Mixing
Abelian Group In The Middle Of A Short Exact Sequence
$begingroup$
Let $p$ be a prime number. Determine all isomorphism classes of abelian groups $A$ that can appear as the middle term of a short exact sequence: $$0 rightarrow mathbb{Z}/(p^a) rightarrow A rightarrow mathbb{Z}/(p^b) rightarrow 0$$
Here is my attempt at this problem:
By the exactness of the sequence, $f: mathbb{Z}/(p^a) hookrightarrow A$, $g: A twoheadrightarrow mathbb{Z}/(p^b)$, and $Im(f) = Ker(g)$, so $Im(f) cong mathbb{Z}/(p^a)$, and thus by the First Isomorphism Theorem, $A/Ker(g) = A/Im(f) cong A/(mathbb{Z}/(p^a)) cong Im(g) = mathbb{Z}/(p^b)$. Since $mathbb{Z}/(p^a)$ and $mathbb{Z}/(p^b)$ are finite groups, $A$ is a finite group and hence finitely generated. Thus by Lagrange's Theorem, $|A| = |mathbb{Z}/(p^a)||mathbb{Z}/(p^b)| = p^{a+b}$. By the Fundamental Theorem of Finitely Generated Abelian Groups, $displaystyle A cong mathbb{Z}^r times prod_{i = 1}^s mathbb{Z}/(n_i)$ for some $r,s geq 0$, $n_i geq 2$, and $n_{i+1}|n_i$ for all $1 leq i leq s - 1$. Since $A$ is finite, $r = 0$ and $displaystyle |A| = p^{a+b} = prod_{i = 1}^s |mathbb{Z}/(n_i)| = prod_{i = 1}^s n_i$. Thus, the prime decomposition of each $n_i$ cannot have any primes other than $p$, so each $n_i$ is some power $m_i$ of $p$ so that $displaystyle sum_{i = 1}^s m_i = a+b$.
I want to prove that $A cong mathbb{Z}/(p^{a+b})$ but I am stuck at the last step of this proof. What we can conclude from the Fundamental Theorem is that $A$ is isomorphic to a product of cyclic groups of $p$-power order, but how can we prove that there is only one cyclic group factor in the factorization of $A$?
group-theory abelian-groups cyclic-groups exact-sequence
asked Feb 2 at 2:49
Frederic ChopinFrederic Chopin
362111
$endgroup$
|
show 2 more comments
$begingroup$
Let $p$ be a prime number. Determine all isomorphism classes of abelian groups $A$ that can appear as the middle term of a short exact sequence: $$0 rightarrow mathbb{Z}/(p^a) rightarrow A rightarrow mathbb{Z}/(p^b) rightarrow 0$$
Here is my attempt at this problem:
By the exactness of the sequence, $f: mathbb{Z}/(p^a) hookrightarrow A$, $g: A twoheadrightarrow mathbb{Z}/(p^b)$, and $Im(f) = Ker(g)$, so $Im(f) cong mathbb{Z}/(p^a)$, and thus by the First Isomorphism Theorem, $A/Ker(g) = A/Im(f) cong A/(mathbb{Z}/(p^a)) cong Im(g) = mathbb{Z}/(p^b)$. Since $mathbb{Z}/(p^a)$ and $mathbb{Z}/(p^b)$ are finite groups, $A$ is a finite group and hence finitely generated. Thus by Lagrange's Theorem, $|A| = |mathbb{Z}/(p^a)||mathbb{Z}/(p^b)| = p^{a+b}$. By the Fundamental Theorem of Finitely Generated Abelian Groups, $displaystyle A cong mathbb{Z}^r times prod_{i = 1}^s mathbb{Z}/(n_i)$ for some $r,s geq 0$, $n_i geq 2$, and $n_{i+1}|n_i$ for all $1 leq i leq s - 1$. Since $A$ is finite, $r = 0$ and $displaystyle |A| = p^{a+b} = prod_{i = 1}^s |mathbb{Z}/(n_i)| = prod_{i = 1}^s n_i$. Thus, the prime decomposition of each $n_i$ cannot have any primes other than $p$, so each $n_i$ is some power $m_i$ of $p$ so that $displaystyle sum_{i = 1}^s m_i = a+b$.
