How many group homomorphisms from $Bbb Z_3$ to $text{Aut}(Bbb Z_7)$
$begingroup$
How many group homomorphism from $Bbb Z_3$ to $text{Aut}(Bbb Z_7)$?
My attempt: Since $Bbb Z_3$ is cyclic, hence the homomorphisms are all determined by $phi(overline{1})$. On the other hand $text{Aut}(Bbb Z_7)cong Bbb Z_6$
.
Since $o(phi(overline{1}))mid |Bbb Z_3|$, so $o(phi(overline{1}))$ can only be $1,~3$. If the order is $1$, it means $phi(overline{1})=overline{1}$. It is indeed a homo. If the order is $3$, it means $phi(overline{1})=overline{2}$ or $overline{4}$. So the total number of group homomorphisms from $Bbb Z_3$ to $text{Aut}(Bbb Z_7)$ is three, as I show above. Am I correct?
abstract-algebra
asked Dec 31 '18 at 9:55
EricEric
1,810615
$endgroup$
add a comment |
$begingroup$
How many group homomorphism from $Bbb Z_3$ to $text{Aut}(Bbb Z_7)$?
My attempt: Since $Bbb Z_3$ is cyclic, hence the homomorphisms are all determined by $phi(overline{1})$. On the other hand $text{Aut}(Bbb Z_7)cong Bbb Z_6$
.
Since $o(phi(overline{1}))mid |Bbb Z_3|$, so $o(phi(overline{1}))$ can only be $1,~3$. If the order is $1$, it means $phi(overline{1})=overline{1}$. It is indeed a homo. If the order is $3$, it means $phi(overline{1})=overline{2}$ or $overline{4}$. So the total number of group homomorphisms from $Bbb Z_3$ to $text{Aut}(Bbb Z_7)$ is three, as I show above. Am I correct?
abstract-algebra
asked Dec 31 '18 at 9:55
EricEric
1,810615
$endgroup$
1
$begingroup$
yes, looks right
$endgroup$
– Martín Vacas Vignolo
Dec 31 '18 at 10:00
1
$begingroup$
Yes, it is correct.
$endgroup$
– toric_actions
Dec 31 '18 at 10:05
add a comment |
$begingroup$
How many group homomorphism from $Bbb Z_3$ to $text{Aut}(Bbb Z_7)$?
My attempt: Since $Bbb Z_3$ is cyclic, hence the homomorphisms are all determined by $phi(overline{1})$. On the other hand $text{Aut}(Bbb Z_7)cong Bbb Z_6$
.
Since $o(phi(overline{1}))mid |Bbb Z_3|$, so $o(phi(overline{1}))$ can only be $1,~3$. If the order is $1$, it means $phi(overline{1})=overline{1}$. It is indeed a homo. If the order is $3$, it means $phi(overline{1})=overline{2}$ or $overline{4}$. So the total number of group homomorphisms from $Bbb Z_3$ to $text{Aut}(Bbb Z_7)$ is three, as I show above. Am I correct?
abstract-algebra
asked Dec 31 '18 at 9:55
EricEric
1,810615
$endgroup$
How many group homomorphism from $Bbb Z_3$ to $text{Aut}(Bbb Z_7)$?
My attempt: Since $Bbb Z_3$ is cyclic, hence the homomorphisms are all determined by $phi(overline{1})$. On the other hand $text{Aut}(Bbb Z_7)cong Bbb Z_6$
.
Since $o(phi(overline{1}))mid |Bbb Z_3|$, so $o(phi(overline{1}))$ can only be $1,~3$. If the order is $1$, it means $phi(overline{1})=overline{1}$. It is indeed a homo. If the order is $3$, it means $phi(overline{1})=overline{2}$ or $overline{4}$. So the total number of group homomorphisms from $Bbb Z_3$ to $text{Aut}(Bbb Z_7)$ is three, as I show above. Am I correct?
abstract-algebra
abstract-algebra
asked Dec 31 '18 at 9:55
EricEric
1,810615
asked Dec 31 '18 at 9:55
EricEric
1,810615
asked Dec 31 '18 at 9:55
EricEric
1,810615
asked Dec 31 '18 at 9:55
EricEric
1,810615
asked Dec 31 '18 at 9:55
EricEric
1,810615
1,810615
1
$begingroup$
yes, looks right
$endgroup$
– Martín Vacas Vignolo
Dec 31 '18 at 10:00
1
$begingroup$
Yes, it is correct.
