How to determine if there is a surjective homomorphism from $U(15)$ to $mathbb{Z}_4$?
$begingroup$
How to determine if there is a surjective homomorphism from $U(15)$ to $mathbb{Z}_4$? $U(15)$ is the multiplicative group of positive integers less than $n$ that are coprime with $n$.
I already know $U(15)$ is not cyclic with 3 elements of order 2, 4 elements of order 4. $U(15)={1,2,4,7,8,11,13,14}$. While $mathbb{Z}_4$ is obviously cyclic.
abstract-algebra
asked Dec 6 '18 at 11:37
LOISLOIS
3708
$endgroup$
add a comment |
$begingroup$
How to determine if there is a surjective homomorphism from $U(15)$ to $mathbb{Z}_4$? $U(15)$ is the multiplicative group of positive integers less than $n$ that are coprime with $n$.
I already know $U(15)$ is not cyclic with 3 elements of order 2, 4 elements of order 4. $U(15)={1,2,4,7,8,11,13,14}$. While $mathbb{Z}_4$ is obviously cyclic.
abstract-algebra
asked Dec 6 '18 at 11:37
LOISLOIS
3708
$endgroup$
$begingroup$
@ Arthur yes,zero homomorphism. I lost the condition 'surjective'.
$endgroup$
– LOIS
Dec 6 '18 at 11:42
1
$begingroup$
You solve this by trying. If there is a surjective homomorhism, then some element has to hit $1in Bbb Z_4$. Which elements could do that? Then which elements do you know where they're mapped? Then try the rest of the elements.
$endgroup$
– Arthur
Dec 6 '18 at 11:42
$begingroup$
$1in mathbb{Z}_4$ is of order 4. hence the elements hit 1 should have order 4 .They should be 2,7,8,13. The rest except 1 should hit $3in mathbb{Z}_4$?
$endgroup$
– LOIS
Dec 6 '18 at 12:00
$begingroup$
What order does $3in Bbb Z_4$ have? If $f(2) = 1$, then what must $f(8) = f(2cdot 2cdot 2)$ be? And remember that $2$ and $0$ are in there too.
$endgroup$
– Arthur
Dec 6 '18 at 12:08
add a comment |
$begingroup$
How to determine if there is a surjective homomorphism from $U(15)$ to $mathbb{Z}_4$? $U(15)$ is the multiplicative group of positive integers less than $n$ that are coprime with $n$.
I already know $U(15)$ is not cyclic with 3 elements of order 2, 4 elements of order 4. $U(15)={1,2,4,7,8,11,13,14}$. While $mathbb{Z}_4$ is obviously cyclic.
abstract-algebra
asked Dec 6 '18 at 11:37
LOISLOIS
3708
$endgroup$
How to determine if there is a surjective homomorphism from $U(15)$ to $mathbb{Z}_4$? $U(15)$ is the multiplicative group of positive integers less than $n$ that are coprime with $n$.
I already know $U(15)$ is not cyclic with 3 elements of order 2, 4 elements of order 4. $U(15)={1,2,4,7,8,11,13,14}$. While $mathbb{Z}_4$ is obviously cyclic.
abstract-algebra
abstract-algebra
asked Dec 6 '18 at 11:37
LOISLOIS
3708
asked Dec 6 '18 at 11:37
LOISLOIS
3708
edited Dec 6 '18 at 11:41
LOIS
asked Dec 6 '18 at 11:37
LOISLOIS
3708
asked Dec 6 '18 at 11:37
LOISLOIS
3708
asked Dec 6 '18 at 11:37
LOISLOIS
3708
3708
$begingroup$
@ Arthur yes,zero homomorphism. I lost the condition 'surjective'.
$endgroup$
– LOIS
Dec 6 '18 at 11:42
1
$begingroup$
You solve this by trying. If there is a surjective homomorhism, then some element has to hit $1in Bbb Z_4$. Which elements could do that? Then which elements do you know where they're mapped? Then try the rest of the elements.
