Mixing
Quotient Group Is Finite Implies That Group Is Finite?
$begingroup$
Let $A$ be an Abelian group and suppose $A$ has a subgroup isomorphic to $mathbb{Z}/p^amathbb{Z}$, for some prime $p$ and positive integer $a$. Suppose that $A/(mathbb{Z}/p^amathbb{Z}) cong mathbb{Z}/p^bmathbb{Z}$, for some positive integer $b$.
How can I show that $A$ is a finite group? I added these extra hypotheses for my question, but in general, if $A$ is some group and $B leq A$ is a finite normal subgroup of $A$ and $A/B cong C$ for some finite group $C$, does this imply that $A$ is finite?
group-theory finite-groups quotient-group
asked Feb 1 at 1:51
Frederic ChopinFrederic Chopin
362111
$endgroup$
add a comment |
$begingroup$
Let $A$ be an Abelian group and suppose $A$ has a subgroup isomorphic to $mathbb{Z}/p^amathbb{Z}$, for some prime $p$ and positive integer $a$. Suppose that $A/(mathbb{Z}/p^amathbb{Z}) cong mathbb{Z}/p^bmathbb{Z}$, for some positive integer $b$.
How can I show that $A$ is a finite group? I added these extra hypotheses for my question, but in general, if $A$ is some group and $B leq A$ is a finite normal subgroup of $A$ and $A/B cong C$ for some finite group $C$, does this imply that $A$ is finite?
group-theory finite-groups quotient-group
asked Feb 1 at 1:51
Frederic ChopinFrederic Chopin
362111
$endgroup$
3
$begingroup$
Certainly, with $|A|=|B||C|$.
$endgroup$
– the_fox
Feb 1 at 1:56
$begingroup$
But that assumes that $A$ is a finite group. This is what we want to show.
$endgroup$
– Frederic Chopin
Feb 1 at 1:56
3
$begingroup$
@FredericChopin No it doesn't, that formula is valid regardless of the cardinalities of the groups involved.
$endgroup$
– Noah Schweber
Feb 1 at 2:23
add a comment |
$begingroup$
Let $A$ be an Abelian group and suppose $A$ has a subgroup isomorphic to $mathbb{Z}/p^amathbb{Z}$, for some prime $p$ and positive integer $a$. Suppose that $A/(mathbb{Z}/p^amathbb{Z}) cong mathbb{Z}/p^bmathbb{Z}$, for some positive integer $b$.
How can I show that $A$ is a finite group? I added these extra hypotheses for my question, but in general, if $A$ is some group and $B leq A$ is a finite normal subgroup of $A$ and $A/B cong C$ for some finite group $C$, does this imply that $A$ is finite?
group-theory finite-groups quotient-group
asked Feb 1 at 1:51
Frederic ChopinFrederic Chopin
362111
$endgroup$
Let $A$ be an Abelian group and suppose $A$ has a subgroup isomorphic to $mathbb{Z}/p^amathbb{Z}$, for some prime $p$ and positive integer $a$. Suppose that $A/(mathbb{Z}/p^amathbb{Z}) cong mathbb{Z}/p^bmathbb{Z}$, for some positive integer $b$.
How can I show that $A$ is a finite group? I added these extra hypotheses for my question, but in general, if $A$ is some group and $B leq A$ is a finite normal subgroup of $A$ and $A/B cong C$ for some finite group $C$, does this imply that $A$ is finite?
group-theory finite-groups quotient-group
group-theory finite-groups quotient-group
asked Feb 1 at 1:51
Frederic ChopinFrederic Chopin
362111
asked Feb 1 at 1:51
Frederic ChopinFrederic Chopin
362111
asked Feb 1 at 1:51
Frederic ChopinFrederic Chopin
362111
asked Feb 1 at 1:51
Frederic ChopinFrederic Chopin
362111
asked Feb 1 at 1:51
Frederic ChopinFrederic Chopin
362111
362111
3
$begingroup$
Certainly, with $|A|=|B||C|$.
$endgroup$
– the_fox
Feb 1 at 1:56
$begingroup$
But that assumes that $A$ is a finite group. This is what we want to show.
$endgroup$
– Frederic Chopin
Feb 1 at 1:56
3
$begingroup$
@FredericChopin No it doesn't, that formula is valid regardless of the cardinalities of the groups involved.
