Homomorphism and relative order questions
1) I have to prove that if $exists$ a nontrivial homomorphism $phi:Arightarrow B$, where A and B are finite and Abelian, then $|A|$ and $|B|$ are not relatively prime.
I know that $phi(A)$ is a subgroup of $B$ and $because$ $B$ is Abelian $phi(A)triangleleft B$. So I think $phi(A) | |B|$ but I can't draw a connection between $|A|$ and $|B|$. Figured this out now. $A/Ker(phi) cong phi(A)$, by Lagrange's, $|A/Ker(phi)|$ divides $|A|$ and hence $|phi(A)|$ divides $|A|$. Therefore, $|A|$ and $|B|$ are not relatively prime.
2) Deduce that $exists$ a nontrivial homomorphism $Brightarrow A$
Do we have to prove the converse of (1) because the orders are relatively prime? Figured it out. We can use the same argument as above.
abstract-algebra group-theory finite-groups abelian-groups group-homomorphism
asked Dec 3 '18 at 20:19
manifolded
615
|
show 1 more comment
1) I have to prove that if $exists$ a nontrivial homomorphism $phi:Arightarrow B$, where A and B are finite and Abelian, then $|A|$ and $|B|$ are not relatively prime.
I know that $phi(A)$ is a subgroup of $B$ and $because$ $B$ is Abelian $phi(A)triangleleft B$. So I think $phi(A) | |B|$ but I can't draw a connection between $|A|$ and $|B|$. Figured this out now. $A/Ker(phi) cong phi(A)$, by Lagrange's, $|A/Ker(phi)|$ divides $|A|$ and hence $|phi(A)|$ divides $|A|$. Therefore, $|A|$ and $|B|$ are not relatively prime.
2) Deduce that $exists$ a nontrivial homomorphism $Brightarrow A$
Do we have to prove the converse of (1) because the orders are relatively prime? Figured it out. We can use the same argument as above.
abstract-algebra group-theory finite-groups abelian-groups group-homomorphism
asked Dec 3 '18 at 20:19
manifolded
615
(1): Show that the order of $phi(A)$ is a divisor of $|A|$.
– Arturo Magidin
Dec 3 '18 at 20:29
Made an edit, able to show for elements of A but how can I do that for the whole group A?
– manifolded
Dec 3 '18 at 20:53
You can use the isomorphism theorems and Lagrange’s Theorem.
– Arturo Magidin
Dec 3 '18 at 20:55
And, no, the fact that The order of $phi(a)$ divides the order of $a$ for each $a$ does not “lead” to $|phi(A)|$ dividing $|A|$. It’s actually the other way around.
– Arturo Magidin
Dec 3 '18 at 21:02
True, was able to show $|A/N|$ divides $|A|$ and $A/N cong phi(A)$ and hence $phi(A)$ dividies $|A|$. Thanks for the hint. How about part (2)?
– manifolded
Dec 3 '18 at 21:06
|
show 1 more comment
1) I have to prove that if $exists$ a nontrivial homomorphism $phi:Arightarrow B$, where A and B are finite and Abelian, then $|A|$ and $|B|$ are not relatively prime.
I know that $phi(A)$ is a subgroup of $B$ and $because$ $B$ is Abelian $phi(A)triangleleft B$. So I think $phi(A) | |B|$ but I can't draw a connection between $|A|$ and $|B|$. Figured this out now. $A/Ker(phi) cong phi(A)$, by Lagrange's, $|A/Ker(phi)|$ divides $|A|$ and hence $|phi(A)|$ divides $|A|$. Therefore, $|A|$ and $|B|$ are not relatively prime.
2) Deduce that $exists$ a nontrivial homomorphism $Brightarrow A$
Do we have to prove the converse of (1) because the orders are relatively prime? Figured it out. We can use the same argument as above.
abstract-algebra group-theory finite-groups abelian-groups group-homomorphism
asked Dec 3 '18 at 20:19
manifolded
615
1) I have to prove that if $exists$ a nontrivial homomorphism $phi:Arightarrow B$, where A and B are finite and Abelian, then $|A|$ and $|B|$ are not relatively prime.
I know that $phi(A)$ is a subgroup of $B$ and $because$ $B$ is Abelian $phi(A)triangleleft B$. So I think $phi(A) | |B|$ but I can't draw a connection between $|A|$ and $|B|$. Figured this out now. $A/Ker(phi) cong phi(A)$, by Lagrange's, $|A/Ker(phi)|$ divides $|A|$ and hence $|phi(A)|$ divides $|A|$. Therefore, $|A|$ and $|B|$ are not relatively prime.
