Proving that if $G/Y$ is cyclic then $G/X$ and $X/Y$ are cyclic. [closed]
$begingroup$
Had an exam this week and there was the following question:
Let $X,Y$ be normal subgroups of $G$ so $Yleq X$. Prove that if $G/Y$ is cyclic then $G/X$ and $X/Y$ are cyclic.
After spending a lot of time trying to prove it I gave up. How should I prove this theorem?
abstract-algebra group-theory
asked Jan 12 at 15:39
abuka123abuka123
445
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closed as off-topic by Saad, Arnaud D., Namaste, Paul Frost, Leucippus Jan 13 at 7:31
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Arnaud D., Namaste, Paul Frost, Leucippus
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Had an exam this week and there was the following question:
Let $X,Y$ be normal subgroups of $G$ so $Yleq X$. Prove that if $G/Y$ is cyclic then $G/X$ and $X/Y$ are cyclic.
After spending a lot of time trying to prove it I gave up. How should I prove this theorem?
abstract-algebra group-theory
asked Jan 12 at 15:39
abuka123abuka123
445
$endgroup$
closed as off-topic by Saad, Arnaud D., Namaste, Paul Frost, Leucippus Jan 13 at 7:31
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Arnaud D., Namaste, Paul Frost, Leucippus
If this question can be reworded to fit the rules in the help center, please edit the question.
2
$begingroup$
You need to find a generator. First "guess" a candidate. (Note that $X/Y$ is a subgroup of $G/Y$ and $G/X$ is a quotient.)
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– Yanko
Jan 12 at 15:42
add a comment |
$begingroup$
Had an exam this week and there was the following question:
Let $X,Y$ be normal subgroups of $G$ so $Yleq X$. Prove that if $G/Y$ is cyclic then $G/X$ and $X/Y$ are cyclic.
After spending a lot of time trying to prove it I gave up. How should I prove this theorem?
abstract-algebra group-theory
asked Jan 12 at 15:39
abuka123abuka123
445
$endgroup$
Had an exam this week and there was the following question:
Let $X,Y$ be normal subgroups of $G$ so $Yleq X$. Prove that if $G/Y$ is cyclic then $G/X$ and $X/Y$ are cyclic.
After spending a lot of time trying to prove it I gave up. How should I prove this theorem?
abstract-algebra group-theory
abstract-algebra group-theory
asked Jan 12 at 15:39
abuka123abuka123
445
asked Jan 12 at 15:39
abuka123abuka123
445
asked Jan 12 at 15:39
abuka123abuka123
445
asked Jan 12 at 15:39
abuka123abuka123
445
asked Jan 12 at 15:39
abuka123abuka123
445
445
closed as off-topic by Saad, Arnaud D., Namaste, Paul Frost, Leucippus Jan 13 at 7:31
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Arnaud D., Namaste, Paul Frost, Leucippus
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Saad, Arnaud D., Namaste, Paul Frost, Leucippus Jan 13 at 7:31
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Arnaud D., Namaste, Paul Frost, Leucippus
If this question can be reworded to fit the rules in the help center, please edit the question.
2
$begingroup$
You need to find a generator. First "guess" a candidate. (Note that $X/Y$ is a subgroup of $G/Y$ and $G/X$ is a quotient.)
$endgroup$
– Yanko
Jan 12 at 15:42
add a comment |
2
$begingroup$
You need to find a generator. First "guess" a candidate. (Note that $X/Y$ is a subgroup of $G/Y$ and $G/X$ is a quotient.)
$endgroup$
– Yanko
Jan 12 at 15:42
2
2
$begingroup$
You need to find a generator. First "guess" a candidate. (Note that $X/Y$ is a subgroup of $G/Y$ and $G/X$ is a quotient.)
$endgroup$
– Yanko
Jan 12 at 15:42
$begingroup$
You need to find a generator. First "guess" a candidate. (Note that $X/Y$ is a subgroup of $G/Y$ and $G/X$ is a quotient.)