$endgroup$
– Noah Schweber
Feb 1 at 2:23
add a comment |
3
$begingroup$
Certainly, with $|A|=|B||C|$.
$endgroup$
– the_fox
Feb 1 at 1:56
$begingroup$
But that assumes that $A$ is a finite group. This is what we want to show.
$endgroup$
– Frederic Chopin
Feb 1 at 1:56
3
$begingroup$
@FredericChopin No it doesn't, that formula is valid regardless of the cardinalities of the groups involved.
$endgroup$
– Noah Schweber
Feb 1 at 2:23
3
3
$begingroup$
Certainly, with $|A|=|B||C|$.
$endgroup$
– the_fox
Feb 1 at 1:56
$begingroup$
Certainly, with $|A|=|B||C|$.
$endgroup$
– the_fox
Feb 1 at 1:56
$begingroup$
But that assumes that $A$ is a finite group. This is what we want to show.
$endgroup$
– Frederic Chopin
Feb 1 at 1:56
$begingroup$
But that assumes that $A$ is a finite group. This is what we want to show.
$endgroup$
– Frederic Chopin
Feb 1 at 1:56
3
3
$begingroup$
@FredericChopin No it doesn't, that formula is valid regardless of the cardinalities of the groups involved.
$endgroup$
– Noah Schweber
Feb 1 at 2:23
$begingroup$
@FredericChopin No it doesn't, that formula is valid regardless of the cardinalities of the groups involved.
$endgroup$
– Noah Schweber
Feb 1 at 2:23
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Let $A$ be our general group, and let $B$ be a finite, normal subgroup of $A$. Let’s say that $|B| = k$. Then the following things are true:
For any $xin A$, the coset $xB$ satisfies $|xB| = |B| = k$.
If $x,yin A$, then $xAcap yA = emptyset$ if and only if $xnotin yA$. That is, the cosets form a partition of $A$.
The size of $A/B$ is (by definition) the number of distinct cosets of $B$ in $A$.
If $A/B$ has $m$ elements, that means that $A$ can be partitioned into $m$ disjoint sets, each of which has $k$ elements. Thus $|A| = mk$.
answered Feb 1 at 2:33
Santana AftonSantana Afton
3,0992730
$endgroup$
$begingroup$
These were very helpful hints, thank you. I was able to prove the facts that you claimed were true.
$endgroup$
– Frederic Chopin
Feb 1 at 3:13
add a comment |
$begingroup$
The "slick" way is Lagrange's theorem, which says tells you that the number of elements of $A$ is equal to the number of elements of $B$ times the number of elements of $A/B$. This statement has a rigorous formulation independent of whether the groups are finite or infinite, and it implies that if two of three sets $A, B, A/B$ are finite, then so is the third.
Another way (basically the same thing): let $n$ be the number of elements of $B$. Take an element $a in A$, and consider its image $overline{a}$ in the finite group $A/B = C$. Show that there are exactly $n-1$ other elements of $A$ having the same image. Since there are only finitely many possibilities for $overline{a}$, this shows that there are only finitely many elements of $A$.
answered Feb 1 at 1:59
D_SD_S
14.2k61754
$endgroup$
add a comment |
$begingroup$
Yes. Using the first isomorphism theorem, $mid Cmid=frac{mid Amid}{mid Bmid}$. Just consider the quotient map.
answered Feb 1 at 1:58
Chris CusterChris Custer
14.3k3827
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $A$ be our general group, and let $B$ be a finite, normal subgroup of $A$. Let’s say that $|B| = k$. Then the following things are true:
For any $xin A$, the coset $xB$ satisfies $|xB| = |B| = k$.
If $x,yin A$, then $xAcap yA = emptyset$ if and only if $xnotin yA$. That is, the cosets form a partition of $A$.
The size of $A/B$ is (by definition) the number of distinct cosets of $B$ in $A$.
If $A/B$ has $m$ elements, that means that $A$ can be partitioned into $m$ disjoint sets, each of which has $k$ elements. Thus $|A| = mk$.
answered Feb 1 at 2:33
Santana AftonSantana Afton
3,0992730
$endgroup$
$begingroup$
These were very helpful hints, thank you. I was able to prove the facts that you claimed were true.