$endgroup$
– Arthur
Dec 6 '18 at 11:42
$begingroup$
$1in mathbb{Z}_4$ is of order 4. hence the elements hit 1 should have order 4 .They should be 2,7,8,13. The rest except 1 should hit $3in mathbb{Z}_4$?
$endgroup$
– LOIS
Dec 6 '18 at 12:00
$begingroup$
What order does $3in Bbb Z_4$ have? If $f(2) = 1$, then what must $f(8) = f(2cdot 2cdot 2)$ be? And remember that $2$ and $0$ are in there too.
$endgroup$
– Arthur
Dec 6 '18 at 12:08
add a comment |
$begingroup$
@ Arthur yes,zero homomorphism. I lost the condition 'surjective'.
$endgroup$
– LOIS
Dec 6 '18 at 11:42
1
$begingroup$
You solve this by trying. If there is a surjective homomorhism, then some element has to hit $1in Bbb Z_4$. Which elements could do that? Then which elements do you know where they're mapped? Then try the rest of the elements.
$endgroup$
– Arthur
Dec 6 '18 at 11:42
$begingroup$
$1in mathbb{Z}_4$ is of order 4. hence the elements hit 1 should have order 4 .They should be 2,7,8,13. The rest except 1 should hit $3in mathbb{Z}_4$?
$endgroup$
– LOIS
Dec 6 '18 at 12:00
$begingroup$
What order does $3in Bbb Z_4$ have? If $f(2) = 1$, then what must $f(8) = f(2cdot 2cdot 2)$ be? And remember that $2$ and $0$ are in there too.
$endgroup$
– Arthur
Dec 6 '18 at 12:08
$begingroup$
@ Arthur yes,zero homomorphism. I lost the condition 'surjective'.
$endgroup$
– LOIS
Dec 6 '18 at 11:42
$begingroup$
@ Arthur yes,zero homomorphism. I lost the condition 'surjective'.
$endgroup$
– LOIS
Dec 6 '18 at 11:42
1
1
$begingroup$
You solve this by trying. If there is a surjective homomorhism, then some element has to hit $1in Bbb Z_4$. Which elements could do that? Then which elements do you know where they're mapped? Then try the rest of the elements.
$endgroup$
– Arthur
Dec 6 '18 at 11:42
$begingroup$
You solve this by trying. If there is a surjective homomorhism, then some element has to hit $1in Bbb Z_4$. Which elements could do that? Then which elements do you know where they're mapped? Then try the rest of the elements.
$endgroup$
– Arthur
Dec 6 '18 at 11:42
$begingroup$
$1in mathbb{Z}_4$ is of order 4. hence the elements hit 1 should have order 4 .They should be 2,7,8,13. The rest except 1 should hit $3in mathbb{Z}_4$?
$endgroup$
– LOIS
Dec 6 '18 at 12:00
$begingroup$
$1in mathbb{Z}_4$ is of order 4. hence the elements hit 1 should have order 4 .They should be 2,7,8,13. The rest except 1 should hit $3in mathbb{Z}_4$?
$endgroup$
– LOIS
Dec 6 '18 at 12:00
$begingroup$
What order does $3in Bbb Z_4$ have? If $f(2) = 1$, then what must $f(8) = f(2cdot 2cdot 2)$ be? And remember that $2$ and $0$ are in there too.
$endgroup$
– Arthur
Dec 6 '18 at 12:08
$begingroup$
What order does $3in Bbb Z_4$ have? If $f(2) = 1$, then what must $f(8) = f(2cdot 2cdot 2)$ be? And remember that $2$ and $0$ are in there too.
$endgroup$
– Arthur
Dec 6 '18 at 12:08
add a comment |
4 Answers
4
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oldest
votes
$begingroup$
$f(2)=f(7)=1$
$f(8)=f(13)=3$
Can you finish?
answered Dec 6 '18 at 11:45
ajotatxeajotatxe
53.7k23890
$endgroup$
$begingroup$
$f(4)=f(11)=f(14)=2?$
$endgroup$
– LOIS
Dec 6 '18 at 12:03
add a comment |
$begingroup$
You have got that $U(15)congmathbb Z_4oplus mathbb Z_2$. So the existence of a surjective homomorphism is clear.
