Calculating elements of the $mathbbZ^2/!kervarphi$ groupAn epimorphism from $S_4$ to $S_3$ having the kernel isomorphic to Klein four-groupWhat should be the number of group homomorphism from $mathbb Z_2times mathbb Z_2$ to $S_8$?How many onto group homomorhisms are there from $mathbb Z_2times mathbb Z_4$ onto $mathbb Z_2times mathbb Z_4$?Homomorphism question (group theory)The condition of determining a group homomorphism between two finite abelian groupWhy $ker varphi|_N$ is finitely generated?Contruct a homomorphism $phi$ of $(mathbbQ, +)$ such that $mathrmKer phi = (mathbbZ, +)$the number of the subgroups of a finite group$varphi$ is well defined on $G/N$ if and only if $Nle ker Phi$.Determine the maximal ideals of $mathbb R^2$ by noting $mathbb R^2 cong mathbb R[x]/(x^2-1)$
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Calculating elements of the $mathbbZ^2/!kervarphi$ group
An epimorphism from $S_4$ to $S_3$ having the kernel isomorphic to Klein four-groupWhat should be the number of group homomorphism from $mathbb Z_2times mathbb Z_2$ to $S_8$?How many onto group homomorhisms are there from $mathbb Z_2times mathbb Z_4$ onto $mathbb Z_2times mathbb Z_4$?Homomorphism question (group theory)The condition of determining a group homomorphism between two finite abelian groupWhy $ker varphi|_N$ is finitely generated?Contruct a homomorphism $phi$ of $(mathbbQ, +)$ such that $mathrmKer phi = (mathbbZ, +)$the number of the subgroups of a finite group$varphi$ is well defined on $G/N$ if and only if $Nle ker Phi$.Determine the maximal ideals of $mathbb R^2$ by noting $mathbb R^2 cong mathbb R[x]/(x^2-1)$
$begingroup$
If I know that $varphi: mathbbZ^2 to S_15$ is the homomorphism defined as follows:
beginalign
varphi(1,0)&=(2,7,3)(3,11,5)(12,13)\
varphi(0,1)&=(14,15)(1,4,6,8,9,10)
endalign
I was asked to calculate how many elements $mathbbZ^2/!kervarphi$ has.
Is it true to say that from the first homomorphism theorem we get $mathbbZ^2/!kervarphi cong S_15$. So from $|S_15|=15!$ we can understand that $mathbbZ^2/!kervarphi$ has $15!$ elements. For some reason, it is written in the textbook that is has only $60$ elements. Why is that?
abstract-algebra group-theory permutations
edited Mar 14 at 16:41
egreg
184k1486206
asked Mar 14 at 15:54
abuka123abuka123
445
$endgroup$
|
show 3 more comments
$begingroup$
If I know that $varphi: mathbbZ^2 to S_15$ is the homomorphism defined as follows:
beginalign
varphi(1,0)&=(2,7,3)(3,11,5)(12,13)\
varphi(0,1)&=(14,15)(1,4,6,8,9,10)
endalign
I was asked to calculate how many elements $mathbbZ^2/!kervarphi$ has.
Is it true to say that from the first homomorphism theorem we get $mathbbZ^2/!kervarphi cong S_15$. So from $|S_15|=15!$ we can understand that $mathbbZ^2/!kervarphi$ has $15!$ elements. For some reason, it is written in the textbook that is has only $60$ elements. Why is that?
abstract-algebra group-theory permutations
edited Mar 14 at 16:41
egreg
184k1486206
asked Mar 14 at 15:54
abuka123abuka123
445
$endgroup$
4
$begingroup$
$Bbb Z^2/kervarphi$ is isomorphic to the image of $varphi$, not necessarily the whole codomain.
$endgroup$
– Arthur
Mar 14 at 15:58
1
$begingroup$
To apply the homomorphism theorem the map should be an epimorphism
$endgroup$
– janmarqz
Mar 14 at 15:58
$begingroup$
isn't $varphi$ surjective so $Imvarphi = S_15$. If not, how to count the elements then?
