Difference between revisions of "Classification of fully characteristic subgroups in finitely generated Abelian groups"

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(New page: ==Statement== ===Case of groups of prime power order=== Suppose <math>G</math> is an Abelian group whose order is a power of a prime <math>p</math>. Then, we can write, by the [[structur...)
 
(Case of prime power order)
 
(2 intermediate revisions by the same user not shown)
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where <math>k_1 \le k_2 \le \dots \le k_r</math>.
 
where <math>k_1 \le k_2 \le \dots \le k_r</math>.
  
Then a subgroup <math>H</math> of <math>G</math> is a [[characteristic subgroup]] of <math>G</math> if and only if <math>H</math> is a [[fully characteristic subgroup]] of <math>G</math>, and this happens if and only if the following are satisfied:
+
Then a subgroup <math>H</math> of <math>G</math> is a [[fully characteristic subgroup]] of <math>G</math>, and this happens if and only if the following are satisfied:
  
 
* <math>H</math> is the direct sum of the intersections <math>H \cap G_i</math>.
 
* <math>H</math> is the direct sum of the intersections <math>H \cap G_i</math>.
 
* If the orders of <math>H \cap G_i</math> are <math>p^{l_i}</math>, then <math>l_1 \le l_2 \le \dots \le l_r</math> and <math>k_1 - l_1 \le k_2 - l_2 \le \dots \le k_r - l_r</math>.
 
* If the orders of <math>H \cap G_i</math> are <math>p^{l_i}</math>, then <math>l_1 \le l_2 \le \dots \le l_r</math> and <math>k_1 - l_1 \le k_2 - l_2 \le \dots \le k_r - l_r</math>.
  
===Case of finite groups===
+
===Case of finitely generated groups===
  
Suppose <math>G</math> is a [[finite Abelian group]]. Then, we can write:
+
Suppose <math>G</math> is a [[finitely generated Abelian group]]. Then, we can write:
  
<math>G = \bigoplus_{i=1}^r G_i</math>
+
<math>G = \bigoplus_{i=0}^r G_i</math>
  
where each <math>G_i</math> is an [[Abelian group of prime power order]] for different primes. Then, a subgroup <math>H</math> of <math>G</math> is a [[characteristic subgroup]] if and only if it is a [[fully characteristic subgroup]], if and only if the following are satisfied:
+
where <math>G_0</math> is a torsion-free Abelian group, and each <math>G_i</math> is an [[Abelian group of prime power order]] for different primes. Then, a subgroup <math>H</math> of <math>G</math> is a [[fully characteristic subgroup]], if and only if the following are satisfied:
  
 
* <math>H</math> is the direct sum of the intersections <math>H \cap G_i</math>.
 
* <math>H</math> is the direct sum of the intersections <math>H \cap G_i</math>.
 
* Each <math>H \cap G_i</math> is characteristic (equivalently, fully characteristic) in <math>G_i</math>.
 
* Each <math>H \cap G_i</math> is characteristic (equivalently, fully characteristic) in <math>G_i</math>.
 
===Case of finitely generated groups===
 
 
Suppose <math>G</math> is a finitely generated Abelian group. Then, we can write:
 
 
<math>G = A \oplus B</math>
 
 
where <math>A</math> is a torsion-free group. A subgroup <math>H</math> of <math>G</math> is a [[fully characteristic subgroup]] of <math>G</math> if and only if <math>H = (H \cap A) \oplus (H \cap B)</math> and we have that <math>H \cap A</math> is fully characteristic in <math>A</math> and <math>H \cap B</math> is characteristic in <math>B</math>.
 
  
 
==Proof==
 
==Proof==
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'''To prove''': <math>H</math> is characteristic if and only if it is fully characteristic in <math>G</math>, if and only if <math>H</math> is the direct sum of <math>H \cap G_i</math>, and if the orders of the intersections are <math>p^{l_i}</math>, then <math>l_1 \le l_2 \le \dots \le l_r</math> and <math>k_1 - l_1 \le k_2 - l_2 \le \dots \le k_r - l_r</math>.
 
'''To prove''': <math>H</math> is characteristic if and only if it is fully characteristic in <math>G</math>, if and only if <math>H</math> is the direct sum of <math>H \cap G_i</math>, and if the orders of the intersections are <math>p^{l_i}</math>, then <math>l_1 \le l_2 \le \dots \le l_r</math> and <math>k_1 - l_1 \le k_2 - l_2 \le \dots \le k_r - l_r</math>.
  