2) Deduce that $exists$ a nontrivial homomorphism $Brightarrow A$
Do we have to prove the converse of (1) because the orders are relatively prime? Figured it out. We can use the same argument as above.
abstract-algebra group-theory finite-groups abelian-groups group-homomorphism
abstract-algebra group-theory finite-groups abelian-groups group-homomorphism
asked Dec 3 '18 at 20:19
manifolded
615
asked Dec 3 '18 at 20:19
manifolded
615
edited Dec 3 '18 at 21:35
asked Dec 3 '18 at 20:19
manifolded
615
asked Dec 3 '18 at 20:19
manifolded
615
asked Dec 3 '18 at 20:19
manifolded
615
615
(1): Show that the order of $phi(A)$ is a divisor of $|A|$.
– Arturo Magidin
Dec 3 '18 at 20:29
Made an edit, able to show for elements of A but how can I do that for the whole group A?
– manifolded
Dec 3 '18 at 20:53
You can use the isomorphism theorems and Lagrange’s Theorem.
– Arturo Magidin
Dec 3 '18 at 20:55
And, no, the fact that The order of $phi(a)$ divides the order of $a$ for each $a$ does not “lead” to $|phi(A)|$ dividing $|A|$. It’s actually the other way around.
– Arturo Magidin
Dec 3 '18 at 21:02
True, was able to show $|A/N|$ divides $|A|$ and $A/N cong phi(A)$ and hence $phi(A)$ dividies $|A|$. Thanks for the hint. How about part (2)?
– manifolded
Dec 3 '18 at 21:06
|
show 1 more comment
(1): Show that the order of $phi(A)$ is a divisor of $|A|$.
– Arturo Magidin
Dec 3 '18 at 20:29
Made an edit, able to show for elements of A but how can I do that for the whole group A?
– manifolded
Dec 3 '18 at 20:53
You can use the isomorphism theorems and Lagrange’s Theorem.
– Arturo Magidin
Dec 3 '18 at 20:55
And, no, the fact that The order of $phi(a)$ divides the order of $a$ for each $a$ does not “lead” to $|phi(A)|$ dividing $|A|$. It’s actually the other way around.
– Arturo Magidin
Dec 3 '18 at 21:02
True, was able to show $|A/N|$ divides $|A|$ and $A/N cong phi(A)$ and hence $phi(A)$ dividies $|A|$. Thanks for the hint. How about part (2)?
– manifolded
Dec 3 '18 at 21:06
(1): Show that the order of $phi(A)$ is a divisor of $|A|$.
– Arturo Magidin
Dec 3 '18 at 20:29
(1): Show that the order of $phi(A)$ is a divisor of $|A|$.
– Arturo Magidin
Dec 3 '18 at 20:29
Made an edit, able to show for elements of A but how can I do that for the whole group A?
– manifolded
Dec 3 '18 at 20:53
Made an edit, able to show for elements of A but how can I do that for the whole group A?
– manifolded
Dec 3 '18 at 20:53
You can use the isomorphism theorems and Lagrange’s Theorem.
– Arturo Magidin
Dec 3 '18 at 20:55
You can use the isomorphism theorems and Lagrange’s Theorem.
– Arturo Magidin
Dec 3 '18 at 20:55
And, no, the fact that The order of $phi(a)$ divides the order of $a$ for each $a$ does not “lead” to $|phi(A)|$ dividing $|A|$. It’s actually the other way around.
– Arturo Magidin
Dec 3 '18 at 21:02
And, no, the fact that The order of $phi(a)$ divides the order of $a$ for each $a$ does not “lead” to $|phi(A)|$ dividing $|A|$. It’s actually the other way around.
– Arturo Magidin
Dec 3 '18 at 21:02
True, was able to show $|A/N|$ divides $|A|$ and $A/N cong phi(A)$ and hence $phi(A)$ dividies $|A|$. Thanks for the hint. How about part (2)?
– manifolded
Dec 3 '18 at 21:06
True, was able to show $|A/N|$ divides $|A|$ and $A/N cong phi(A)$ and hence $phi(A)$ dividies $|A|$. Thanks for the hint. How about part (2)?
– manifolded
Dec 3 '18 at 21:06
|
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(1): Show that the order of $phi(A)$ is a divisor of $|A|$.
– Arturo Magidin
Dec 3 '18 at 20:29
Made an edit, able to show for elements of A but how can I do that for the whole group A?
– manifolded
Dec 3 '18 at 20:53
You can use the isomorphism theorems and Lagrange’s Theorem.
– Arturo Magidin
Dec 3 '18 at 20:55
And, no, the fact that The order of $phi(a)$ divides the order of $a$ for each $a$ does not “lead” to $|phi(A)|$ dividing $|A|$. It’s actually the other way around.
– Arturo Magidin
Dec 3 '18 at 21:02
True, was able to show $|A/N|$ divides $|A|$ and $A/N cong phi(A)$ and hence $phi(A)$ dividies $|A|$. Thanks for the hint. How about part (2)?
– manifolded
Dec 3 '18 at 21:06