answered Dec 6 '18 at 14:59
Chris CusterChris Custer
11.4k3824
$endgroup$
add a comment |
$begingroup$
Recall the Chinese remainder theorem:
let $A$ be an arbitrary ring, $Lambda neq emptyset$ a finite set (of indices) and $I in (mathcal{Id}_b(A))^{Lambda}$ a family of pairwise comaximal bilateral (two-sided) ideals of $A$, which specifically means that $$lambda, mu in Lambda, lambda neq mu implies I_{lambda}+I_{mu}=A$$ Under these conditions the following isomorphism of rings takes place:
$${A}Big{/}bigcap_{lambda in Lambda} I_{lambda} approx prod_{lambda in Lambda} (A big{/}I_{lambda})$$
As a side-note, $mathcal{Id}$ stands for ideals and the subscript $b$ for bilateral; for sets $M$ and $J$, $M^J$ denotes the cartesian product $prod limits_{j in J} M$, in other words the set of all families of elements of $M$ indexed by $J$.
Applying the general form of the theorem to the ring of integers, you have the following formulation:
let $I$ be finite and $m in mathbb{Z}^I$ such that $i, j in I, i neq j implies m_i mathbb{Z}+ m_jmathbb{Z}=mathbb{Z}$. Then $mathbb{Z}_{prodlimits_{i in I} m_i} approx prodlimits_{i in I}mathbb{Z}_{m_i}$ as rings.
In the instance you are treating, with prime factor decomposition $15=3cdot5$ the theorem yields the following isomorphism: $$mathbb{Z}_{15} approx mathbb{Z}_3 times mathbb{Z}_5$$
This yields an isomorphism between the corresponding groups of units (invertible elements) of the two rings:
$$mathrm{U}(mathbb{Z}_{15})approx mathrm{U}(mathbb{Z}_3)times mathrm{U}(mathbb{Z}_5)$$
However the quotient rings appearing as factors in the direct product on the right-hand side are actually fields and it is easily seen that $(mathrm{U}(mathbb{Z}_3), cdot) approx (mathbb{Z}_2, +)$ (cyclic of prime order, so any nontrivial element is a generator) and $(mathrm{U}(mathbb{Z}_5), cdot) approx (mathbb{Z}_4, +)$ (where either $overline{2}$ or $overline{3}$ are generators, being the only elements of order 4). The operations of the respective groups are appended for clarity. You thus eventually obtain the isomorphism:
$$mathrm{U}(mathbb{Z}_{15}) approx mathbb{Z}_2 times mathbb{Z}_4$$
exhibiting the invariant factors of $mathrm{U}(mathbb{Z}_{15})$ as a finitely generated torsion $mathbb{Z}$-module (cnf. the theory of finitely generated modules over principal ideal domains). At this point you can easily determine all the subgroups and quotient groups of $mathrm{U}(mathbb{Z}_{15})$.
answered Dec 6 '18 at 12:32
ΑΘΩΑΘΩ
2463
$endgroup$
add a comment |
$begingroup$
Hint: Consider the canonical projection $U(15) twoheadrightarrow U(5)$ given by $x bmod 15 mapsto x bmod 5$.
answered Dec 6 '18 at 11:46
lhflhf
163k10168392
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
$f(2)=f(7)=1$
$f(8)=f(13)=3$
Can you finish?
answered Dec 6 '18 at 11:45
ajotatxeajotatxe
53.7k23890
$endgroup$
$begingroup$
$f(4)=f(11)=f(14)=2?$
$endgroup$
– LOIS
Dec 6 '18 at 12:03
add a comment |
$begingroup$
$f(2)=f(7)=1$
$f(8)=f(13)=3$
Can you finish?
answered Dec 6 '18 at 11:45
ajotatxeajotatxe
53.7k23890
$endgroup$
$begingroup$
$f(4)=f(11)=f(14)=2?$
$endgroup$
– LOIS
Dec 6 '18 at 12:03
add a comment |
$begingroup$
$f(2)=f(7)=1$
$f(8)=f(13)=3$
Can you finish?