'''Proof''': The proof relies on two important homomorphisms. For <math>a < b</math>, there is an injective homomorphism:
+
'''Proof''': The proof relies on two important homomorphisms. For <math>a le b</math>, there is an injective homomorphism:
  
 
<math>\nu_{a,b}: \mathbb{Z}/p^a\mathbb{Z} \to \mathbb{Z}/p^b\mathbb{Z}</math>
 
<math>\nu_{a,b}: \mathbb{Z}/p^a\mathbb{Z} \to \mathbb{Z}/p^b\mathbb{Z}</math>
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and is zero elsewhere.
 
and is zero elsewhere.
  
Define <math>\sigma_{i,j}</math> as the sum of the endomorphism <math>\alpha_{i,j}</math> and the identity map on <math>G</math>. Clearly, <math>\sigma_{i,j}</math> is an automorphism.
+
Similarly, define <math>\beta_{j,i}</math> as the endomorphism of <math>G</math> that sends <math>G_j</math> to <math>G_i</math> via the map <math>\pi_{p^{k_j},p^{k_i}}</math> and is zero elsewhere.
 
 
Similarly, define <math>\beta_{j,i}</math> as the endomorphism of <math>G</math> that sends <math>G_j</math> to <math>G_i</math> via the map <math>\pi_{p^{k_j},p^{k_i}}</math> and is zero elsewhere. Define <math>\varphi_{j,i}</math> as the automorphism obtained as the sum of <math>\beta_{j,i}</math> and the identity map.
 
  
 
We are now in a position to prove the main result.
 
We are now in a position to prove the main result.
  
Suppose <math>H</math> is a characteristic subgroup of <math>G</math>. We first show that <math>H</math> is a direct sum of <math>H \cap G_i</math>. For this, suppose the element:
+
Suppose <math>H</math> is a fully characteristic subgroup of <math>G</math>. We first show that <math>H</math> is a direct sum of <math>H \cap G_i</math>. For this, suppose the element:
  
 
<math>(g_1,g_2, \dots, g_r) \in H</math>.
 
<math>(g_1,g_2, \dots, g_r) \in H</math>.
 +
 +
Since <math>H</math> is fully characteristic, it is invariant under the projections to the direct factors, which are endomorphisms. Thus, each <math>(0,0, \dots, 0, g_i,0, \dots, 0)</math> is in <math>H</math>. Thus, every element of <math>H</math> can be expressed as a sum of elements in <math>H \cap G_i</math>, and we get that <math>H</math> is a direct sum of the <math>H \cap G_i</math>s.
 +
 +
Finally, we need to show the conditions on the <math>l_i</math>s. For this, observe that for <math>i < j</math>, we have:
 +
 +
* The endomorphism <math>\alpha_{i,j}</math> sends <math>H</math> to itself: Thus, <math>H \cap G_i</math> injects into <math>H \cap G_j</math> via <math>\alpha_{i,j}</math>, so <math>l_i \le l_j</math>.
 +
* The endomorphism <math>\beta_{j,i}</math> sends <math>H</math> to itself: Thus, <math>\beta_{j,i}</math> induces a surjective endomorphism from <math>G_j/(H \cap G_j)</math> to <math>G_i/(H \cap G_i)</math>, forcing <math>k_i - l_i \le k_j - l_j</math>.
 +
 +
Now, we show that if <math>H</math> satisfies the conditions described above, then <math>H</math> is fully characteristic in <math>G</math>.
 +
 +
Suppose <math>\rho:G \to G</math> is an endomorphism. Since <math>H</math> is a direct sum of <math>H \cap G_i</math>, it suffices to show that <math>\rho(H \cap G_i) \le H</math>. For this, in turn, it suffices to show that the <math>j^{th}</math> coordinate of <math>\rho(H \cap G_i)</math> is contained in <math>H \cap G_j</math>. This is easily done in three cases:
 +
 +
* <math>j = i</math>: In this case, it is direct since <math>H \cap G_i</math> is a fully characteristic subgroup of the cyclic group <math>G_i</math> (all subgroups of cyclic groups are fully characteristic).
 +
* <math>j > i</math>: In this case, <math>k_i \le k_j</math>. Consider the homomorphism from <math>H \cap G_i</math> to <math>G_j</math> obtained by composing the <math>j^{th}</math> projection with <math>\rho</math>. This map must send <math>H \cap G_i</math> to a subgroup of size at most <math>p^{l_i}</math> in <math>G_j</math>. But since <math>l_i \le l_j</math>, this subgroup of size <math>p^{l_i}</math> is contained in the subgroup <math>H \cap G_j</math> that has order <math>p^j</math>.
 +
* <math>j < i</math>: In this case, <math>k_j \le k_i</math>. COnsider the homomorphism from <math>G_i</math> to <math>G_j</math> obtained by composing the <math>j^{th}</math> projection with <math>\rho</math>. The index of the image of <math>H \cap G_i</math> is at least the index of <math>H \cap G_i</math> in <math>G_i</math>, which is <math>p^{k_i - l_i}</math>. Thus, the image has size at most <math>p^{k_j - (k_i - l_i)} \le p^{l_j}</math> because <math>k_j - l_j \le k_i - l_i</math>. Thus, it is in <math>H \cap G_j</math>.
 +
 +
===General case===
 +
 +
This is direct. {{fillin}}

