Mixing
Find the order of $operatorname{Aut}(G)$.
$begingroup$
Let $G$ be a non-abelian group of order $20$. What will be the order of $operatorname{Aut}(G)$?
$(a)$ $1$.
$(b)$ $10$.
$(c)$ $30$.
$(d)$ $40$.
If I take $G = D_{10}$ then I find that the order of $operatorname{Aut}(G)$ is a multiple of $10$. In this case I use the fact that $G/Z(G) cong operatorname{Inn}(G)$. So I think $(d)$ is the correct option but I don't know the general result as to why it is true for any non-abelian group of order $20$. Please help me in this regard.
Thank you very much.
group-theory finite-groups automorphism-group
edited Jan 27 at 17:07
the_fox
2,90031538
asked Jan 27 at 14:24
Dbchatto67Dbchatto67
2,166320
$endgroup$
add a comment |
$begingroup$
Let $G$ be a non-abelian group of order $20$. What will be the order of $operatorname{Aut}(G)$?
$(a)$ $1$.
$(b)$ $10$.
$(c)$ $30$.
$(d)$ $40$.
If I take $G = D_{10}$ then I find that the order of $operatorname{Aut}(G)$ is a multiple of $10$. In this case I use the fact that $G/Z(G) cong operatorname{Inn}(G)$. So I think $(d)$ is the correct option but I don't know the general result as to why it is true for any non-abelian group of order $20$. Please help me in this regard.
Thank you very much.
group-theory finite-groups automorphism-group
edited Jan 27 at 17:07
the_fox
2,90031538
asked Jan 27 at 14:24
Dbchatto67Dbchatto67
2,166320
$endgroup$
add a comment |
$begingroup$
Let $G$ be a non-abelian group of order $20$. What will be the order of $operatorname{Aut}(G)$?
$(a)$ $1$.
$(b)$ $10$.
$(c)$ $30$.
$(d)$ $40$.
If I take $G = D_{10}$ then I find that the order of $operatorname{Aut}(G)$ is a multiple of $10$. In this case I use the fact that $G/Z(G) cong operatorname{Inn}(G)$. So I think $(d)$ is the correct option but I don't know the general result as to why it is true for any non-abelian group of order $20$. Please help me in this regard.
Thank you very much.
group-theory finite-groups automorphism-group
edited Jan 27 at 17:07
the_fox
2,90031538
asked Jan 27 at 14:24
Dbchatto67Dbchatto67
2,166320
$endgroup$
Let $G$ be a non-abelian group of order $20$. What will be the order of $operatorname{Aut}(G)$?
$(a)$ $1$.
$(b)$ $10$.
$(c)$ $30$.
$(d)$ $40$.
If I take $G = D_{10}$ then I find that the order of $operatorname{Aut}(G)$ is a multiple of $10$. In this case I use the fact that $G/Z(G) cong operatorname{Inn}(G)$. So I think $(d)$ is the correct option but I don't know the general result as to why it is true for any non-abelian group of order $20$. Please help me in this regard.
Thank you very much.
group-theory finite-groups automorphism-group
group-theory finite-groups automorphism-group
edited Jan 27 at 17:07
the_fox
2,90031538
asked Jan 27 at 14:24
Dbchatto67Dbchatto67
2,166320
edited Jan 27 at 17:07
the_fox
2,90031538
asked Jan 27 at 14:24
Dbchatto67Dbchatto67
2,166320
edited Jan 27 at 17:07
the_fox
2,90031538
edited Jan 27 at 17:07
the_fox
2,90031538
edited Jan 27 at 17:07
the_fox
2,90031538
2,90031538
asked Jan 27 at 14:24
Dbchatto67Dbchatto67
2,166320
asked Jan 27 at 14:24
Dbchatto67Dbchatto67
2,166320
asked Jan 27 at 14:24
Dbchatto67Dbchatto67
2,166320
2,166320
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
This is not a well-formulated problem. There is a non-abelian group of order $20$ with an automorphism group of order $20$ and one with automorphism group of order $40$.
The group $G_1$ of shape $C_5 rtimes C_4$, given by the presentation $langle a,b mid a^5 = b^4 = 1, bab^{-1} = a^2 rangle$, is complete (i.e. it has trivial centre and all automorphisms are inner). Thus $operatorname{Aut}(G_1) cong G_1$ has order $20$.
On the other hand, $G_2 = D_{20}$ has centre of order $2$, so $operatorname{Inn}(G_2) cong D_{20}/Z(D_{20}) cong D_{10}$ has size $10$, but $operatorname{Out}(G_2)$ has size $4$ and is isomorphic to the Klein four-group, so $lvert operatorname{Aut}(G_2) rvert =40$.
answered Jan 27 at 14:46
the_foxthe_fox
2,90031538
$endgroup$
add a comment |
$begingroup$
Are you allowed to assume that at least one of the options is correct?
Recall that $G/Z(G)$ cyclic $Rightarrow$ G abelian. But $|G/Z(G)|$ must be a divisor of $|G|=20$. Hence the only non-cyclic options are $|G/Z(G)| = 4$, $10$ or $20$ (since prime order $Rightarrow$ cyclic).