$endgroup$
– abuka123
Mar 14 at 16:04
2
$begingroup$
Also, the image of an abelian group under an homomorphism is always abelian, so the image of $mathbb Z^2$ is abelian, and all of $S_15$ is not abelian.
$endgroup$
– Thomas Andrews
Mar 14 at 16:08
1
$begingroup$
so how to count the elements?
$endgroup$
– abuka123
Mar 14 at 16:09
|
show 3 more comments
$begingroup$
If I know that $varphi: mathbbZ^2 to S_15$ is the homomorphism defined as follows:
beginalign
varphi(1,0)&=(2,7,3)(3,11,5)(12,13)\
varphi(0,1)&=(14,15)(1,4,6,8,9,10)
endalign
I was asked to calculate how many elements $mathbbZ^2/!kervarphi$ has.
Is it true to say that from the first homomorphism theorem we get $mathbbZ^2/!kervarphi cong S_15$. So from $|S_15|=15!$ we can understand that $mathbbZ^2/!kervarphi$ has $15!$ elements. For some reason, it is written in the textbook that is has only $60$ elements. Why is that?
abstract-algebra group-theory permutations
edited Mar 14 at 16:41
egreg
184k1486206
asked Mar 14 at 15:54
abuka123abuka123
445
$endgroup$
If I know that $varphi: mathbbZ^2 to S_15$ is the homomorphism defined as follows:
beginalign
varphi(1,0)&=(2,7,3)(3,11,5)(12,13)\
varphi(0,1)&=(14,15)(1,4,6,8,9,10)
endalign
I was asked to calculate how many elements $mathbbZ^2/!kervarphi$ has.
Is it true to say that from the first homomorphism theorem we get $mathbbZ^2/!kervarphi cong S_15$. So from $|S_15|=15!$ we can understand that $mathbbZ^2/!kervarphi$ has $15!$ elements. For some reason, it is written in the textbook that is has only $60$ elements. Why is that?
abstract-algebra group-theory permutations
abstract-algebra group-theory permutations
edited Mar 14 at 16:41
egreg
184k1486206
asked Mar 14 at 15:54
abuka123abuka123
445
edited Mar 14 at 16:41
egreg
184k1486206
asked Mar 14 at 15:54
abuka123abuka123
445
edited Mar 14 at 16:41
egreg
184k1486206
edited Mar 14 at 16:41
egreg
184k1486206
edited Mar 14 at 16:41
egreg
184k1486206
184k1486206
asked Mar 14 at 15:54
abuka123abuka123
445
asked Mar 14 at 15:54
abuka123abuka123
445
asked Mar 14 at 15:54
abuka123abuka123
445
445
4
$begingroup$
$Bbb Z^2/kervarphi$ is isomorphic to the image of $varphi$, not necessarily the whole codomain.
$endgroup$
– Arthur
Mar 14 at 15:58
1
$begingroup$
To apply the homomorphism theorem the map should be an epimorphism
$endgroup$
– janmarqz
Mar 14 at 15:58
$begingroup$
isn't $varphi$ surjective so $Imvarphi = S_15$. If not, how to count the elements then?
$endgroup$
– abuka123
Mar 14 at 16:04
2
$begingroup$
Also, the image of an abelian group under an homomorphism is always abelian, so the image of $mathbb Z^2$ is abelian, and all of $S_15$ is not abelian.
$endgroup$
– Thomas Andrews
Mar 14 at 16:08
1
$begingroup$
so how to count the elements?
$endgroup$
– abuka123
Mar 14 at 16:09
|
show 3 more comments
4
$begingroup$
$Bbb Z^2/kervarphi$ is isomorphic to the image of $varphi$, not necessarily the whole codomain.
$endgroup$
– Arthur
Mar 14 at 15:58
1
$begingroup$
To apply the homomorphism theorem the map should be an epimorphism
$endgroup$
– janmarqz
Mar 14 at 15:58
$begingroup$
isn't $varphi$ surjective so $Imvarphi = S_15$. If not, how to count the elements then?
$endgroup$
– abuka123
Mar 14 at 16:04
2
$begingroup$
Also, the image of an abelian group under an homomorphism is always abelian, so the image of $mathbb Z^2$ is abelian, and all of $S_15$ is not abelian.