$endgroup$
– toric_actions
Dec 31 '18 at 10:05
add a comment |
1
$begingroup$
yes, looks right
$endgroup$
– Martín Vacas Vignolo
Dec 31 '18 at 10:00
1
$begingroup$
Yes, it is correct.
$endgroup$
– toric_actions
Dec 31 '18 at 10:05
1
1
$begingroup$
yes, looks right
$endgroup$
– Martín Vacas Vignolo
Dec 31 '18 at 10:00
$begingroup$
yes, looks right
$endgroup$
– Martín Vacas Vignolo
Dec 31 '18 at 10:00
1
1
$begingroup$
Yes, it is correct.
$endgroup$
– toric_actions
Dec 31 '18 at 10:05
$begingroup$
Yes, it is correct.
$endgroup$
– toric_actions
Dec 31 '18 at 10:05
add a comment |
1 Answer
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$begingroup$
$$operatorname{Hom}(Bbb Z/3Bbb Z, operatorname{Aut}(Bbb Z/7Bbb Z)) = operatorname{Hom}(Bbb Z/3Bbb Z, C_6) = C_6[3] = {e, g^2, g^4}$$
answered Dec 31 '18 at 10:04
Kenny LauKenny Lau
19.9k2160
$endgroup$
add a comment |
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1 Answer
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$begingroup$
$$operatorname{Hom}(Bbb Z/3Bbb Z, operatorname{Aut}(Bbb Z/7Bbb Z)) = operatorname{Hom}(Bbb Z/3Bbb Z, C_6) = C_6[3] = {e, g^2, g^4}$$
answered Dec 31 '18 at 10:04
Kenny LauKenny Lau
19.9k2160
$endgroup$
add a comment |
$begingroup$
$$operatorname{Hom}(Bbb Z/3Bbb Z, operatorname{Aut}(Bbb Z/7Bbb Z)) = operatorname{Hom}(Bbb Z/3Bbb Z, C_6) = C_6[3] = {e, g^2, g^4}$$
answered Dec 31 '18 at 10:04
Kenny LauKenny Lau
19.9k2160
$endgroup$
add a comment |
$begingroup$
$$operatorname{Hom}(Bbb Z/3Bbb Z, operatorname{Aut}(Bbb Z/7Bbb Z)) = operatorname{Hom}(Bbb Z/3Bbb Z, C_6) = C_6[3] = {e, g^2, g^4}$$
answered Dec 31 '18 at 10:04
Kenny LauKenny Lau
19.9k2160
$endgroup$
$$operatorname{Hom}(Bbb Z/3Bbb Z, operatorname{Aut}(Bbb Z/7Bbb Z)) = operatorname{Hom}(Bbb Z/3Bbb Z, C_6) = C_6[3] = {e, g^2, g^4}$$
answered Dec 31 '18 at 10:04
Kenny LauKenny Lau
19.9k2160
answered Dec 31 '18 at 10:04
Kenny LauKenny Lau
19.9k2160
answered Dec 31 '18 at 10:04
Kenny LauKenny Lau
19.9k2160
answered Dec 31 '18 at 10:04
Kenny LauKenny Lau
19.9k2160
19.9k2160
add a comment |
add a comment |
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1
$begingroup$
yes, looks right
$endgroup$
– Martín Vacas Vignolo
Dec 31 '18 at 10:00
1
$begingroup$
Yes, it is correct.
$endgroup$
– toric_actions
Dec 31 '18 at 10:05