Automorphism of $D_8$ [duplicate]
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$mathrm{Aut}(D_4)$ is isomorphic to $D_4$
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I am trying to prove that $Aut(D_8) equiv D_8$. It is not hard to see that $lvert Aut(D_8)rvert = 8$. Indeed, it is at most $8$ as $r$ (canonical rotation) has order $4$ and $s$ (canonical reflection) has order $2$. On the other hand, $D_8$ is normal in $D_{16}$ which acts by conjugation on $D_8$. So, the order is exactly $8$. However, I am having troubles to show that $Aut(D_8) equiv D_8$ (I could check all possible solutions and stop once I have 8 different automorphisms but I wonder if there is a simpler solution).
abstract-algebra group-theory finite-groups dihedral-groups automorphism-group
asked Dec 12 '18 at 0:50
PeterPeter
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Dec 12 '18 at 9:30
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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$begingroup$
This question already has an answer here:
$mathrm{Aut}(D_4)$ is isomorphic to $D_4$
3 answers
I am trying to prove that $Aut(D_8) equiv D_8$. It is not hard to see that $lvert Aut(D_8)rvert = 8$. Indeed, it is at most $8$ as $r$ (canonical rotation) has order $4$ and $s$ (canonical reflection) has order $2$. On the other hand, $D_8$ is normal in $D_{16}$ which acts by conjugation on $D_8$. So, the order is exactly $8$. However, I am having troubles to show that $Aut(D_8) equiv D_8$ (I could check all possible solutions and stop once I have 8 different automorphisms but I wonder if there is a simpler solution).
abstract-algebra group-theory finite-groups dihedral-groups automorphism-group
asked Dec 12 '18 at 0:50
PeterPeter
934
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marked as duplicate by Dietrich Burde abstract-algebra
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Dec 12 '18 at 9:30
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You can do it by showing that there are $f,gin text{Aut}(D_8)$ such that $f^2=1,g^4=1$ and $fgf^{-1}=g^{-1}$.
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– user9077
Dec 12 '18 at 0:55
add a comment |
$begingroup$
This question already has an answer here:
$mathrm{Aut}(D_4)$ is isomorphic to $D_4$
3 answers
I am trying to prove that $Aut(D_8) equiv D_8$. It is not hard to see that $lvert Aut(D_8)rvert = 8$. Indeed, it is at most $8$ as $r$ (canonical rotation) has order $4$ and $s$ (canonical reflection) has order $2$. On the other hand, $D_8$ is normal in $D_{16}$ which acts by conjugation on $D_8$. So, the order is exactly $8$. However, I am having troubles to show that $Aut(D_8) equiv D_8$ (I could check all possible solutions and stop once I have 8 different automorphisms but I wonder if there is a simpler solution).
abstract-algebra group-theory finite-groups dihedral-groups automorphism-group
asked Dec 12 '18 at 0:50
PeterPeter
934
$endgroup$
This question already has an answer here:
$mathrm{Aut}(D_4)$ is isomorphic to $D_4$
3 answers
I am trying to prove that $Aut(D_8) equiv D_8$. It is not hard to see that $lvert Aut(D_8)rvert = 8$. Indeed, it is at most $8$ as $r$ (canonical rotation) has order $4$ and $s$ (canonical reflection) has order $2$. On the other hand, $D_8$ is normal in $D_{16}$ which acts by conjugation on $D_8$. So, the order is exactly $8$. However, I am having troubles to show that $Aut(D_8) equiv D_8$ (I could check all possible solutions and stop once I have 8 different automorphisms but I wonder if there is a simpler solution).
This question already has an answer here:
$mathrm{Aut}(D_4)$ is isomorphic to $D_4$
3 answers
abstract-algebra group-theory finite-groups dihedral-groups automorphism-group
abstract-algebra group-theory finite-groups dihedral-groups automorphism-group
asked Dec 12 '18 at 0:50
PeterPeter
934
asked Dec 12 '18 at 0:50
PeterPeter
934
asked Dec 12 '18 at 0:50
PeterPeter
934
asked Dec 12 '18 at 0:50
PeterPeter
934
asked Dec 12 '18 at 0:50
PeterPeter
934
934
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Dec 12 '18 at 9:30
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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Dec 12 '18 at 9:30
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
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You can do it by showing that there are $f,gin text{Aut}(D_8)$ such that $f^2=1,g^4=1$ and $fgf^{-1}=g^{-1}$.
$endgroup$
– user9077
Dec 12 '18 at 0:55
add a comment |
1
$begingroup$
You can do it by showing that there are $f,gin text{Aut}(D_8)$ such that $f^2=1,g^4=1$ and $fgf^{-1}=g^{-1}$.
$endgroup$
– user9077
Dec 12 '18 at 0:55
1
1
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You can do it by showing that there are $f,gin text{Aut}(D_8)$ such that $f^2=1,g^4=1$ and $fgf^{-1}=g^{-1}$.
$endgroup$
– user9077
Dec 12 '18 at 0:55
$begingroup$
You can do it by showing that there are $f,gin text{Aut}(D_8)$ such that $f^2=1,g^4=1$ and $fgf^{-1}=g^{-1}$.
$endgroup$
– user9077
Dec 12 '18 at 0:55
add a comment |
1 Answer
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You already know it's an order 8 factor group given by $D_{16}$ mod its center. Clearly that has 2 elements of order 4 (images of the order four order 8 elements) and 5 of order 2. The only such group is $D_8$....just list all 5 groups of order 8 and how many elements of each order they each have.
answered Dec 12 '18 at 1:01
C MonsourC Monsour
6,2191325
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1 Answer
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1 Answer
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$begingroup$
You already know it's an order 8 factor group given by $D_{16}$ mod its center. Clearly that has 2 elements of order 4 (images of the order four order 8 elements) and 5 of order 2. The only such group is $D_8$....just list all 5 groups of order 8 and how many elements of each order they each have.
answered Dec 12 '18 at 1:01
C MonsourC Monsour
6,2191325
$endgroup$
add a comment |
$begingroup$
You already know it's an order 8 factor group given by $D_{16}$ mod its center. Clearly that has 2 elements of order 4 (images of the order four order 8 elements) and 5 of order 2. The only such group is $D_8$....just list all 5 groups of order 8 and how many elements of each order they each have.
answered Dec 12 '18 at 1:01
C MonsourC Monsour
6,2191325
$endgroup$
add a comment |
$begingroup$
You already know it's an order 8 factor group given by $D_{16}$ mod its center. Clearly that has 2 elements of order 4 (images of the order four order 8 elements) and 5 of order 2. The only such group is $D_8$....just list all 5 groups of order 8 and how many elements of each order they each have.
answered Dec 12 '18 at 1:01
C MonsourC Monsour
6,2191325
$endgroup$
You already know it's an order 8 factor group given by $D_{16}$ mod its center. Clearly that has 2 elements of order 4 (images of the order four order 8 elements) and 5 of order 2. The only such group is $D_8$....just list all 5 groups of order 8 and how many elements of each order they each have.
answered Dec 12 '18 at 1:01
C MonsourC Monsour
6,2191325
answered Dec 12 '18 at 1:01
C MonsourC Monsour
6,2191325
answered Dec 12 '18 at 1:01
C MonsourC Monsour
6,2191325
answered Dec 12 '18 at 1:01
C MonsourC Monsour
6,2191325
6,2191325
add a comment |
add a comment |
1
$begingroup$
You can do it by showing that there are $f,gin text{Aut}(D_8)$ such that $f^2=1,g^4=1$ and $fgf^{-1}=g^{-1}$.
$endgroup$
– user9077
Dec 12 '18 at 0:55