$endgroup$
– Thomas Andrews
Mar 14 at 16:08
1
$begingroup$
so how to count the elements?
$endgroup$
– abuka123
Mar 14 at 16:09
4
4
$begingroup$
$Bbb Z^2/kervarphi$ is isomorphic to the image of $varphi$, not necessarily the whole codomain.
$endgroup$
– Arthur
Mar 14 at 15:58
$begingroup$
$Bbb Z^2/kervarphi$ is isomorphic to the image of $varphi$, not necessarily the whole codomain.
$endgroup$
– Arthur
Mar 14 at 15:58
1
1
$begingroup$
To apply the homomorphism theorem the map should be an epimorphism
$endgroup$
– janmarqz
Mar 14 at 15:58
$begingroup$
To apply the homomorphism theorem the map should be an epimorphism
$endgroup$
– janmarqz
Mar 14 at 15:58
$begingroup$
isn't $varphi$ surjective so $Imvarphi = S_15$. If not, how to count the elements then?
$endgroup$
– abuka123
Mar 14 at 16:04
$begingroup$
isn't $varphi$ surjective so $Imvarphi = S_15$. If not, how to count the elements then?
$endgroup$
– abuka123
Mar 14 at 16:04
2
2
$begingroup$
Also, the image of an abelian group under an homomorphism is always abelian, so the image of $mathbb Z^2$ is abelian, and all of $S_15$ is not abelian.
$endgroup$
– Thomas Andrews
Mar 14 at 16:08
$begingroup$
Also, the image of an abelian group under an homomorphism is always abelian, so the image of $mathbb Z^2$ is abelian, and all of $S_15$ is not abelian.
$endgroup$
– Thomas Andrews
Mar 14 at 16:08
1
1
$begingroup$
so how to count the elements?
$endgroup$
– abuka123
Mar 14 at 16:09
$begingroup$
so how to count the elements?
$endgroup$
– abuka123
Mar 14 at 16:09
|
show 3 more comments
2 Answers
2
active
oldest
votes
$begingroup$
Set, for simplicity, $sigma=varphi(1,0)$ and $tau=varphi(0,1)$. Then
$$
sigma=(2,7,3,11,5)(12,13)
$$
and so $sigma$ and $tau$ are disjoint, so $sigmatau=tausigma$ and the homomorphism is indeed well defined.
The image of $varphi$ is an abelian subgroup of $S_15$, so it certainly isn't the whole $S_15$.
Set, for simplicity, $sigma=varphi(1,0)$ and $tau=varphi(0,1)$. Then
$$
varphi(m,n)=sigma^mtau^n
$$
and so the image of $varphi$ is the subgroup generated by $sigma$ and $tau$. Now you should be able to finish, using the fact that there is an obvious surjective homomorphism
$$
langlesigmarangletimeslangletaurangleto
langlesigmaranglelangletaurangle=operatornameimvarphi
$$
What's the kernel of this homomorphism?
answered Mar 14 at 16:46
egregegreg
184k1486206
$endgroup$
$begingroup$
$(2,7,3)(3,11,5)$ aren't disjoint
$endgroup$
– janmarqz
Mar 14 at 17:24
1
$begingroup$
@janmarqz I meant $sigma$ and $tau$, but I have clarified.
$endgroup$
– egreg
Mar 14 at 18:04
$begingroup$
your explaination is certainly true and I would upvote you if you criticized positively and upvote mine, greets!
$endgroup$
– janmarqz
Mar 15 at 18:14
add a comment |
$begingroup$
To apply the homomorphism theorem the map should be an epimorphism.
In the other hand to count how many elements in the image are, you just got to see is the orders of $varphi(1,0)=(2,7,3,11,5)(12,13)$ and $varphi(0,1)$ which happen to be ten and six respectively. Then $|rm imvarphi|$ is $60$.
