b_2))$$ Is it the correct representation or am I missing something?
group-theory semidirect-product
asked Dec 7 '18 at 11:44
Mittal GMittal G
1,193516
$endgroup$
I want to verify that structure of group $Q_8 rtimes C_2$, i.e. semi direct product of Quaternion group of order $8$ and $C_2 = {1, a}$ (cyclic group of order $2$) can be defined like:
If we write $$Q_8 = {pm 1, pm x, pm y, pm xy}$$ Define $$h:C_2 to Aut(Q_8): a mapsto g(a) = begin{cases}x to y\ y to x\ -1to -1end{cases} $$
The define product in above group as $$(d_1, b_1)*(d_2, b_2) = (d_1d_2, b_1(g(d_1)(b_2))$$ Is it the correct representation or am I missing something?
group-theory semidirect-product
group-theory semidirect-product
asked Dec 7 '18 at 11:44
Mittal GMittal G
1,193516
asked Dec 7 '18 at 11:44
Mittal GMittal G
1,193516
asked Dec 7 '18 at 11:44
Mittal GMittal G
1,193516
asked Dec 7 '18 at 11:44
Mittal GMittal G
1,193516
asked Dec 7 '18 at 11:44
Mittal GMittal G
1,193516
1,193516
$begingroup$
It's think it might be a little unclear what your question is. Did you have some particular semi-direct product of $Q_8$ and $C_2$ in mind and you are wondering you have the right one?
$endgroup$
– Ben
Dec 7 '18 at 12:42
$begingroup$
@Ben yes exactly.
$endgroup$
– Mittal G
Dec 7 '18 at 13:08
add a comment |
$begingroup$
It's think it might be a little unclear what your question is. Did you have some particular semi-direct product of $Q_8$ and $C_2$ in mind and you are wondering you have the right one?
$endgroup$
– Ben
Dec 7 '18 at 12:42
$begingroup$
@Ben yes exactly.
$endgroup$
– Mittal G
Dec 7 '18 at 13:08
$begingroup$
It's think it might be a little unclear what your question is. Did you have some particular semi-direct product of $Q_8$ and $C_2$ in mind and you are wondering you have the right one?
$endgroup$
– Ben
Dec 7 '18 at 12:42
$begingroup$
It's think it might be a little unclear what your question is. Did you have some particular semi-direct product of $Q_8$ and $C_2$ in mind and you are wondering you have the right one?
$endgroup$
– Ben
Dec 7 '18 at 12:42
$begingroup$
@Ben yes exactly.
$endgroup$
– Mittal G
Dec 7 '18 at 13:08
$begingroup$
@Ben yes exactly.
$endgroup$
– Mittal G
Dec 7 '18 at 13:08
add a comment |
1 Answer
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$begingroup$
Your group is a nontrivial semidirect product $Q_8rtimes C_2$ isomorphic to the semidihedral group
$$SD_{16} = langle a,b | a^8 = b^2 = 1, bab^{-1} = a^3rangle$$
There is an isomorphism sending $a mapsto (x,z)$ and $bmapsto (1,z)$, where $z$ is the generator of $C_2$. Just check the relations:
$(x,z)^2 = (x.h_z(x),e) = (xy,e)$ so $(x,z)$ has order 8.- $(1,z)(x,z)(1,z) = (1,z)(x,1) = (h_z(x),z) = (y,z)$
- $(x,z)^3 = (xy,e)(x,z) = (xyx,z) = (y,z)$
If you instead wanted $Q_{16} = Q_8rtimes C_2$ to be the generalized quaternion group I think you should take $h_z(x) = y, h_z(y) = -x$ instead.
