Onto homomorphism from G to Z21 Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Abelian group admitting a surjective homomorphism onto an infinite cyclic groupSurjective Homomorphism to $mathbbZ$ -> pre-image has normal Subgroup of index nApplication of Correspondence TheoremInterpretation of First Isomorphism TheoremThe Fundamental Homomorphism TheoremThere is no homomorphism from $mathbbZ_8 times mathbbZ_2 times mathbbZ_2$ onto $mathbbZ_4 times mathbbZ_4$Find all groups such that there is a surjective homomorphismSubgroup Correspondence preserves indexNo homomorphism from $Z_16oplus Z_2$ onto $Z_4oplus Z_4$.$f colon G_1 to G_2$ group homomorphism, then $G_1 / N_1 cong G_2 / N_2$
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Onto homomorphism from G to Z21
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Abelian group admitting a surjective homomorphism onto an infinite cyclic groupSurjective Homomorphism to $mathbbZ$ -> pre-image has normal Subgroup of index nApplication of Correspondence TheoremInterpretation of First Isomorphism TheoremThe Fundamental Homomorphism TheoremThere is no homomorphism from $mathbbZ_8 times mathbbZ_2 times mathbbZ_2$ onto $mathbbZ_4 times mathbbZ_4$Find all groups such that there is a surjective homomorphismSubgroup Correspondence preserves indexNo homomorphism from $Z_16oplus Z_2$ onto $Z_4oplus Z_4$.$f colon G_1 to G_2$ group homomorphism, then $G_1 / N_1 cong G_2 / N_2$
$begingroup$
Let $G$ be a group such that a surjective homomorphism from $G$ to $mathbbZ/21mathbbZ$ exists. Prove that $G$ contains normal subgroups of index 3 and 7
I have seen a proof using the Correspondence theorem. However, we have no learned this theorem and cannot use it. I understand that I need to use the First Isomorphism theorem.
Thank you!
abstract-algebra group-theory
edited Apr 1 at 0:39
Santana Afton
3,0992730
asked Apr 1 at 0:30
lj_growllj_growl
627
$endgroup$
add a comment |
$begingroup$
Let $G$ be a group such that a surjective homomorphism from $G$ to $mathbbZ/21mathbbZ$ exists. Prove that $G$ contains normal subgroups of index 3 and 7
I have seen a proof using the Correspondence theorem. However, we have no learned this theorem and cannot use it. I understand that I need to use the First Isomorphism theorem.
Thank you!
abstract-algebra group-theory
edited Apr 1 at 0:39
Santana Afton
3,0992730
asked Apr 1 at 0:30
lj_growllj_growl
627
$endgroup$
add a comment |
$begingroup$
Let $G$ be a group such that a surjective homomorphism from $G$ to $mathbbZ/21mathbbZ$ exists. Prove that $G$ contains normal subgroups of index 3 and 7
I have seen a proof using the Correspondence theorem. However, we have no learned this theorem and cannot use it. I understand that I need to use the First Isomorphism theorem.
Thank you!
abstract-algebra group-theory
edited Apr 1 at 0:39
Santana Afton
3,0992730
asked Apr 1 at 0:30
lj_growllj_growl
627
$endgroup$
Let $G$ be a group such that a surjective homomorphism from $G$ to $mathbbZ/21mathbbZ$ exists. Prove that $G$ contains normal subgroups of index 3 and 7
I have seen a proof using the Correspondence theorem. However, we have no learned this theorem and cannot use it. I understand that I need to use the First Isomorphism theorem.
Thank you!
abstract-algebra group-theory
abstract-algebra group-theory
edited Apr 1 at 0:39
Santana Afton
3,0992730
asked Apr 1 at 0:30
lj_growllj_growl
627
edited Apr 1 at 0:39
Santana Afton
3,0992730
asked Apr 1 at 0:30
lj_growllj_growl
627
edited Apr 1 at 0:39
Santana Afton
3,0992730
edited Apr 1 at 0:39
Santana Afton
3,0992730
edited Apr 1 at 0:39
Santana Afton
3,0992730
3,0992730
asked Apr 1 at 0:30
lj_growllj_growl
627
asked Apr 1 at 0:30
lj_growllj_growl
627
asked Apr 1 at 0:30
lj_growllj_growl
627
627
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Hint:
Construct surjective maps:
$$beginarray ~mathbbZ/21mathbbZ to mathbbZ/3mathbbZ \ mathbbZ/21mathbbZ to mathbbZ/7mathbbZendarray$$
answered Apr 1 at 0:37
Santana AftonSantana Afton
3,0992730
$endgroup$
add a comment |
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$begingroup$
Hint:
Construct surjective maps:
$$beginarray ~mathbbZ/21mathbbZ to mathbbZ/3mathbbZ \ mathbbZ/21mathbbZ to mathbbZ/7mathbbZendarray$$
answered Apr 1 at 0:37
Santana AftonSantana Afton
3,0992730
$endgroup$
add a comment |
$begingroup$
Hint:
Construct surjective maps:
$$beginarray ~mathbbZ/21mathbbZ to mathbbZ/3mathbbZ \ mathbbZ/21mathbbZ to mathbbZ/7mathbbZendarray$$
answered Apr 1 at 0:37
Santana AftonSantana Afton
3,0992730
$endgroup$
add a comment |
$begingroup$
Hint:
Construct surjective maps:
$$beginarray ~mathbbZ/21mathbbZ to mathbbZ/3mathbbZ \ mathbbZ/21mathbbZ to mathbbZ/7mathbbZendarray$$
answered Apr 1 at 0:37
Santana AftonSantana Afton
3,0992730
$endgroup$
Hint:
Construct surjective maps:
$$beginarray ~mathbbZ/21mathbbZ to mathbbZ/3mathbbZ \ mathbbZ/21mathbbZ to mathbbZ/7mathbbZendarray$$
answered Apr 1 at 0:37
Santana AftonSantana Afton
3,0992730
answered Apr 1 at 0:37
Santana AftonSantana Afton
3,0992730
answered Apr 1 at 0:37
Santana AftonSantana Afton
3,0992730
answered Apr 1 at 0:37
Santana AftonSantana Afton
3,0992730
3,0992730
add a comment |
add a comment |
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