$endgroup$
– Frederic Chopin
Feb 1 at 3:13
add a comment |
$begingroup$
Let $A$ be our general group, and let $B$ be a finite, normal subgroup of $A$. Let’s say that $|B| = k$. Then the following things are true:
For any $xin A$, the coset $xB$ satisfies $|xB| = |B| = k$.
If $x,yin A$, then $xAcap yA = emptyset$ if and only if $xnotin yA$. That is, the cosets form a partition of $A$.
The size of $A/B$ is (by definition) the number of distinct cosets of $B$ in $A$.
If $A/B$ has $m$ elements, that means that $A$ can be partitioned into $m$ disjoint sets, each of which has $k$ elements. Thus $|A| = mk$.
answered Feb 1 at 2:33
Santana AftonSantana Afton
3,0992730
$endgroup$
$begingroup$
These were very helpful hints, thank you. I was able to prove the facts that you claimed were true.
$endgroup$
– Frederic Chopin
Feb 1 at 3:13
add a comment |
$begingroup$
Let $A$ be our general group, and let $B$ be a finite, normal subgroup of $A$. Let’s say that $|B| = k$. Then the following things are true:
For any $xin A$, the coset $xB$ satisfies $|xB| = |B| = k$.
If $x,yin A$, then $xAcap yA = emptyset$ if and only if $xnotin yA$. That is, the cosets form a partition of $A$.
The size of $A/B$ is (by definition) the number of distinct cosets of $B$ in $A$.
If $A/B$ has $m$ elements, that means that $A$ can be partitioned into $m$ disjoint sets, each of which has $k$ elements. Thus $|A| = mk$.
answered Feb 1 at 2:33
Santana AftonSantana Afton
3,0992730
$endgroup$
Let $A$ be our general group, and let $B$ be a finite, normal subgroup of $A$. Let’s say that $|B| = k$. Then the following things are true:
For any $xin A$, the coset $xB$ satisfies $|xB| = |B| = k$.
If $x,yin A$, then $xAcap yA = emptyset$ if and only if $xnotin yA$. That is, the cosets form a partition of $A$.
The size of $A/B$ is (by definition) the number of distinct cosets of $B$ in $A$.
If $A/B$ has $m$ elements, that means that $A$ can be partitioned into $m$ disjoint sets, each of which has $k$ elements. Thus $|A| = mk$.
answered Feb 1 at 2:33
Santana AftonSantana Afton
3,0992730
answered Feb 1 at 2:33
Santana AftonSantana Afton
3,0992730
answered Feb 1 at 2:33
Santana AftonSantana Afton
3,0992730
answered Feb 1 at 2:33
Santana AftonSantana Afton
3,0992730
3,0992730
$begingroup$
These were very helpful hints, thank you. I was able to prove the facts that you claimed were true.
$endgroup$
– Frederic Chopin
Feb 1 at 3:13
add a comment |
$begingroup$
These were very helpful hints, thank you. I was able to prove the facts that you claimed were true.
$endgroup$
– Frederic Chopin
Feb 1 at 3:13
$begingroup$
These were very helpful hints, thank you. I was able to prove the facts that you claimed were true.
$endgroup$
– Frederic Chopin
Feb 1 at 3:13
$begingroup$
These were very helpful hints, thank you. I was able to prove the facts that you claimed were true.
$endgroup$
– Frederic Chopin
Feb 1 at 3:13
add a comment |
$begingroup$
The "slick" way is Lagrange's theorem, which says tells you that the number of elements of $A$ is equal to the number of elements of $B$ times the number of elements of $A/B$. This statement has a rigorous formulation independent of whether the groups are finite or infinite, and it implies that if two of three sets $A, B, A/B$ are finite, then so is the third.
Another way (basically the same thing): let $n$ be the number of elements of $B$. Take an element $a in A$, and consider its image $overline{a}$ in the finite group $A/B = C$. Show that there are exactly $n-1$ other elements of $A$ having the same image. Since there are only finitely many possibilities for $overline{a}$, this shows that there are only finitely many elements of $A$.
answered Feb 1 at 1:59
D_SD_S
14.2k61754
$endgroup$
add a comment |
$begingroup$
The "slick" way is Lagrange's theorem, which says tells you that the number of elements of $A$ is equal to the number of elements of $B$ times the number of elements of $A/B$. This statement has a rigorous formulation independent of whether the groups are finite or infinite, and it implies that if two of three sets $A, B, A/B$ are finite, then so is the third.
