Mixing
Function that is homomorphism and bijection
$begingroup$
Find a function from a group
to a permutation group
{1,2,3} that this function be
homomorphism and bijection
abstract-algebra
edited Feb 2 at 10:07
Bernard
124k741117
asked Feb 2 at 7:15
S.gulS.gul
112
$endgroup$
add a comment |
$begingroup$
Find a function from a group
to a permutation group
{1,2,3} that this function be
homomorphism and bijection
abstract-algebra
edited Feb 2 at 10:07
Bernard
124k741117
asked Feb 2 at 7:15
S.gulS.gul
112
$endgroup$
add a comment |
$begingroup$
Find a function from a group
to a permutation group
{1,2,3} that this function be
homomorphism and bijection
abstract-algebra
edited Feb 2 at 10:07
Bernard
124k741117
asked Feb 2 at 7:15
S.gulS.gul
112
$endgroup$
Find a function from a group
to a permutation group
{1,2,3} that this function be
homomorphism and bijection
abstract-algebra
abstract-algebra
edited Feb 2 at 10:07
Bernard
124k741117
asked Feb 2 at 7:15
S.gulS.gul
112
edited Feb 2 at 10:07
Bernard
124k741117
asked Feb 2 at 7:15
S.gulS.gul
112
edited Feb 2 at 10:07
Bernard
124k741117
edited Feb 2 at 10:07
Bernard
124k741117
edited Feb 2 at 10:07
Bernard
124k741117
124k741117
asked Feb 2 at 7:15
S.gulS.gul
112
asked Feb 2 at 7:15
S.gulS.gul
112
asked Feb 2 at 7:15
S.gulS.gul
112
112
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
One could just take the identity on $S_3$, the group of permutations of ${1,2,3}$.
For a slightly less trivial example, how about $h:D_3to S_3$ determined by $h(sigma)=(12)$ and $h(rho)=(123)$.
Here $D_3=langle sigma, rhomidsigma^2,rho^3,(sigmarho)^2rangle $.
So $D_3cong S_3$.
answered Feb 2 at 8:43
Chris CusterChris Custer
14.4k3827
$endgroup$
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
One could just take the identity on $S_3$, the group of permutations of ${1,2,3}$.
For a slightly less trivial example, how about $h:D_3to S_3$ determined by $h(sigma)=(12)$ and $h(rho)=(123)$.
Here $D_3=langle sigma, rhomidsigma^2,rho^3,(sigmarho)^2rangle $.
So $D_3cong S_3$.
answered Feb 2 at 8:43
Chris CusterChris Custer
14.4k3827
$endgroup$
add a comment |
$begingroup$
One could just take the identity on $S_3$, the group of permutations of ${1,2,3}$.
For a slightly less trivial example, how about $h:D_3to S_3$ determined by $h(sigma)=(12)$ and $h(rho)=(123)$.
Here $D_3=langle sigma, rhomidsigma^2,rho^3,(sigmarho)^2rangle $.
So $D_3cong S_3$.
answered Feb 2 at 8:43
Chris CusterChris Custer
14.4k3827
$endgroup$
add a comment |
$begingroup$
One could just take the identity on $S_3$, the group of permutations of ${1,2,3}$.
For a slightly less trivial example, how about $h:D_3to S_3$ determined by $h(sigma)=(12)$ and $h(rho)=(123)$.
Here $D_3=langle sigma, rhomidsigma^2,rho^3,(sigmarho)^2rangle $.
So $D_3cong S_3$.
answered Feb 2 at 8:43
Chris CusterChris Custer
14.4k3827
$endgroup$
One could just take the identity on $S_3$, the group of permutations of ${1,2,3}$.
For a slightly less trivial example, how about $h:D_3to S_3$ determined by $h(sigma)=(12)$ and $h(rho)=(123)$.
Here $D_3=langle sigma, rhomidsigma^2,rho^3,(sigmarho)^2rangle $.
So $D_3cong S_3$.
answered Feb 2 at 8:43
Chris CusterChris Custer
14.4k3827
answered Feb 2 at 8:43
Chris CusterChris Custer
14.4k3827
answered Feb 2 at 8:43
Chris CusterChris Custer
14.4k3827
answered Feb 2 at 8:43
Chris CusterChris Custer
14.4k3827
14.4k3827
add a comment |
add a comment |
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