$endgroup$
– Yanko
Jan 12 at 15:42
add a comment |
1 Answer
1
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$begingroup$
Notice that $X/Y leq G/Y$ since $X leq G$, and since a subgroup of a cyclic group is cyclic, we have that $X/Y$ is cyclic.
Now, if $G/Y$ is cyclic then it is generated by an element, say $gY in G/Y$ for some $g in G$. I claim that $gX$ generates $G/X$. Indeed, if $hX$ is any element in $G/X$, we have that $g^kY = hY$ for some $k in mathbb{Z}$, that is, $g^k h^{-1} in Y$. But $Y leq X$ and so $g^kh^{-1} in X$ which means that $g^kX = hX$.
answered Jan 12 at 15:49
matt stokesmatt stokes
580310
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add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Notice that $X/Y leq G/Y$ since $X leq G$, and since a subgroup of a cyclic group is cyclic, we have that $X/Y$ is cyclic.
Now, if $G/Y$ is cyclic then it is generated by an element, say $gY in G/Y$ for some $g in G$. I claim that $gX$ generates $G/X$. Indeed, if $hX$ is any element in $G/X$, we have that $g^kY = hY$ for some $k in mathbb{Z}$, that is, $g^k h^{-1} in Y$. But $Y leq X$ and so $g^kh^{-1} in X$ which means that $g^kX = hX$.
answered Jan 12 at 15:49
matt stokesmatt stokes
580310
$endgroup$
add a comment |
$begingroup$
Notice that $X/Y leq G/Y$ since $X leq G$, and since a subgroup of a cyclic group is cyclic, we have that $X/Y$ is cyclic.
Now, if $G/Y$ is cyclic then it is generated by an element, say $gY in G/Y$ for some $g in G$. I claim that $gX$ generates $G/X$. Indeed, if $hX$ is any element in $G/X$, we have that $g^kY = hY$ for some $k in mathbb{Z}$, that is, $g^k h^{-1} in Y$. But $Y leq X$ and so $g^kh^{-1} in X$ which means that $g^kX = hX$.
answered Jan 12 at 15:49
matt stokesmatt stokes
580310
$endgroup$
add a comment |
$begingroup$
Notice that $X/Y leq G/Y$ since $X leq G$, and since a subgroup of a cyclic group is cyclic, we have that $X/Y$ is cyclic.
Now, if $G/Y$ is cyclic then it is generated by an element, say $gY in G/Y$ for some $g in G$. I claim that $gX$ generates $G/X$. Indeed, if $hX$ is any element in $G/X$, we have that $g^kY = hY$ for some $k in mathbb{Z}$, that is, $g^k h^{-1} in Y$. But $Y leq X$ and so $g^kh^{-1} in X$ which means that $g^kX = hX$.
answered Jan 12 at 15:49
matt stokesmatt stokes
580310
$endgroup$
Notice that $X/Y leq G/Y$ since $X leq G$, and since a subgroup of a cyclic group is cyclic, we have that $X/Y$ is cyclic.
Now, if $G/Y$ is cyclic then it is generated by an element, say $gY in G/Y$ for some $g in G$. I claim that $gX$ generates $G/X$. Indeed, if $hX$ is any element in $G/X$, we have that $g^kY = hY$ for some $k in mathbb{Z}$, that is, $g^k h^{-1} in Y$. But $Y leq X$ and so $g^kh^{-1} in X$ which means that $g^kX = hX$.
answered Jan 12 at 15:49
matt stokesmatt stokes
580310
edited Jan 12 at 15:54
answered Jan 12 at 15:49
matt stokesmatt stokes
580310
answered Jan 12 at 15:49
matt stokesmatt stokes
580310
answered Jan 12 at 15:49
matt stokesmatt stokes
580310
580310
add a comment |
add a comment |
2
$begingroup$
You need to find a generator. First "guess" a candidate. (Note that $X/Y$ is a subgroup of $G/Y$ and $G/X$ is a quotient.)
$endgroup$
– Yanko
Jan 12 at 15:42