$phi in operatornameAut(Bbb Z_50)$ via $phi(11) = 3$ Then $phi(x) = $? For any $x inBbb Z_50$ The Next CEO of Stack OverflowInterpretations of the first cohomology groupIs there a nontrivial semidirect product of two groups isomorphic to their direct product?An exact sequence of unit groupsConstructing a group automorphism making a diagram commuteExistence of an isomorphism $BbbZ_n^times rightarrow BbbZ_phi(n)^+$Question about split short exact sequences of groupsTo complete the proof that $operatornamePSL(2,Bbb F_5)cong A_5$By analyzing the map $G rightarrow$ Aut(G) given by $g rightarrow phi_g$, identify Inn(G) as a quotient of G.How to calculate the generators for the multiplicative group $Bbb Z_4^*$Proving Characteristic Subgroups are Transitive
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$phi in operatornameAut(Bbb Z_50)$ via $phi(11) = 3$ Then $phi(x) = $? For any $x inBbb Z_50$
The Next CEO of Stack OverflowInterpretations of the first cohomology groupIs there a nontrivial semidirect product of two groups isomorphic to their direct product?An exact sequence of unit groupsConstructing a group automorphism making a diagram commuteExistence of an isomorphism $BbbZ_n^times rightarrow BbbZ_phi(n)^+$Question about split short exact sequences of groupsTo complete the proof that $operatornamePSL(2,Bbb F_5)cong A_5$By analyzing the map $G rightarrow$ Aut(G) given by $g rightarrow phi_g$, identify Inn(G) as a quotient of G.How to calculate the generators for the multiplicative group $Bbb Z_4^*$Proving Characteristic Subgroups are Transitive
$begingroup$
$phi in operatornameAut(Bbb Z_50)$ via $phi(11) = 3$ Then $phi(x) = $? For any $x in Bbb Z_50$
The answer is $23x$ but I'm not quite sure how to figure that out. Here's what I did:
$phi(11) = 3 Rightarrow$ The function maps $11x rightarrow 3x$
Adding $(11)^-1 = 39$ (mod 50) on both sides gives us
$x rightarrow 42x$
What am I not doing correct?
group-theory
edited Mar 18 at 22:38
Bernard
123k741117
asked Mar 18 at 22:12
user2965071user2965071
1887
$endgroup$
add a comment |
$begingroup$
$phi in operatornameAut(Bbb Z_50)$ via $phi(11) = 3$ Then $phi(x) = $? For any $x in Bbb Z_50$
The answer is $23x$ but I'm not quite sure how to figure that out. Here's what I did:
$phi(11) = 3 Rightarrow$ The function maps $11x rightarrow 3x$
Adding $(11)^-1 = 39$ (mod 50) on both sides gives us
$x rightarrow 42x$
What am I not doing correct?
group-theory
edited Mar 18 at 22:38
Bernard
123k741117
asked Mar 18 at 22:12
user2965071user2965071
1887
$endgroup$
add a comment |
$begingroup$
$phi in operatornameAut(Bbb Z_50)$ via $phi(11) = 3$ Then $phi(x) = $? For any $x in Bbb Z_50$
The answer is $23x$ but I'm not quite sure how to figure that out. Here's what I did:
$phi(11) = 3 Rightarrow$ The function maps $11x rightarrow 3x$
Adding $(11)^-1 = 39$ (mod 50) on both sides gives us
$x rightarrow 42x$
What am I not doing correct?
group-theory
edited Mar 18 at 22:38
Bernard
123k741117
asked Mar 18 at 22:12
user2965071user2965071
1887
$endgroup$
$phi in operatornameAut(Bbb Z_50)$ via $phi(11) = 3$ Then $phi(x) = $? For any $x in Bbb Z_50$
The answer is $23x$ but I'm not quite sure how to figure that out. Here's what I did:
$phi(11) = 3 Rightarrow$ The function maps $11x rightarrow 3x$
Adding $(11)^-1 = 39$ (mod 50) on both sides gives us
$x rightarrow 42x$
What am I not doing correct?
group-theory
group-theory
edited Mar 18 at 22:38
Bernard
123k741117
asked Mar 18 at 22:12
user2965071user2965071
1887
edited Mar 18 at 22:38
Bernard
123k741117
asked Mar 18 at 22:12
user2965071user2965071
1887
edited Mar 18 at 22:38
Bernard
123k741117
edited Mar 18 at 22:38
Bernard
123k741117
edited Mar 18 at 22:38
Bernard
123k741117
123k741117
asked Mar 18 at 22:12
user2965071user2965071
1887
asked Mar 18 at 22:12
user2965071user2965071
1887
asked Mar 18 at 22:12
user2965071user2965071
1887
1887
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
We have
$$1equiv -9cdot11pmod50 .$$
Therefore,
$$phi(x)equivphi(-9xcdot11)equiv-9xcdotphi(11)equiv-27xequiv23xpmod50 .$$
In the second equivalence we used that $phi(nx) = phi(x)+cdots+phi(x) = nphi(x)$.