answered Dec 6 '18 at 11:45
ajotatxeajotatxe
53.7k23890
$endgroup$
$f(2)=f(7)=1$
$f(8)=f(13)=3$
Can you finish?
answered Dec 6 '18 at 11:45
ajotatxeajotatxe
53.7k23890
answered Dec 6 '18 at 11:45
ajotatxeajotatxe
53.7k23890
answered Dec 6 '18 at 11:45
ajotatxeajotatxe
53.7k23890
answered Dec 6 '18 at 11:45
ajotatxeajotatxe
53.7k23890
53.7k23890
$begingroup$
$f(4)=f(11)=f(14)=2?$
$endgroup$
– LOIS
Dec 6 '18 at 12:03
add a comment |
$begingroup$
$f(4)=f(11)=f(14)=2?$
$endgroup$
– LOIS
Dec 6 '18 at 12:03
$begingroup$
$f(4)=f(11)=f(14)=2?$
$endgroup$
– LOIS
Dec 6 '18 at 12:03
$begingroup$
$f(4)=f(11)=f(14)=2?$
$endgroup$
– LOIS
Dec 6 '18 at 12:03
add a comment |
$begingroup$
You have got that $U(15)congmathbb Z_4oplus mathbb Z_2$. So the existence of a surjective homomorphism is clear.
answered Dec 6 '18 at 14:59
Chris CusterChris Custer
11.4k3824
$endgroup$
add a comment |
$begingroup$
You have got that $U(15)congmathbb Z_4oplus mathbb Z_2$. So the existence of a surjective homomorphism is clear.
answered Dec 6 '18 at 14:59
Chris CusterChris Custer
11.4k3824
$endgroup$
add a comment |
$begingroup$
You have got that $U(15)congmathbb Z_4oplus mathbb Z_2$. So the existence of a surjective homomorphism is clear.
answered Dec 6 '18 at 14:59
Chris CusterChris Custer
11.4k3824
$endgroup$
You have got that $U(15)congmathbb Z_4oplus mathbb Z_2$. So the existence of a surjective homomorphism is clear.
answered Dec 6 '18 at 14:59
Chris CusterChris Custer
11.4k3824
answered Dec 6 '18 at 14:59
Chris CusterChris Custer
11.4k3824
answered Dec 6 '18 at 14:59
Chris CusterChris Custer
11.4k3824
answered Dec 6 '18 at 14:59
Chris CusterChris Custer
11.4k3824
11.4k3824
add a comment |
add a comment |
$begingroup$
Recall the Chinese remainder theorem:
let $A$ be an arbitrary ring, $Lambda neq emptyset$ a finite set (of indices) and $I in (mathcal{Id}_b(A))^{Lambda}$ a family of pairwise comaximal bilateral (two-sided) ideals of $A$, which specifically means that $$lambda, mu in Lambda, lambda neq mu implies I_{lambda}+I_{mu}=A$$ Under these conditions the following isomorphism of rings takes place:
$${A}Big{/}bigcap_{lambda in Lambda} I_{lambda} approx prod_{lambda in Lambda} (A big{/}I_{lambda})$$
As a side-note, $mathcal{Id}$ stands for ideals and the subscript $b$ for bilateral; for sets $M$ and $J$, $M^J$ denotes the cartesian product $prod limits_{j in J} M$, in other words the set of all families of elements of $M$ indexed by $J$.
Applying the general form of the theorem to the ring of integers, you have the following formulation:
let $I$ be finite and $m in mathbb{Z}^I$ such that $i, j in I, i neq j implies m_i mathbb{Z}+ m_jmathbb{Z}=mathbb{Z}$. Then $mathbb{Z}_{prodlimits_{i in I} m_i} approx prodlimits_{i in I}mathbb{Z}_{m_i}$ as rings.