I want to prove that $A cong mathbb{Z}/(p^{a+b})$ but I am stuck at the last step of this proof. What we can conclude from the Fundamental Theorem is that $A$ is isomorphic to a product of cyclic groups of $p$-power order, but how can we prove that there is only one cyclic group factor in the factorization of $A$?
group-theory abelian-groups cyclic-groups exact-sequence
asked Feb 2 at 2:49
Frederic ChopinFrederic Chopin
362111
$endgroup$
$begingroup$
$A$ need not be cyclic. Consider, for example, the case that $p=2$ and $a=b=1$.
$endgroup$
– Andreas Blass
Feb 2 at 3:25
$begingroup$
So I guess all we can conclude is that $A$ is isomorphic to a product of cyclic groups whose orders are powers of $p$.
$endgroup$
– Frederic Chopin
Feb 2 at 5:33
1
$begingroup$
In future, please do not rely on pictures of text and instead type up or copy & paste such things here. It makes them easier to search for and, for some people, much easier to see.
$endgroup$
– Shaun
Feb 2 at 15:17
1
$begingroup$
@Shaun I edited my post so that it contains the text.
$endgroup$
– Frederic Chopin
Feb 2 at 20:14
1
$begingroup$
Since $A$ has a normal subgroup $N$ such that both $N$ and $A/N$ are cyclic, $A$ can be generated by two elements and so it can have at most two nontrivial direct factors.
$endgroup$
– Derek Holt
Feb 2 at 20:45
|
show 2 more comments
$begingroup$
Let $p$ be a prime number. Determine all isomorphism classes of abelian groups $A$ that can appear as the middle term of a short exact sequence: $$0 rightarrow mathbb{Z}/(p^a) rightarrow A rightarrow mathbb{Z}/(p^b) rightarrow 0$$
Here is my attempt at this problem:
By the exactness of the sequence, $f: mathbb{Z}/(p^a) hookrightarrow A$, $g: A twoheadrightarrow mathbb{Z}/(p^b)$, and $Im(f) = Ker(g)$, so $Im(f) cong mathbb{Z}/(p^a)$, and thus by the First Isomorphism Theorem, $A/Ker(g) = A/Im(f) cong A/(mathbb{Z}/(p^a)) cong Im(g) = mathbb{Z}/(p^b)$. Since $mathbb{Z}/(p^a)$ and $mathbb{Z}/(p^b)$ are finite groups, $A$ is a finite group and hence finitely generated. Thus by Lagrange's Theorem, $|A| = |mathbb{Z}/(p^a)||mathbb{Z}/(p^b)| = p^{a+b}$. By the Fundamental Theorem of Finitely Generated Abelian Groups, $displaystyle A cong mathbb{Z}^r times prod_{i = 1}^s mathbb{Z}/(n_i)$ for some $r,s geq 0$, $n_i geq 2$, and $n_{i+1}|n_i$ for all $1 leq i leq s - 1$. Since $A$ is finite, $r = 0$ and $displaystyle |A| = p^{a+b} = prod_{i = 1}^s |mathbb{Z}/(n_i)| = prod_{i = 1}^s n_i$. Thus, the prime decomposition of each $n_i$ cannot have any primes other than $p$, so each $n_i$ is some power $m_i$ of $p$ so that $displaystyle sum_{i = 1}^s m_i = a+b$.