Adding $30equiv-11pmod50$ to both sides just tells you that
$$phi(11x)-11equiv3x+39 ,$$
which is no help at all.
answered Mar 18 at 22:17
Daniel Robert-NicoudDaniel Robert-Nicoud
20.5k33797
$endgroup$
$begingroup$
Shouldn't the second equivalence be $-9 phi(x) cdot phi(11)$?
$endgroup$
– Robert Shore
Mar 18 at 22:38
1
$begingroup$
@RobertShore No, the group structure is given by addition, not multiplication.
$endgroup$
– Daniel Robert-Nicoud
Mar 19 at 6:20
add a comment |
$begingroup$
Any automorphism $phi$ of $Bbb Z / n Bbb Z$ is determined by $phi(1)$ because $phi(n) = n phi(1)$, so it suffices to determine $phi(1)$. As noted by Daniel Robert-Nicoud, $1 equiv -9 cdot 11 pmod50$, so $phi(1) = -9 phi(11) = -9 cdot 3 = -27 ~(mod 50) = 23$ and $phi(n) = n phi(1) = 23n$.
answered Mar 18 at 22:45
Robert ShoreRobert Shore
3,603324
$endgroup$
add a comment |
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2 Answers
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2 Answers
2
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votes
$begingroup$
We have
$$1equiv -9cdot11pmod50 .$$
Therefore,
$$phi(x)equivphi(-9xcdot11)equiv-9xcdotphi(11)equiv-27xequiv23xpmod50 .$$
In the second equivalence we used that $phi(nx) = phi(x)+cdots+phi(x) = nphi(x)$.
Adding $30equiv-11pmod50$ to both sides just tells you that
$$phi(11x)-11equiv3x+39 ,$$
which is no help at all.
answered Mar 18 at 22:17
Daniel Robert-NicoudDaniel Robert-Nicoud
20.5k33797
$endgroup$
$begingroup$
Shouldn't the second equivalence be $-9 phi(x) cdot phi(11)$?
$endgroup$
– Robert Shore
Mar 18 at 22:38
1
$begingroup$
@RobertShore No, the group structure is given by addition, not multiplication.
$endgroup$
– Daniel Robert-Nicoud
Mar 19 at 6:20
add a comment |
$begingroup$
We have
$$1equiv -9cdot11pmod50 .$$
Therefore,
$$phi(x)equivphi(-9xcdot11)equiv-9xcdotphi(11)equiv-27xequiv23xpmod50 .$$
In the second equivalence we used that $phi(nx) = phi(x)+cdots+phi(x) = nphi(x)$.
Adding $30equiv-11pmod50$ to both sides just tells you that
$$phi(11x)-11equiv3x+39 ,$$
which is no help at all.
answered Mar 18 at 22:17
Daniel Robert-NicoudDaniel Robert-Nicoud
20.5k33797
$endgroup$
$begingroup$
Shouldn't the second equivalence be $-9 phi(x) cdot phi(11)$?
$endgroup$
– Robert Shore
Mar 18 at 22:38
1
$begingroup$
@RobertShore No, the group structure is given by addition, not multiplication.
$endgroup$
– Daniel Robert-Nicoud
Mar 19 at 6:20
add a comment |
$begingroup$
We have
$$1equiv -9cdot11pmod50 .$$
Therefore,
$$phi(x)equivphi(-9xcdot11)equiv-9xcdotphi(11)equiv-27xequiv23xpmod50 .$$
In the second equivalence we used that $phi(nx) = phi(x)+cdots+phi(x) = nphi(x)$.
Adding $30equiv-11pmod50$ to both sides just tells you that
$$phi(11x)-11equiv3x+39 ,$$
which is no help at all.
answered Mar 18 at 22:17
Daniel Robert-NicoudDaniel Robert-Nicoud
20.5k33797
$endgroup$
We have
$$1equiv -9cdot11pmod50 .$$
Therefore,
$$phi(x)equivphi(-9xcdot11)equiv-9xcdotphi(11)equiv-27xequiv23xpmod50 .$$
In the second equivalence we used that $phi(nx) = phi(x)+cdots+phi(x) = nphi(x)$.