In the instance you are treating, with prime factor decomposition $15=3cdot5$ the theorem yields the following isomorphism: $$mathbb{Z}_{15} approx mathbb{Z}_3 times mathbb{Z}_5$$
This yields an isomorphism between the corresponding groups of units (invertible elements) of the two rings:
$$mathrm{U}(mathbb{Z}_{15})approx mathrm{U}(mathbb{Z}_3)times mathrm{U}(mathbb{Z}_5)$$
However the quotient rings appearing as factors in the direct product on the right-hand side are actually fields and it is easily seen that $(mathrm{U}(mathbb{Z}_3), cdot) approx (mathbb{Z}_2, +)$ (cyclic of prime order, so any nontrivial element is a generator) and $(mathrm{U}(mathbb{Z}_5), cdot) approx (mathbb{Z}_4, +)$ (where either $overline{2}$ or $overline{3}$ are generators, being the only elements of order 4). The operations of the respective groups are appended for clarity. You thus eventually obtain the isomorphism:
$$mathrm{U}(mathbb{Z}_{15}) approx mathbb{Z}_2 times mathbb{Z}_4$$
exhibiting the invariant factors of $mathrm{U}(mathbb{Z}_{15})$ as a finitely generated torsion $mathbb{Z}$-module (cnf. the theory of finitely generated modules over principal ideal domains). At this point you can easily determine all the subgroups and quotient groups of $mathrm{U}(mathbb{Z}_{15})$.
answered Dec 6 '18 at 12:32
ΑΘΩΑΘΩ
2463
$endgroup$
add a comment |
$begingroup$
Recall the Chinese remainder theorem:
let $A$ be an arbitrary ring, $Lambda neq emptyset$ a finite set (of indices) and $I in (mathcal{Id}_b(A))^{Lambda}$ a family of pairwise comaximal bilateral (two-sided) ideals of $A$, which specifically means that $$lambda, mu in Lambda, lambda neq mu implies I_{lambda}+I_{mu}=A$$ Under these conditions the following isomorphism of rings takes place:
$${A}Big{/}bigcap_{lambda in Lambda} I_{lambda} approx prod_{lambda in Lambda} (A big{/}I_{lambda})$$
As a side-note, $mathcal{Id}$ stands for ideals and the subscript $b$ for bilateral; for sets $M$ and $J$, $M^J$ denotes the cartesian product $prod limits_{j in J} M$, in other words the set of all families of elements of $M$ indexed by $J$.
Applying the general form of the theorem to the ring of integers, you have the following formulation:
let $I$ be finite and $m in mathbb{Z}^I$ such that $i, j in I, i neq j implies m_i mathbb{Z}+ m_jmathbb{Z}=mathbb{Z}$. Then $mathbb{Z}_{prodlimits_{i in I} m_i} approx prodlimits_{i in I}mathbb{Z}_{m_i}$ as rings.
In the instance you are treating, with prime factor decomposition $15=3cdot5$ the theorem yields the following isomorphism: $$mathbb{Z}_{15} approx mathbb{Z}_3 times mathbb{Z}_5$$
This yields an isomorphism between the corresponding groups of units (invertible elements) of the two rings:
$$mathrm{U}(mathbb{Z}_{15})approx mathrm{U}(mathbb{Z}_3)times mathrm{U}(mathbb{Z}_5)$$
However the quotient rings appearing as factors in the direct product on the right-hand side are actually fields and it is easily seen that $(mathrm{U}(mathbb{Z}_3), cdot) approx (mathbb{Z}_2, +)$ (cyclic of prime order, so any nontrivial element is a generator) and $(mathrm{U}(mathbb{Z}_5), cdot) approx (mathbb{Z}_4, +)$ (where either $overline{2}$ or $overline{3}$ are generators, being the only elements of order 4). The operations of the respective groups are appended for clarity. You thus eventually obtain the isomorphism:
$$mathrm{U}(mathbb{Z}_{15}) approx mathbb{Z}_2 times mathbb{Z}_4$$
exhibiting the invariant factors of $mathrm{U}(mathbb{Z}_{15})$ as a finitely generated torsion $mathbb{Z}$-module (cnf. the theory of finitely generated modules over principal ideal domains). At this point you can easily determine all the subgroups and quotient groups of $mathrm{U}(mathbb{Z}_{15})$.