Update:
The next calculations might help you to understand
begineqnarray*
varphi(2,0)&=&((2,7,3,11,5)(12,13))^2,\
&=&(2,3,5,7,11).\
&&\
varphi(3,0)&=&((2,7,3,11,5)(12,13))^3,\
&=&(2,11,7,5,3)(12,13).\
&&\
varphi(4,0)&=&(2,5,11,3,7).\
&&\
varphi(5,0)&=&(12,13).\
&&\
&...&\
&&\
varphi(10,0)&=&e.
endeqnarray*
answered Mar 14 at 16:12
janmarqzjanmarqz
6,22741630
$endgroup$
1
$begingroup$
$phi(1,0)$ only has order $4$ (the "$3$" in the formula for $phi(1,0)$ occurs twice).
$endgroup$
– Lee Mosher
Mar 14 at 16:14
$begingroup$
ah! yes I'll correct
$endgroup$
– janmarqz
Mar 14 at 16:15
1
$begingroup$
why calculating $o(varphi(1,0))$ and $o(varphi(0,1))$ gives the solution?
$endgroup$
– abuka123
Mar 14 at 16:28
1
$begingroup$
Observe that $text Im (varphi)$ is generated by $varphi (1,0)$ and $varphi(0,1)$ since $(1,0)$ and $(0,1)$ generate $Bbb Z^2.$
$endgroup$
– Dbchatto67
Mar 14 at 16:43
1
$begingroup$
Also observe that the elements of a finite group $G$ generated by $a,b in G$ are of the form $a^ib^j$ where $0 leq i leq n$ and $0 leq j leq m$ where $o(a)=n$ and $o(b)=m.$ But then $|G| = mn.$
$endgroup$
– Dbchatto67
Mar 14 at 16:47
|
show 1 more comment
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Set, for simplicity, $sigma=varphi(1,0)$ and $tau=varphi(0,1)$. Then
$$
sigma=(2,7,3,11,5)(12,13)
$$
and so $sigma$ and $tau$ are disjoint, so $sigmatau=tausigma$ and the homomorphism is indeed well defined.
The image of $varphi$ is an abelian subgroup of $S_15$, so it certainly isn't the whole $S_15$.
Set, for simplicity, $sigma=varphi(1,0)$ and $tau=varphi(0,1)$. Then
$$
varphi(m,n)=sigma^mtau^n
$$
and so the image of $varphi$ is the subgroup generated by $sigma$ and $tau$. Now you should be able to finish, using the fact that there is an obvious surjective homomorphism
$$
langlesigmarangletimeslangletaurangleto
langlesigmaranglelangletaurangle=operatornameimvarphi
$$
What's the kernel of this homomorphism?
answered Mar 14 at 16:46
egregegreg
184k1486206
$endgroup$
$begingroup$
$(2,7,3)(3,11,5)$ aren't disjoint
$endgroup$
– janmarqz
Mar 14 at 17:24
1
$begingroup$
@janmarqz I meant $sigma$ and $tau$, but I have clarified.
$endgroup$
– egreg
Mar 14 at 18:04
$begingroup$
your explaination is certainly true and I would upvote you if you criticized positively and upvote mine, greets!
$endgroup$
– janmarqz
Mar 15 at 18:14
add a comment |
$begingroup$
Set, for simplicity, $sigma=varphi(1,0)$ and $tau=varphi(0,1)$. Then
$$
sigma=(2,7,3,11,5)(12,13)
$$
and so $sigma$ and $tau$ are disjoint, so $sigmatau=tausigma$ and the homomorphism is indeed well defined.
The image of $varphi$ is an abelian subgroup of $S_15$, so it certainly isn't the whole $S_15$.
Set, for simplicity, $sigma=varphi(1,0)$ and $tau=varphi(0,1)$. Then
$$
varphi(m,n)=sigma^mtau^n
$$
and so the image of $varphi$ is the subgroup generated by $sigma$ and $tau$. Now you should be able to finish, using the fact that there is an obvious surjective homomorphism
$$
langlesigmarangletimeslangletaurangleto
langlesigmaranglelangletaurangle=operatornameimvarphi
$$
What's the kernel of this homomorphism?
answered Mar 14 at 16:46
egregegreg
184k1486206
$endgroup$
$begingroup$
$(2,7,3)(3,11,5)$ aren't disjoint
$endgroup$
– janmarqz
Mar 14 at 17:24
1
$begingroup$
@janmarqz I meant $sigma$ and $tau$, but I have clarified.