If $|G/Z(G)| = 4$ or $20$, then we are done. I do not know how to deal with the $|G/Z(G)| = 10$ case.
answered Jan 27 at 14:35
o.h.o.h.
6917
$endgroup$
1
$begingroup$
What about $|G/Z(G)|=10$?
$endgroup$
– the_fox
Jan 27 at 14:37
$begingroup$
Oh, thanks! Forgot that, don't know how to salvage the proof
$endgroup$
– o.h.
Jan 27 at 14:38
3
$begingroup$
You can't show that $|G/Z(G)| neq 10$ because $|Z(D_{20})|=2$.
$endgroup$
– the_fox
Jan 27 at 14:42
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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active
oldest
votes
$begingroup$
This is not a well-formulated problem. There is a non-abelian group of order $20$ with an automorphism group of order $20$ and one with automorphism group of order $40$.
The group $G_1$ of shape $C_5 rtimes C_4$, given by the presentation $langle a,b mid a^5 = b^4 = 1, bab^{-1} = a^2 rangle$, is complete (i.e. it has trivial centre and all automorphisms are inner). Thus $operatorname{Aut}(G_1) cong G_1$ has order $20$.
On the other hand, $G_2 = D_{20}$ has centre of order $2$, so $operatorname{Inn}(G_2) cong D_{20}/Z(D_{20}) cong D_{10}$ has size $10$, but $operatorname{Out}(G_2)$ has size $4$ and is isomorphic to the Klein four-group, so $lvert operatorname{Aut}(G_2) rvert =40$.
answered Jan 27 at 14:46
the_foxthe_fox
2,90031538
$endgroup$
add a comment |
$begingroup$
This is not a well-formulated problem. There is a non-abelian group of order $20$ with an automorphism group of order $20$ and one with automorphism group of order $40$.
The group $G_1$ of shape $C_5 rtimes C_4$, given by the presentation $langle a,b mid a^5 = b^4 = 1, bab^{-1} = a^2 rangle$, is complete (i.e. it has trivial centre and all automorphisms are inner). Thus $operatorname{Aut}(G_1) cong G_1$ has order $20$.
On the other hand, $G_2 = D_{20}$ has centre of order $2$, so $operatorname{Inn}(G_2) cong D_{20}/Z(D_{20}) cong D_{10}$ has size $10$, but $operatorname{Out}(G_2)$ has size $4$ and is isomorphic to the Klein four-group, so $lvert operatorname{Aut}(G_2) rvert =40$.
answered Jan 27 at 14:46
the_foxthe_fox
2,90031538
$endgroup$
add a comment |
$begingroup$
This is not a well-formulated problem. There is a non-abelian group of order $20$ with an automorphism group of order $20$ and one with automorphism group of order $40$.
The group $G_1$ of shape $C_5 rtimes C_4$, given by the presentation $langle a,b mid a^5 = b^4 = 1, bab^{-1} = a^2 rangle$, is complete (i.e. it has trivial centre and all automorphisms are inner). Thus $operatorname{Aut}(G_1) cong G_1$ has order $20$.
On the other hand, $G_2 = D_{20}$ has centre of order $2$, so $operatorname{Inn}(G_2) cong D_{20}/Z(D_{20}) cong D_{10}$ has size $10$, but $operatorname{Out}(G_2)$ has size $4$ and is isomorphic to the Klein four-group, so $lvert operatorname{Aut}(G_2) rvert =40$.
answered Jan 27 at 14:46
the_foxthe_fox
2,90031538
$endgroup$
This is not a well-formulated problem. There is a non-abelian group of order $20$ with an automorphism group of order $20$ and one with automorphism group of order $40$.
The group $G_1$ of shape $C_5 rtimes C_4$, given by the presentation $langle a,b mid a^5 = b^4 = 1, bab^{-1} = a^2 rangle$, is complete (i.e. it has trivial centre and all automorphisms are inner). Thus $operatorname{Aut}(G_1) cong G_1$ has order $20$.
On the other hand, $G_2 = D_{20}$ has centre of order $2$, so $operatorname{Inn}(G_2) cong D_{20}/Z(D_{20}) cong D_{10}$ has size $10$, but $operatorname{Out}(G_2)$ has size $4$ and is isomorphic to the Klein four-group, so $lvert operatorname{Aut}(G_2) rvert =40$.
answered Jan 27 at 14:46
the_foxthe_fox
2,90031538
edited Jan 30 at 17:24
answered Jan 27 at 14:46
the_foxthe_fox
2,90031538
answered Jan 27 at 14:46
the_foxthe_fox
2,90031538
answered Jan 27 at 14:46
the_foxthe_fox
2,90031538
2,90031538
add a comment |
add a comment |
$begingroup$
Are you allowed to assume that at least one of the options is correct?
Recall that $G/Z(G)$ cyclic $Rightarrow$ G abelian. But $|G/Z(G)|$ must be a divisor of $|G|=20$. Hence the only non-cyclic options are $|G/Z(G)| = 4$, $10$ or $20$ (since prime order $Rightarrow$ cyclic).