I want to prove that $A cong mathbb{Z}/(p^{a+b})$ but I am stuck at the last step of this proof. What we can conclude from the Fundamental Theorem is that $A$ is isomorphic to a product of cyclic groups of $p$-power order, but how can we prove that there is only one cyclic group factor in the factorization of $A$?
group-theory abelian-groups cyclic-groups exact-sequence
asked Feb 2 at 2:49
Frederic ChopinFrederic Chopin
362111
$endgroup$
Let $p$ be a prime number. Determine all isomorphism classes of abelian groups $A$ that can appear as the middle term of a short exact sequence: $$0 rightarrow mathbb{Z}/(p^a) rightarrow A rightarrow mathbb{Z}/(p^b) rightarrow 0$$
Here is my attempt at this problem:
By the exactness of the sequence, $f: mathbb{Z}/(p^a) hookrightarrow A$, $g: A twoheadrightarrow mathbb{Z}/(p^b)$, and $Im(f) = Ker(g)$, so $Im(f) cong mathbb{Z}/(p^a)$, and thus by the First Isomorphism Theorem, $A/Ker(g) = A/Im(f) cong A/(mathbb{Z}/(p^a)) cong Im(g) = mathbb{Z}/(p^b)$. Since $mathbb{Z}/(p^a)$ and $mathbb{Z}/(p^b)$ are finite groups, $A$ is a finite group and hence finitely generated. Thus by Lagrange's Theorem, $|A| = |mathbb{Z}/(p^a)||mathbb{Z}/(p^b)| = p^{a+b}$. By the Fundamental Theorem of Finitely Generated Abelian Groups, $displaystyle A cong mathbb{Z}^r times prod_{i = 1}^s mathbb{Z}/(n_i)$ for some $r,s geq 0$, $n_i geq 2$, and $n_{i+1}|n_i$ for all $1 leq i leq s - 1$. Since $A$ is finite, $r = 0$ and $displaystyle |A| = p^{a+b} = prod_{i = 1}^s |mathbb{Z}/(n_i)| = prod_{i = 1}^s n_i$. Thus, the prime decomposition of each $n_i$ cannot have any primes other than $p$, so each $n_i$ is some power $m_i$ of $p$ so that $displaystyle sum_{i = 1}^s m_i = a+b$.
I want to prove that $A cong mathbb{Z}/(p^{a+b})$ but I am stuck at the last step of this proof. What we can conclude from the Fundamental Theorem is that $A$ is isomorphic to a product of cyclic groups of $p$-power order, but how can we prove that there is only one cyclic group factor in the factorization of $A$?
group-theory abelian-groups cyclic-groups exact-sequence
group-theory abelian-groups cyclic-groups exact-sequence
asked Feb 2 at 2:49
Frederic ChopinFrederic Chopin
362111
asked Feb 2 at 2:49
Frederic ChopinFrederic Chopin
362111
edited Feb 2 at 20:14
Frederic Chopin
asked Feb 2 at 2:49
Frederic ChopinFrederic Chopin
362111
asked Feb 2 at 2:49
Frederic ChopinFrederic Chopin
362111
asked Feb 2 at 2:49
Frederic ChopinFrederic Chopin
362111
362111
$begingroup$
$A$ need not be cyclic. Consider, for example, the case that $p=2$ and $a=b=1$.
$endgroup$
– Andreas Blass
Feb 2 at 3:25
$begingroup$
So I guess all we can conclude is that $A$ is isomorphic to a product of cyclic groups whose orders are powers of $p$.
$endgroup$
– Frederic Chopin
Feb 2 at 5:33
1
$begingroup$
In future, please do not rely on pictures of text and instead type up or copy & paste such things here. It makes them easier to search for and, for some people, much easier to see.
$endgroup$
– Shaun
Feb 2 at 15:17
1
$begingroup$
@Shaun I edited my post so that it contains the text.
$endgroup$
– Frederic Chopin
Feb 2 at 20:14
1
$begingroup$
Since $A$ has a normal subgroup $N$ such that both $N$ and $A/N$ are cyclic, $A$ can be generated by two elements and so it can have at most two nontrivial direct factors.
$endgroup$
– Derek Holt
Feb 2 at 20:45
|
show 2 more comments
$begingroup$
$A$ need not be cyclic. Consider, for example, the case that $p=2$ and $a=b=1$.