$endgroup$
– egreg
Mar 14 at 18:04
$begingroup$
your explaination is certainly true and I would upvote you if you criticized positively and upvote mine, greets!
$endgroup$
– janmarqz
Mar 15 at 18:14
add a comment |
$begingroup$
Set, for simplicity, $sigma=varphi(1,0)$ and $tau=varphi(0,1)$. Then
$$
sigma=(2,7,3,11,5)(12,13)
$$
and so $sigma$ and $tau$ are disjoint, so $sigmatau=tausigma$ and the homomorphism is indeed well defined.
The image of $varphi$ is an abelian subgroup of $S_15$, so it certainly isn't the whole $S_15$.
Set, for simplicity, $sigma=varphi(1,0)$ and $tau=varphi(0,1)$. Then
$$
varphi(m,n)=sigma^mtau^n
$$
and so the image of $varphi$ is the subgroup generated by $sigma$ and $tau$. Now you should be able to finish, using the fact that there is an obvious surjective homomorphism
$$
langlesigmarangletimeslangletaurangleto
langlesigmaranglelangletaurangle=operatornameimvarphi
$$
What's the kernel of this homomorphism?
answered Mar 14 at 16:46
egregegreg
184k1486206
$endgroup$
Set, for simplicity, $sigma=varphi(1,0)$ and $tau=varphi(0,1)$. Then
$$
sigma=(2,7,3,11,5)(12,13)
$$
and so $sigma$ and $tau$ are disjoint, so $sigmatau=tausigma$ and the homomorphism is indeed well defined.
The image of $varphi$ is an abelian subgroup of $S_15$, so it certainly isn't the whole $S_15$.
Set, for simplicity, $sigma=varphi(1,0)$ and $tau=varphi(0,1)$. Then
$$
varphi(m,n)=sigma^mtau^n
$$
and so the image of $varphi$ is the subgroup generated by $sigma$ and $tau$. Now you should be able to finish, using the fact that there is an obvious surjective homomorphism
$$
langlesigmarangletimeslangletaurangleto
langlesigmaranglelangletaurangle=operatornameimvarphi
$$
What's the kernel of this homomorphism?
answered Mar 14 at 16:46
egregegreg
184k1486206
edited Mar 14 at 18:04
answered Mar 14 at 16:46
egregegreg
184k1486206
answered Mar 14 at 16:46
egregegreg
184k1486206
answered Mar 14 at 16:46
egregegreg
184k1486206
184k1486206
$begingroup$
$(2,7,3)(3,11,5)$ aren't disjoint
$endgroup$
– janmarqz
Mar 14 at 17:24
1
$begingroup$
@janmarqz I meant $sigma$ and $tau$, but I have clarified.
$endgroup$
– egreg
Mar 14 at 18:04
$begingroup$
your explaination is certainly true and I would upvote you if you criticized positively and upvote mine, greets!
$endgroup$
– janmarqz
Mar 15 at 18:14
add a comment |
$begingroup$
$(2,7,3)(3,11,5)$ aren't disjoint
$endgroup$
– janmarqz
Mar 14 at 17:24
1
$begingroup$
@janmarqz I meant $sigma$ and $tau$, but I have clarified.
$endgroup$
– egreg
Mar 14 at 18:04
$begingroup$
your explaination is certainly true and I would upvote you if you criticized positively and upvote mine, greets!
$endgroup$
– janmarqz
Mar 15 at 18:14
$begingroup$
$(2,7,3)(3,11,5)$ aren't disjoint
$endgroup$
– janmarqz
Mar 14 at 17:24
$begingroup$
$(2,7,3)(3,11,5)$ aren't disjoint
$endgroup$
– janmarqz
Mar 14 at 17:24
1
1
$begingroup$
@janmarqz I meant $sigma$ and $tau$, but I have clarified.
$endgroup$
– egreg
Mar 14 at 18:04
$begingroup$
@janmarqz I meant $sigma$ and $tau$, but I have clarified.