If $|G/Z(G)| = 4$ or $20$, then we are done. I do not know how to deal with the $|G/Z(G)| = 10$ case.
answered Jan 27 at 14:35
o.h.o.h.
6917
$endgroup$
1
$begingroup$
What about $|G/Z(G)|=10$?
$endgroup$
– the_fox
Jan 27 at 14:37
$begingroup$
Oh, thanks! Forgot that, don't know how to salvage the proof
$endgroup$
– o.h.
Jan 27 at 14:38
3
$begingroup$
You can't show that $|G/Z(G)| neq 10$ because $|Z(D_{20})|=2$.
$endgroup$
– the_fox
Jan 27 at 14:42
add a comment |
$begingroup$
Are you allowed to assume that at least one of the options is correct?
Recall that $G/Z(G)$ cyclic $Rightarrow$ G abelian. But $|G/Z(G)|$ must be a divisor of $|G|=20$. Hence the only non-cyclic options are $|G/Z(G)| = 4$, $10$ or $20$ (since prime order $Rightarrow$ cyclic).
If $|G/Z(G)| = 4$ or $20$, then we are done. I do not know how to deal with the $|G/Z(G)| = 10$ case.
answered Jan 27 at 14:35
o.h.o.h.
6917
$endgroup$
1
$begingroup$
What about $|G/Z(G)|=10$?
$endgroup$
– the_fox
Jan 27 at 14:37
$begingroup$
Oh, thanks! Forgot that, don't know how to salvage the proof
$endgroup$
– o.h.
Jan 27 at 14:38
3
$begingroup$
You can't show that $|G/Z(G)| neq 10$ because $|Z(D_{20})|=2$.
$endgroup$
– the_fox
Jan 27 at 14:42
add a comment |
$begingroup$
Are you allowed to assume that at least one of the options is correct?
Recall that $G/Z(G)$ cyclic $Rightarrow$ G abelian. But $|G/Z(G)|$ must be a divisor of $|G|=20$. Hence the only non-cyclic options are $|G/Z(G)| = 4$, $10$ or $20$ (since prime order $Rightarrow$ cyclic).
If $|G/Z(G)| = 4$ or $20$, then we are done. I do not know how to deal with the $|G/Z(G)| = 10$ case.
answered Jan 27 at 14:35
o.h.o.h.
6917
$endgroup$
Are you allowed to assume that at least one of the options is correct?
Recall that $G/Z(G)$ cyclic $Rightarrow$ G abelian. But $|G/Z(G)|$ must be a divisor of $|G|=20$. Hence the only non-cyclic options are $|G/Z(G)| = 4$, $10$ or $20$ (since prime order $Rightarrow$ cyclic).
If $|G/Z(G)| = 4$ or $20$, then we are done. I do not know how to deal with the $|G/Z(G)| = 10$ case.
answered Jan 27 at 14:35
o.h.o.h.
6917
edited Jan 27 at 14:44
answered Jan 27 at 14:35
o.h.o.h.
6917
answered Jan 27 at 14:35
o.h.o.h.
6917
answered Jan 27 at 14:35
o.h.o.h.
6917
6917
1
$begingroup$
What about $|G/Z(G)|=10$?
$endgroup$
– the_fox
Jan 27 at 14:37
$begingroup$
Oh, thanks! Forgot that, don't know how to salvage the proof
$endgroup$
– o.h.
Jan 27 at 14:38
3
$begingroup$
You can't show that $|G/Z(G)| neq 10$ because $|Z(D_{20})|=2$.
$endgroup$
– the_fox
Jan 27 at 14:42
add a comment |
1
$begingroup$
What about $|G/Z(G)|=10$?
$endgroup$
– the_fox
Jan 27 at 14:37
$begingroup$
Oh, thanks! Forgot that, don't know how to salvage the proof
$endgroup$
– o.h.
Jan 27 at 14:38
3
$begingroup$
You can't show that $|G/Z(G)| neq 10$ because $|Z(D_{20})|=2$.
$endgroup$
– the_fox
Jan 27 at 14:42
1
1
$begingroup$
What about $|G/Z(G)|=10$?
$endgroup$
– the_fox
Jan 27 at 14:37
$begingroup$
What about $|G/Z(G)|=10$?
$endgroup$
– the_fox
Jan 27 at 14:37
$begingroup$
Oh, thanks! Forgot that, don't know how to salvage the proof
$endgroup$
– o.h.
Jan 27 at 14:38
$begingroup$
Oh, thanks! Forgot that, don't know how to salvage the proof
$endgroup$
– o.h.
Jan 27 at 14:38
3
3
$begingroup$
You can't show that $|G/Z(G)| neq 10$ because $|Z(D_{20})|=2$.
$endgroup$
– the_fox
Jan 27 at 14:42
$begingroup$
You can't show that $|G/Z(G)| neq 10$ because $|Z(D_{20})|=2$.
$endgroup$
– the_fox
Jan 27 at 14:42
add a comment |
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