answered Dec 6 '18 at 12:32
ΑΘΩΑΘΩ
2463
$endgroup$
add a comment |
$begingroup$
Recall the Chinese remainder theorem:
let $A$ be an arbitrary ring, $Lambda neq emptyset$ a finite set (of indices) and $I in (mathcal{Id}_b(A))^{Lambda}$ a family of pairwise comaximal bilateral (two-sided) ideals of $A$, which specifically means that $$lambda, mu in Lambda, lambda neq mu implies I_{lambda}+I_{mu}=A$$ Under these conditions the following isomorphism of rings takes place:
$${A}Big{/}bigcap_{lambda in Lambda} I_{lambda} approx prod_{lambda in Lambda} (A big{/}I_{lambda})$$
As a side-note, $mathcal{Id}$ stands for ideals and the subscript $b$ for bilateral; for sets $M$ and $J$, $M^J$ denotes the cartesian product $prod limits_{j in J} M$, in other words the set of all families of elements of $M$ indexed by $J$.
Applying the general form of the theorem to the ring of integers, you have the following formulation:
let $I$ be finite and $m in mathbb{Z}^I$ such that $i, j in I, i neq j implies m_i mathbb{Z}+ m_jmathbb{Z}=mathbb{Z}$. Then $mathbb{Z}_{prodlimits_{i in I} m_i} approx prodlimits_{i in I}mathbb{Z}_{m_i}$ as rings.
In the instance you are treating, with prime factor decomposition $15=3cdot5$ the theorem yields the following isomorphism: $$mathbb{Z}_{15} approx mathbb{Z}_3 times mathbb{Z}_5$$
This yields an isomorphism between the corresponding groups of units (invertible elements) of the two rings:
$$mathrm{U}(mathbb{Z}_{15})approx mathrm{U}(mathbb{Z}_3)times mathrm{U}(mathbb{Z}_5)$$
However the quotient rings appearing as factors in the direct product on the right-hand side are actually fields and it is easily seen that $(mathrm{U}(mathbb{Z}_3), cdot) approx (mathbb{Z}_2, +)$ (cyclic of prime order, so any nontrivial element is a generator) and $(mathrm{U}(mathbb{Z}_5), cdot) approx (mathbb{Z}_4, +)$ (where either $overline{2}$ or $overline{3}$ are generators, being the only elements of order 4). The operations of the respective groups are appended for clarity. You thus eventually obtain the isomorphism:
$$mathrm{U}(mathbb{Z}_{15}) approx mathbb{Z}_2 times mathbb{Z}_4$$
exhibiting the invariant factors of $mathrm{U}(mathbb{Z}_{15})$ as a finitely generated torsion $mathbb{Z}$-module (cnf. the theory of finitely generated modules over principal ideal domains). At this point you can easily determine all the subgroups and quotient groups of $mathrm{U}(mathbb{Z}_{15})$.
answered Dec 6 '18 at 12:32
ΑΘΩΑΘΩ
2463
$endgroup$
Recall the Chinese remainder theorem:
let $A$ be an arbitrary ring, $Lambda neq emptyset$ a finite set (of indices) and $I in (mathcal{Id}_b(A))^{Lambda}$ a family of pairwise comaximal bilateral (two-sided) ideals of $A$, which specifically means that $$lambda, mu in Lambda, lambda neq mu implies I_{lambda}+I_{mu}=A$$ Under these conditions the following isomorphism of rings takes place:
$${A}Big{/}bigcap_{lambda in Lambda} I_{lambda} approx prod_{lambda in Lambda} (A big{/}I_{lambda})$$
As a side-note, $mathcal{Id}$ stands for ideals and the subscript $b$ for bilateral; for sets $M$ and $J$, $M^J$ denotes the cartesian product $prod limits_{j in J} M$, in other words the set of all families of elements of $M$ indexed by $J$.
Applying the general form of the theorem to the ring of integers, you have the following formulation:
let $I$ be finite and $m in mathbb{Z}^I$ such that $i, j in I, i neq j implies m_i mathbb{Z}+ m_jmathbb{Z}=mathbb{Z}$. Then $mathbb{Z}_{prodlimits_{i in I} m_i} approx prodlimits_{i in I}mathbb{Z}_{m_i}$ as rings.