$endgroup$
– egreg
Mar 14 at 18:04
$begingroup$
your explaination is certainly true and I would upvote you if you criticized positively and upvote mine, greets!
$endgroup$
– janmarqz
Mar 15 at 18:14
$begingroup$
your explaination is certainly true and I would upvote you if you criticized positively and upvote mine, greets!
$endgroup$
– janmarqz
Mar 15 at 18:14
add a comment |
$begingroup$
To apply the homomorphism theorem the map should be an epimorphism.
In the other hand to count how many elements in the image are, you just got to see is the orders of $varphi(1,0)=(2,7,3,11,5)(12,13)$ and $varphi(0,1)$ which happen to be ten and six respectively. Then $|rm imvarphi|$ is $60$.
Update:
The next calculations might help you to understand
begineqnarray*
varphi(2,0)&=&((2,7,3,11,5)(12,13))^2,\
&=&(2,3,5,7,11).\
&&\
varphi(3,0)&=&((2,7,3,11,5)(12,13))^3,\
&=&(2,11,7,5,3)(12,13).\
&&\
varphi(4,0)&=&(2,5,11,3,7).\
&&\
varphi(5,0)&=&(12,13).\
&&\
&...&\
&&\
varphi(10,0)&=&e.
endeqnarray*
answered Mar 14 at 16:12
janmarqzjanmarqz
6,22741630
$endgroup$
1
$begingroup$
$phi(1,0)$ only has order $4$ (the "$3$" in the formula for $phi(1,0)$ occurs twice).
$endgroup$
– Lee Mosher
Mar 14 at 16:14
$begingroup$
ah! yes I'll correct
$endgroup$
– janmarqz
Mar 14 at 16:15
1
$begingroup$
why calculating $o(varphi(1,0))$ and $o(varphi(0,1))$ gives the solution?
$endgroup$
– abuka123
Mar 14 at 16:28
1
$begingroup$
Observe that $text Im (varphi)$ is generated by $varphi (1,0)$ and $varphi(0,1)$ since $(1,0)$ and $(0,1)$ generate $Bbb Z^2.$
$endgroup$
– Dbchatto67
Mar 14 at 16:43
1
$begingroup$
Also observe that the elements of a finite group $G$ generated by $a,b in G$ are of the form $a^ib^j$ where $0 leq i leq n$ and $0 leq j leq m$ where $o(a)=n$ and $o(b)=m.$ But then $|G| = mn.$
$endgroup$
– Dbchatto67
Mar 14 at 16:47
|
show 1 more comment
$begingroup$
To apply the homomorphism theorem the map should be an epimorphism.
In the other hand to count how many elements in the image are, you just got to see is the orders of $varphi(1,0)=(2,7,3,11,5)(12,13)$ and $varphi(0,1)$ which happen to be ten and six respectively. Then $|rm imvarphi|$ is $60$.
Update:
The next calculations might help you to understand
begineqnarray*
varphi(2,0)&=&((2,7,3,11,5)(12,13))^2,\
&=&(2,3,5,7,11).\
&&\
varphi(3,0)&=&((2,7,3,11,5)(12,13))^3,\
&=&(2,11,7,5,3)(12,13).\
&&\
varphi(4,0)&=&(2,5,11,3,7).\
&&\
varphi(5,0)&=&(12,13).\
&&\
&...&\
&&\
varphi(10,0)&=&e.
endeqnarray*
answered Mar 14 at 16:12
janmarqzjanmarqz
6,22741630
$endgroup$
1
$begingroup$
$phi(1,0)$ only has order $4$ (the "$3$" in the formula for $phi(1,0)$ occurs twice).
$endgroup$
– Lee Mosher
Mar 14 at 16:14
$begingroup$
ah! yes I'll correct
$endgroup$
– janmarqz
Mar 14 at 16:15
1
$begingroup$
why calculating $o(varphi(1,0))$ and $o(varphi(0,1))$ gives the solution?