Adding $30equiv-11pmod50$ to both sides just tells you that
$$phi(11x)-11equiv3x+39 ,$$
which is no help at all.
answered Mar 18 at 22:17
Daniel Robert-NicoudDaniel Robert-Nicoud
20.5k33797
answered Mar 18 at 22:17
Daniel Robert-NicoudDaniel Robert-Nicoud
20.5k33797
answered Mar 18 at 22:17
Daniel Robert-NicoudDaniel Robert-Nicoud
20.5k33797
answered Mar 18 at 22:17
Daniel Robert-NicoudDaniel Robert-Nicoud
20.5k33797
20.5k33797
$begingroup$
Shouldn't the second equivalence be $-9 phi(x) cdot phi(11)$?
$endgroup$
– Robert Shore
Mar 18 at 22:38
1
$begingroup$
@RobertShore No, the group structure is given by addition, not multiplication.
$endgroup$
– Daniel Robert-Nicoud
Mar 19 at 6:20
add a comment |
$begingroup$
Shouldn't the second equivalence be $-9 phi(x) cdot phi(11)$?
$endgroup$
– Robert Shore
Mar 18 at 22:38
1
$begingroup$
@RobertShore No, the group structure is given by addition, not multiplication.
$endgroup$
– Daniel Robert-Nicoud
Mar 19 at 6:20
$begingroup$
Shouldn't the second equivalence be $-9 phi(x) cdot phi(11)$?
$endgroup$
– Robert Shore
Mar 18 at 22:38
$begingroup$
Shouldn't the second equivalence be $-9 phi(x) cdot phi(11)$?
$endgroup$
– Robert Shore
Mar 18 at 22:38
1
1
$begingroup$
@RobertShore No, the group structure is given by addition, not multiplication.
$endgroup$
– Daniel Robert-Nicoud
Mar 19 at 6:20
$begingroup$
@RobertShore No, the group structure is given by addition, not multiplication.
$endgroup$
– Daniel Robert-Nicoud
Mar 19 at 6:20
add a comment |
$begingroup$
Any automorphism $phi$ of $Bbb Z / n Bbb Z$ is determined by $phi(1)$ because $phi(n) = n phi(1)$, so it suffices to determine $phi(1)$. As noted by Daniel Robert-Nicoud, $1 equiv -9 cdot 11 pmod50$, so $phi(1) = -9 phi(11) = -9 cdot 3 = -27 ~(mod 50) = 23$ and $phi(n) = n phi(1) = 23n$.
answered Mar 18 at 22:45
Robert ShoreRobert Shore
3,603324
$endgroup$
add a comment |
$begingroup$
Any automorphism $phi$ of $Bbb Z / n Bbb Z$ is determined by $phi(1)$ because $phi(n) = n phi(1)$, so it suffices to determine $phi(1)$. As noted by Daniel Robert-Nicoud, $1 equiv -9 cdot 11 pmod50$, so $phi(1) = -9 phi(11) = -9 cdot 3 = -27 ~(mod 50) = 23$ and $phi(n) = n phi(1) = 23n$.
answered Mar 18 at 22:45
Robert ShoreRobert Shore
3,603324
$endgroup$
add a comment |
$begingroup$
Any automorphism $phi$ of $Bbb Z / n Bbb Z$ is determined by $phi(1)$ because $phi(n) = n phi(1)$, so it suffices to determine $phi(1)$. As noted by Daniel Robert-Nicoud, $1 equiv -9 cdot 11 pmod50$, so $phi(1) = -9 phi(11) = -9 cdot 3 = -27 ~(mod 50) = 23$ and $phi(n) = n phi(1) = 23n$.
answered Mar 18 at 22:45
Robert ShoreRobert Shore
3,603324
$endgroup$
Any automorphism $phi$ of $Bbb Z / n Bbb Z$ is determined by $phi(1)$ because $phi(n) = n phi(1)$, so it suffices to determine $phi(1)$. As noted by Daniel Robert-Nicoud, $1 equiv -9 cdot 11 pmod50$, so $phi(1) = -9 phi(11) = -9 cdot 3 = -27 ~(mod 50) = 23$ and $phi(n) = n phi(1) = 23n$.
answered Mar 18 at 22:45
Robert ShoreRobert Shore
3,603324
answered Mar 18 at 22:45
Robert ShoreRobert Shore
3,603324
answered Mar 18 at 22:45
Robert ShoreRobert Shore
3,603324
answered Mar 18 at 22:45
Robert ShoreRobert Shore
3,603324
3,603324
add a comment |
add a comment |
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