Another way (basically the same thing): let $n$ be the number of elements of $B$. Take an element $a in A$, and consider its image $overline{a}$ in the finite group $A/B = C$. Show that there are exactly $n-1$ other elements of $A$ having the same image. Since there are only finitely many possibilities for $overline{a}$, this shows that there are only finitely many elements of $A$.
answered Feb 1 at 1:59
D_SD_S
14.2k61754
$endgroup$
add a comment |
$begingroup$
The "slick" way is Lagrange's theorem, which says tells you that the number of elements of $A$ is equal to the number of elements of $B$ times the number of elements of $A/B$. This statement has a rigorous formulation independent of whether the groups are finite or infinite, and it implies that if two of three sets $A, B, A/B$ are finite, then so is the third.
Another way (basically the same thing): let $n$ be the number of elements of $B$. Take an element $a in A$, and consider its image $overline{a}$ in the finite group $A/B = C$. Show that there are exactly $n-1$ other elements of $A$ having the same image. Since there are only finitely many possibilities for $overline{a}$, this shows that there are only finitely many elements of $A$.
answered Feb 1 at 1:59
D_SD_S
14.2k61754
$endgroup$
The "slick" way is Lagrange's theorem, which says tells you that the number of elements of $A$ is equal to the number of elements of $B$ times the number of elements of $A/B$. This statement has a rigorous formulation independent of whether the groups are finite or infinite, and it implies that if two of three sets $A, B, A/B$ are finite, then so is the third.
Another way (basically the same thing): let $n$ be the number of elements of $B$. Take an element $a in A$, and consider its image $overline{a}$ in the finite group $A/B = C$. Show that there are exactly $n-1$ other elements of $A$ having the same image. Since there are only finitely many possibilities for $overline{a}$, this shows that there are only finitely many elements of $A$.
answered Feb 1 at 1:59
D_SD_S
14.2k61754
answered Feb 1 at 1:59
D_SD_S
14.2k61754
answered Feb 1 at 1:59
D_SD_S
14.2k61754
answered Feb 1 at 1:59
D_SD_S
14.2k61754
14.2k61754
add a comment |
add a comment |
$begingroup$
Yes. Using the first isomorphism theorem, $mid Cmid=frac{mid Amid}{mid Bmid}$. Just consider the quotient map.
answered Feb 1 at 1:58
Chris CusterChris Custer
14.3k3827
$endgroup$
add a comment |
$begingroup$
Yes. Using the first isomorphism theorem, $mid Cmid=frac{mid Amid}{mid Bmid}$. Just consider the quotient map.
answered Feb 1 at 1:58
Chris CusterChris Custer
14.3k3827
$endgroup$
add a comment |
$begingroup$
Yes. Using the first isomorphism theorem, $mid Cmid=frac{mid Amid}{mid Bmid}$. Just consider the quotient map.
answered Feb 1 at 1:58
Chris CusterChris Custer
14.3k3827
$endgroup$
Yes. Using the first isomorphism theorem, $mid Cmid=frac{mid Amid}{mid Bmid}$. Just consider the quotient map.
answered Feb 1 at 1:58
Chris CusterChris Custer
14.3k3827
answered Feb 1 at 1:58
Chris CusterChris Custer
14.3k3827
answered Feb 1 at 1:58
Chris CusterChris Custer
14.3k3827
answered Feb 1 at 1:58
Chris CusterChris Custer
14.3k3827
14.3k3827
add a comment |
add a comment |
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3
$begingroup$
Certainly, with $|A|=|B||C|$.
$endgroup$
– the_fox
Feb 1 at 1:56
$begingroup$
But that assumes that $A$ is a finite group. This is what we want to show.
$endgroup$
– Frederic Chopin
Feb 1 at 1:56
3
$begingroup$
@FredericChopin No it doesn't, that formula is valid regardless of the cardinalities of the groups involved.
$endgroup$
– Noah Schweber
Feb 1 at 2:23