$endgroup$
– abuka123
Mar 14 at 16:28
1
$begingroup$
Observe that $text Im (varphi)$ is generated by $varphi (1,0)$ and $varphi(0,1)$ since $(1,0)$ and $(0,1)$ generate $Bbb Z^2.$
$endgroup$
– Dbchatto67
Mar 14 at 16:43
1
$begingroup$
Also observe that the elements of a finite group $G$ generated by $a,b in G$ are of the form $a^ib^j$ where $0 leq i leq n$ and $0 leq j leq m$ where $o(a)=n$ and $o(b)=m.$ But then $|G| = mn.$
$endgroup$
– Dbchatto67
Mar 14 at 16:47
|
show 1 more comment
$begingroup$
To apply the homomorphism theorem the map should be an epimorphism.
In the other hand to count how many elements in the image are, you just got to see is the orders of $varphi(1,0)=(2,7,3,11,5)(12,13)$ and $varphi(0,1)$ which happen to be ten and six respectively. Then $|rm imvarphi|$ is $60$.
Update:
The next calculations might help you to understand
begineqnarray*
varphi(2,0)&=&((2,7,3,11,5)(12,13))^2,\
&=&(2,3,5,7,11).\
&&\
varphi(3,0)&=&((2,7,3,11,5)(12,13))^3,\
&=&(2,11,7,5,3)(12,13).\
&&\
varphi(4,0)&=&(2,5,11,3,7).\
&&\
varphi(5,0)&=&(12,13).\
&&\
&...&\
&&\
varphi(10,0)&=&e.
endeqnarray*
answered Mar 14 at 16:12
janmarqzjanmarqz
6,22741630
$endgroup$
To apply the homomorphism theorem the map should be an epimorphism.
In the other hand to count how many elements in the image are, you just got to see is the orders of $varphi(1,0)=(2,7,3,11,5)(12,13)$ and $varphi(0,1)$ which happen to be ten and six respectively. Then $|rm imvarphi|$ is $60$.
Update:
The next calculations might help you to understand
begineqnarray*
varphi(2,0)&=&((2,7,3,11,5)(12,13))^2,\
&=&(2,3,5,7,11).\
&&\
varphi(3,0)&=&((2,7,3,11,5)(12,13))^3,\
&=&(2,11,7,5,3)(12,13).\
&&\
varphi(4,0)&=&(2,5,11,3,7).\
&&\
varphi(5,0)&=&(12,13).\
&&\
&...&\
&&\
varphi(10,0)&=&e.
endeqnarray*
answered Mar 14 at 16:12
janmarqzjanmarqz
6,22741630
edited Mar 15 at 20:46
answered Mar 14 at 16:12
janmarqzjanmarqz
6,22741630
answered Mar 14 at 16:12
janmarqzjanmarqz
6,22741630
answered Mar 14 at 16:12
janmarqzjanmarqz
6,22741630
6,22741630
1
$begingroup$
$phi(1,0)$ only has order $4$ (the "$3$" in the formula for $phi(1,0)$ occurs twice).
$endgroup$
– Lee Mosher
Mar 14 at 16:14
$begingroup$
ah! yes I'll correct
$endgroup$
– janmarqz
Mar 14 at 16:15
1
$begingroup$
why calculating $o(varphi(1,0))$ and $o(varphi(0,1))$ gives the solution?
$endgroup$
– abuka123
Mar 14 at 16:28
1
$begingroup$
Observe that $text Im (varphi)$ is generated by $varphi (1,0)$ and $varphi(0,1)$ since $(1,0)$ and $(0,1)$ generate $Bbb Z^2.$
$endgroup$
– Dbchatto67
Mar 14 at 16:43
1
$begingroup$
Also observe that the elements of a finite group $G$ generated by $a,b in G$ are of the form $a^ib^j$ where $0 leq i leq n$ and $0 leq j leq m$ where $o(a)=n$ and $o(b)=m.$ But then $|G| = mn.$
$endgroup$
– Dbchatto67
Mar 14 at 16:47
|
show 1 more comment
1
$begingroup$
$phi(1,0)$ only has order $4$ (the "$3$" in the formula for $phi(1,0)$ occurs twice).