In the instance you are treating, with prime factor decomposition $15=3cdot5$ the theorem yields the following isomorphism: $$mathbb{Z}_{15} approx mathbb{Z}_3 times mathbb{Z}_5$$
This yields an isomorphism between the corresponding groups of units (invertible elements) of the two rings:
$$mathrm{U}(mathbb{Z}_{15})approx mathrm{U}(mathbb{Z}_3)times mathrm{U}(mathbb{Z}_5)$$
However the quotient rings appearing as factors in the direct product on the right-hand side are actually fields and it is easily seen that $(mathrm{U}(mathbb{Z}_3), cdot) approx (mathbb{Z}_2, +)$ (cyclic of prime order, so any nontrivial element is a generator) and $(mathrm{U}(mathbb{Z}_5), cdot) approx (mathbb{Z}_4, +)$ (where either $overline{2}$ or $overline{3}$ are generators, being the only elements of order 4). The operations of the respective groups are appended for clarity. You thus eventually obtain the isomorphism:
$$mathrm{U}(mathbb{Z}_{15}) approx mathbb{Z}_2 times mathbb{Z}_4$$
exhibiting the invariant factors of $mathrm{U}(mathbb{Z}_{15})$ as a finitely generated torsion $mathbb{Z}$-module (cnf. the theory of finitely generated modules over principal ideal domains). At this point you can easily determine all the subgroups and quotient groups of $mathrm{U}(mathbb{Z}_{15})$.
answered Dec 6 '18 at 12:32
ΑΘΩΑΘΩ
2463
edited Dec 6 '18 at 16:16
answered Dec 6 '18 at 12:32
ΑΘΩΑΘΩ
2463
answered Dec 6 '18 at 12:32
ΑΘΩΑΘΩ
2463
answered Dec 6 '18 at 12:32
ΑΘΩΑΘΩ
2463
2463
add a comment |
add a comment |
$begingroup$
Hint: Consider the canonical projection $U(15) twoheadrightarrow U(5)$ given by $x bmod 15 mapsto x bmod 5$.
answered Dec 6 '18 at 11:46
lhflhf
163k10168392
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$begingroup$
Hint: Consider the canonical projection $U(15) twoheadrightarrow U(5)$ given by $x bmod 15 mapsto x bmod 5$.
answered Dec 6 '18 at 11:46
lhflhf
163k10168392
$endgroup$
add a comment |
$begingroup$
Hint: Consider the canonical projection $U(15) twoheadrightarrow U(5)$ given by $x bmod 15 mapsto x bmod 5$.
answered Dec 6 '18 at 11:46
lhflhf
163k10168392
$endgroup$
Hint: Consider the canonical projection $U(15) twoheadrightarrow U(5)$ given by $x bmod 15 mapsto x bmod 5$.
answered Dec 6 '18 at 11:46
lhflhf
163k10168392
edited Dec 7 '18 at 14:32
answered Dec 6 '18 at 11:46
lhflhf
163k10168392
answered Dec 6 '18 at 11:46
lhflhf
163k10168392
answered Dec 6 '18 at 11:46
lhflhf
163k10168392
163k10168392
add a comment |
add a comment |
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$begingroup$
@ Arthur yes,zero homomorphism. I lost the condition 'surjective'.
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– LOIS
Dec 6 '18 at 11:42
1
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You solve this by trying. If there is a surjective homomorhism, then some element has to hit $1in Bbb Z_4$. Which elements could do that? Then which elements do you know where they're mapped? Then try the rest of the elements.
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– Arthur
Dec 6 '18 at 11:42
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$1in mathbb{Z}_4$ is of order 4. hence the elements hit 1 should have order 4 .They should be 2,7,8,13. The rest except 1 should hit $3in mathbb{Z}_4$?
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– LOIS
Dec 6 '18 at 12:00
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What order does $3in Bbb Z_4$ have? If $f(2) = 1$, then what must $f(8) = f(2cdot 2cdot 2)$ be? And remember that $2$ and $0$ are in there too.
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– Arthur
Dec 6 '18 at 12:08