answered Dec 7 '18 at 14:35
BenBen
3,218616
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1 Answer
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1 Answer
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active
oldest
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$begingroup$
Your group is a nontrivial semidirect product $Q_8rtimes C_2$ isomorphic to the semidihedral group
$$SD_{16} = langle a,b | a^8 = b^2 = 1, bab^{-1} = a^3rangle$$
There is an isomorphism sending $a mapsto (x,z)$ and $bmapsto (1,z)$, where $z$ is the generator of $C_2$. Just check the relations:
$(x,z)^2 = (x.h_z(x),e) = (xy,e)$ so $(x,z)$ has order 8.- $(1,z)(x,z)(1,z) = (1,z)(x,1) = (h_z(x),z) = (y,z)$
- $(x,z)^3 = (xy,e)(x,z) = (xyx,z) = (y,z)$
If you instead wanted $Q_{16} = Q_8rtimes C_2$ to be the generalized quaternion group I think you should take $h_z(x) = y, h_z(y) = -x$ instead.
answered Dec 7 '18 at 14:35
BenBen
3,218616
$endgroup$
add a comment |
$begingroup$
Your group is a nontrivial semidirect product $Q_8rtimes C_2$ isomorphic to the semidihedral group
$$SD_{16} = langle a,b | a^8 = b^2 = 1, bab^{-1} = a^3rangle$$
There is an isomorphism sending $a mapsto (x,z)$ and $bmapsto (1,z)$, where $z$ is the generator of $C_2$. Just check the relations:
$(x,z)^2 = (x.h_z(x),e) = (xy,e)$ so $(x,z)$ has order 8.- $(1,z)(x,z)(1,z) = (1,z)(x,1) = (h_z(x),z) = (y,z)$
- $(x,z)^3 = (xy,e)(x,z) = (xyx,z) = (y,z)$
If you instead wanted $Q_{16} = Q_8rtimes C_2$ to be the generalized quaternion group I think you should take $h_z(x) = y, h_z(y) = -x$ instead.
answered Dec 7 '18 at 14:35
BenBen
3,218616
$endgroup$
add a comment |
$begingroup$
Your group is a nontrivial semidirect product $Q_8rtimes C_2$ isomorphic to the semidihedral group
$$SD_{16} = langle a,b | a^8 = b^2 = 1, bab^{-1} = a^3rangle$$
There is an isomorphism sending $a mapsto (x,z)$ and $bmapsto (1,z)$, where $z$ is the generator of $C_2$. Just check the relations:
$(x,z)^2 = (x.h_z(x),e) = (xy,e)$ so $(x,z)$ has order 8.- $(1,z)(x,z)(1,z) = (1,z)(x,1) = (h_z(x),z) = (y,z)$
- $(x,z)^3 = (xy,e)(x,z) = (xyx,z) = (y,z)$
If you instead wanted $Q_{16} = Q_8rtimes C_2$ to be the generalized quaternion group I think you should take $h_z(x) = y, h_z(y) = -x$ instead.
answered Dec 7 '18 at 14:35
BenBen
3,218616
$endgroup$
Your group is a nontrivial semidirect product $Q_8rtimes C_2$ isomorphic to the semidihedral group
$$SD_{16} = langle a,b | a^8 = b^2 = 1, bab^{-1} = a^3rangle$$
There is an isomorphism sending $a mapsto (x,z)$ and $bmapsto (1,z)$, where $z$ is the generator of $C_2$. Just check the relations:
$(x,z)^2 = (x.h_z(x),e) = (xy,e)$ so $(x,z)$ has order 8.- $(1,z)(x,z)(1,z) = (1,z)(x,1) = (h_z(x),z) = (y,z)$
- $(x,z)^3 = (xy,e)(x,z) = (xyx,z) = (y,z)$
If you instead wanted $Q_{16} = Q_8rtimes C_2$ to be the generalized quaternion group I think you should take $h_z(x) = y, h_z(y) = -x$ instead.
answered Dec 7 '18 at 14:35
BenBen
3,218616
answered Dec 7 '18 at 14:35
BenBen
3,218616
answered Dec 7 '18 at 14:35
BenBen
3,218616
answered Dec 7 '18 at 14:35
BenBen
3,218616
3,218616
add a comment |
add a comment |
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$begingroup$
It's think it might be a little unclear what your question is. Did you have some particular semi-direct product of $Q_8$ and $C_2$ in mind and you are wondering you have the right one?
$endgroup$
– Ben
Dec 7 '18 at 12:42
$begingroup$
@Ben yes exactly.
$endgroup$
– Mittal G
Dec 7 '18 at 13:08