$endgroup$
– Lee Mosher
Mar 14 at 16:14
$begingroup$
ah! yes I'll correct
$endgroup$
– janmarqz
Mar 14 at 16:15
1
$begingroup$
why calculating $o(varphi(1,0))$ and $o(varphi(0,1))$ gives the solution?
$endgroup$
– abuka123
Mar 14 at 16:28
1
$begingroup$
Observe that $text Im (varphi)$ is generated by $varphi (1,0)$ and $varphi(0,1)$ since $(1,0)$ and $(0,1)$ generate $Bbb Z^2.$
$endgroup$
– Dbchatto67
Mar 14 at 16:43
1
$begingroup$
Also observe that the elements of a finite group $G$ generated by $a,b in G$ are of the form $a^ib^j$ where $0 leq i leq n$ and $0 leq j leq m$ where $o(a)=n$ and $o(b)=m.$ But then $|G| = mn.$
$endgroup$
– Dbchatto67
Mar 14 at 16:47
1
1
$begingroup$
$phi(1,0)$ only has order $4$ (the "$3$" in the formula for $phi(1,0)$ occurs twice).
$endgroup$
– Lee Mosher
Mar 14 at 16:14
$begingroup$
$phi(1,0)$ only has order $4$ (the "$3$" in the formula for $phi(1,0)$ occurs twice).
$endgroup$
– Lee Mosher
Mar 14 at 16:14
$begingroup$
ah! yes I'll correct
$endgroup$
– janmarqz
Mar 14 at 16:15
$begingroup$
ah! yes I'll correct
$endgroup$
– janmarqz
Mar 14 at 16:15
1
1
$begingroup$
why calculating $o(varphi(1,0))$ and $o(varphi(0,1))$ gives the solution?
$endgroup$
– abuka123
Mar 14 at 16:28
$begingroup$
why calculating $o(varphi(1,0))$ and $o(varphi(0,1))$ gives the solution?
$endgroup$
– abuka123
Mar 14 at 16:28
1
1
$begingroup$
Observe that $text Im (varphi)$ is generated by $varphi (1,0)$ and $varphi(0,1)$ since $(1,0)$ and $(0,1)$ generate $Bbb Z^2.$
$endgroup$
– Dbchatto67
Mar 14 at 16:43
$begingroup$
Observe that $text Im (varphi)$ is generated by $varphi (1,0)$ and $varphi(0,1)$ since $(1,0)$ and $(0,1)$ generate $Bbb Z^2.$
$endgroup$
– Dbchatto67
Mar 14 at 16:43
1
1
$begingroup$
Also observe that the elements of a finite group $G$ generated by $a,b in G$ are of the form $a^ib^j$ where $0 leq i leq n$ and $0 leq j leq m$ where $o(a)=n$ and $o(b)=m.$ But then $|G| = mn.$
$endgroup$
– Dbchatto67
Mar 14 at 16:47
$begingroup$
Also observe that the elements of a finite group $G$ generated by $a,b in G$ are of the form $a^ib^j$ where $0 leq i leq n$ and $0 leq j leq m$ where $o(a)=n$ and $o(b)=m.$ But then $|G| = mn.$
$endgroup$
– Dbchatto67
Mar 14 at 16:47
|
show 1 more comment
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4
$begingroup$
$Bbb Z^2/kervarphi$ is isomorphic to the image of $varphi$, not necessarily the whole codomain.
$endgroup$
– Arthur
Mar 14 at 15:58
1
$begingroup$
To apply the homomorphism theorem the map should be an epimorphism
$endgroup$
– janmarqz
Mar 14 at 15:58
$begingroup$
isn't $varphi$ surjective so $Imvarphi = S_15$. If not, how to count the elements then?
$endgroup$
– abuka123
Mar 14 at 16:04
2
$begingroup$
Also, the image of an abelian group under an homomorphism is always abelian, so the image of $mathbb Z^2$ is abelian, and all of $S_15$ is not abelian.
$endgroup$
– Thomas Andrews
Mar 14 at 16:08
1
$begingroup$
so how to count the elements?
$endgroup$
– abuka123
Mar 14 at 16:09