Let $phi$,$psi : K_1 to K_2$ be surjective homomorphisms between $2$ cyclic groups. Then $psi=phi circ...
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Past year paper question
I have managed to do the part (a). For part (b) I am having trouble with (ii) and (iii).
Attempt
Pick some generator $g$ of $K_1$. Suppose $psi(g)=h$. Then $h$ must be a generator of $K_2$. We also have $psi(g^{y|K_2|})=e$. So for all $y in mathbb{Z}$, $psi(g^{1+y|K_2|})=h$.
For part (i):
We consider $phi(g)$. We must have $phi(g)=h'$ where $h'$ is also a generator of $|K_2|$. So we can find some integer $m$ such that $(h')^m=h$. $m$ must be coprime to $|K_2|$ also, since for a cyclic group generated by $g$, the generators are $g^k$ where $k$ coprime $ord(g)$. This solves part (i).
(For part (ii), I am not sure if the $g$ is $psi(g)$ refers to any particular $g$. Since I already used $g$ to denote a generator of $K_1$, I will substitute the $g$ in part (ii) with $x$)
So let $x in K_1$. Then $x=g^r$. Then $psi(x)=psi(g^r)=psi(g)^r=phi(g)^{rm}=phi(g)^{rm + y|K_2|}$. So I think the $b$ that I am supposed to find is $rm$. Is this correct? I also need to verify that $b$ coprime to $|K_2|$. But I am not sure how to see this, so I'm not sure if my $b$ is correct.
For part (iii), my hunch is that $theta$ sends $x$ to $x^b$. ($b$ being the $b$ in part (ii)) So then for the generator $g$ of $K_1$, we have: $(phi circ theta)(g)=phi(g^b)=psi(g)$. And since they agree on a generator, they agree on all of $K_1$. Is this correct? I didn't even use part (a)(i), so I'm not sure.
abstract-algebra group-theory
asked Nov 23 at 11:05
eatfood
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Past year paper question
I have managed to do the part (a). For part (b) I am having trouble with (ii) and (iii).
Attempt
Pick some generator $g$ of $K_1$. Suppose $psi(g)=h$. Then $h$ must be a generator of $K_2$. We also have $psi(g^{y|K_2|})=e$. So for all $y in mathbb{Z}$, $psi(g^{1+y|K_2|})=h$.
For part (i):
We consider $phi(g)$. We must have $phi(g)=h'$ where $h'$ is also a generator of $|K_2|$. So we can find some integer $m$ such that $(h')^m=h$. $m$ must be coprime to $|K_2|$ also, since for a cyclic group generated by $g$, the generators are $g^k$ where $k$ coprime $ord(g)$. This solves part (i).
(For part (ii), I am not sure if the $g$ is $psi(g)$ refers to any particular $g$. Since I already used $g$ to denote a generator of $K_1$, I will substitute the $g$ in part (ii) with $x$)
So let $x in K_1$. Then $x=g^r$. Then $psi(x)=psi(g^r)=psi(g)^r=phi(g)^{rm}=phi(g)^{rm + y|K_2|}$. So I think the $b$ that I am supposed to find is $rm$. Is this correct? I also need to verify that $b$ coprime to $|K_2|$. But I am not sure how to see this, so I'm not sure if my $b$ is correct.
For part (iii), my hunch is that $theta$ sends $x$ to $x^b$. ($b$ being the $b$ in part (ii)) So then for the generator $g$ of $K_1$, we have: $(phi circ theta)(g)=phi(g^b)=psi(g)$. And since they agree on a generator, they agree on all of $K_1$. Is this correct? I didn't even use part (a)(i), so I'm not sure.
abstract-algebra group-theory
asked Nov 23 at 11:05
eatfood
1827
add a comment |
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0
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up vote
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Past year paper question
I have managed to do the part (a). For part (b) I am having trouble with (ii) and (iii).
Attempt
Pick some generator $g$ of $K_1$. Suppose $psi(g)=h$. Then $h$ must be a generator of $K_2$. We also have $psi(g^{y|K_2|})=e$. So for all $y in mathbb{Z}$, $psi(g^{1+y|K_2|})=h$.
For part (i):
We consider $phi(g)$. We must have $phi(g)=h'$ where $h'$ is also a generator of $|K_2|$. So we can find some integer $m$ such that $(h')^m=h$. $m$ must be coprime to $|K_2|$ also, since for a cyclic group generated by $g$, the generators are $g^k$ where $k$ coprime $ord(g)$. This solves part (i).
(For part (ii), I am not sure if the $g$ is $psi(g)$ refers to any particular $g$. Since I already used $g$ to denote a generator of $K_1$, I will substitute the $g$ in part (ii) with $x$)
So let $x in K_1$. Then $x=g^r$. Then $psi(x)=psi(g^r)=psi(g)^r=phi(g)^{rm}=phi(g)^{rm + y|K_2|}$. So I think the $b$ that I am supposed to find is $rm$. Is this correct? I also need to verify that $b$ coprime to $|K_2|$. But I am not sure how to see this, so I'm not sure if my $b$ is correct.
For part (iii), my hunch is that $theta$ sends $x$ to $x^b$. ($b$ being the $b$ in part (ii)) So then for the generator $g$ of $K_1$, we have: $(phi circ theta)(g)=phi(g^b)=psi(g)$. And since they agree on a generator, they agree on all of $K_1$. Is this correct? I didn't even use part (a)(i), so I'm not sure.
abstract-algebra group-theory
asked Nov 23 at 11:05
eatfood
1827
Past year paper question
I have managed to do the part (a). For part (b) I am having trouble with (ii) and (iii).
Attempt
Pick some generator $g$ of $K_1$. Suppose $psi(g)=h$. Then $h$ must be a generator of $K_2$. We also have $psi(g^{y|K_2|})=e$. So for all $y in mathbb{Z}$, $psi(g^{1+y|K_2|})=h$.
For part (i):
We consider $phi(g)$. We must have $phi(g)=h'$ where $h'$ is also a generator of $|K_2|$. So we can find some integer $m$ such that $(h')^m=h$. $m$ must be coprime to $|K_2|$ also, since for a cyclic group generated by $g$, the generators are $g^k$ where $k$ coprime $ord(g)$. This solves part (i).
(For part (ii), I am not sure if the $g$ is $psi(g)$ refers to any particular $g$. Since I already used $g$ to denote a generator of $K_1$, I will substitute the $g$ in part (ii) with $x$)
So let $x in K_1$. Then $x=g^r$. Then $psi(x)=psi(g^r)=psi(g)^r=phi(g)^{rm}=phi(g)^{rm + y|K_2|}$. So I think the $b$ that I am supposed to find is $rm$. Is this correct? I also need to verify that $b$ coprime to $|K_2|$. But I am not sure how to see this, so I'm not sure if my $b$ is correct.
For part (iii), my hunch is that $theta$ sends $x$ to $x^b$. ($b$ being the $b$ in part (ii)) So then for the generator $g$ of $K_1$, we have: $(phi circ theta)(g)=phi(g^b)=psi(g)$. And since they agree on a generator, they agree on all of $K_1$. Is this correct? I didn't even use part (a)(i), so I'm not sure.
abstract-algebra group-theory
abstract-algebra group-theory
asked Nov 23 at 11:05
eatfood
1827
asked Nov 23 at 11:05
eatfood
1827
asked Nov 23 at 11:05
eatfood
1827
asked Nov 23 at 11:05
eatfood
1827
asked Nov 23 at 11:05
eatfood
1827
1827
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