$endgroup$
– Andreas Blass
Feb 2 at 3:25
$begingroup$
So I guess all we can conclude is that $A$ is isomorphic to a product of cyclic groups whose orders are powers of $p$.
$endgroup$
– Frederic Chopin
Feb 2 at 5:33
1
$begingroup$
In future, please do not rely on pictures of text and instead type up or copy & paste such things here. It makes them easier to search for and, for some people, much easier to see.
$endgroup$
– Shaun
Feb 2 at 15:17
1
$begingroup$
@Shaun I edited my post so that it contains the text.
$endgroup$
– Frederic Chopin
Feb 2 at 20:14
1
$begingroup$
Since $A$ has a normal subgroup $N$ such that both $N$ and $A/N$ are cyclic, $A$ can be generated by two elements and so it can have at most two nontrivial direct factors.
$endgroup$
– Derek Holt
Feb 2 at 20:45
$begingroup$
$A$ need not be cyclic. Consider, for example, the case that $p=2$ and $a=b=1$.
$endgroup$
– Andreas Blass
Feb 2 at 3:25
$begingroup$
$A$ need not be cyclic. Consider, for example, the case that $p=2$ and $a=b=1$.
$endgroup$
– Andreas Blass
Feb 2 at 3:25
$begingroup$
So I guess all we can conclude is that $A$ is isomorphic to a product of cyclic groups whose orders are powers of $p$.
$endgroup$
– Frederic Chopin
Feb 2 at 5:33
$begingroup$
So I guess all we can conclude is that $A$ is isomorphic to a product of cyclic groups whose orders are powers of $p$.
$endgroup$
– Frederic Chopin
Feb 2 at 5:33
1
1
$begingroup$
In future, please do not rely on pictures of text and instead type up or copy & paste such things here. It makes them easier to search for and, for some people, much easier to see.
$endgroup$
– Shaun
Feb 2 at 15:17
$begingroup$
In future, please do not rely on pictures of text and instead type up or copy & paste such things here. It makes them easier to search for and, for some people, much easier to see.
$endgroup$
– Shaun
Feb 2 at 15:17
1
1
$begingroup$
@Shaun I edited my post so that it contains the text.
$endgroup$
– Frederic Chopin
Feb 2 at 20:14
$begingroup$
@Shaun I edited my post so that it contains the text.
$endgroup$
– Frederic Chopin
Feb 2 at 20:14
1
1
$begingroup$
Since $A$ has a normal subgroup $N$ such that both $N$ and $A/N$ are cyclic, $A$ can be generated by two elements and so it can have at most two nontrivial direct factors.
$endgroup$
– Derek Holt
Feb 2 at 20:45
$begingroup$
Since $A$ has a normal subgroup $N$ such that both $N$ and $A/N$ are cyclic, $A$ can be generated by two elements and so it can have at most two nontrivial direct factors.
$endgroup$
– Derek Holt
Feb 2 at 20:45
|
show 2 more comments
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$begingroup$
$A$ need not be cyclic. Consider, for example, the case that $p=2$ and $a=b=1$.
$endgroup$
– Andreas Blass
Feb 2 at 3:25
$begingroup$
So I guess all we can conclude is that $A$ is isomorphic to a product of cyclic groups whose orders are powers of $p$.
$endgroup$
– Frederic Chopin
Feb 2 at 5:33
1
$begingroup$
In future, please do not rely on pictures of text and instead type up or copy & paste such things here. It makes them easier to search for and, for some people, much easier to see.
$endgroup$
– Shaun
Feb 2 at 15:17
1
$begingroup$
@Shaun I edited my post so that it contains the text.
$endgroup$
– Frederic Chopin
Feb 2 at 20:14
1
$begingroup$
Since $A$ has a normal subgroup $N$ such that both $N$ and $A/N$ are cyclic, $A$ can be generated by two elements and so it can have at most two nontrivial direct factors.
$endgroup$
– Derek Holt
Feb 2 at 20:45