Latest revision as of 21:57, 4 January 2009

Statement

Case of groups of prime power order

Suppose G is an Abelian group whose order is a power of a prime p. Then, we can write, by the structure theorem for finitely generated Abelian groups:

G = \bigoplus_{i=1}^r G_i

where:

G_i \cong \mathbb{Z}/p^{k_i}\mathbb{Z}.

where k_1 \le k_2 \le \dots \le k_r.

Then a subgroup H of G is a fully characteristic subgroup of G, and this happens if and only if the following are satisfied:

  • H is the direct sum of the intersections H \cap G_i.
  • If the orders of H \cap G_i are p^{l_i}, then l_1 \le l_2 \le \dots \le l_r and k_1 - l_1 \le k_2 - l_2 \le \dots \le k_r - l_r.

Case of finitely generated groups

Suppose G is a finitely generated Abelian group. Then, we can write:

G = \bigoplus_{i=0}^r G_i

where G_0 is a torsion-free Abelian group, and each G_i is an Abelian group of prime power order for different primes. Then, a subgroup H of G is a fully characteristic subgroup, if and only if the following are satisfied:

  • H is the direct sum of the intersections H \cap G_i.
  • Each H \cap G_i is characteristic (equivalently, fully characteristic) in G_i.

Proof

Case of prime power order

Given: G is an Abelian group whose order is a power of a prime p. Then, we can write, by the structure theorem for finitely generated Abelian groups:

G = \bigoplus_{i=1}^r G_i

where:

G_i \cong \mathbb{Z}/p^{k_i}\mathbb{Z}.

where k_1 \le k_2 \le \dots \le k_r.

A subgroup H of G.

To prove: H is characteristic if and only if it is fully characteristic in G, if and only if H is the direct sum of H \cap G_i, and if the orders of the intersections are p^{l_i}, then l_1 \le l_2 \le \dots \le l_r and k_1 - l_1 \le k_2 - l_2 \le \dots \le k_r - l_r.

Proof: The proof relies on two important homomorphisms. For a le b, there is an injective homomorphism:

\nu_{a,b}: \mathbb{Z}/p^a\mathbb{Z} \to \mathbb{Z}/p^b\mathbb{Z}

that sends the generator on the left to p^{b-a} times the generator on the right.

There is also a surjective homomorphism:

\pi_{b,a}: \mathbb{Z}/p^b\mathbb{Z} \to \mathbb{Z}/p^a\mathbb{Z}

that sends the generator to the generator.

Recall that we have:

G = \bigoplus_{i=1}^r G_i.

For 1 \le i < j \le r, define \alpha_{i,j} as the endomorphism of G that sends G_i to G_j via the map:

\nu_{p^{k_i},p^{k_j}}: G_i \to G_j.

and is zero elsewhere.

Similarly, define \beta_{j,i} as the endomorphism of G that sends G_j to G_i via the map \pi_{p^{k_j},p^{k_i}} and is zero elsewhere.

We are now in a position to prove the main result.

Suppose H is a fully characteristic subgroup of G. We first show that H is a direct sum of H \cap G_i. For this, suppose the element:

(g_1,g_2, \dots, g_r) \in H.

Since H is fully characteristic, it is invariant under the projections to the direct factors, which are endomorphisms. Thus, each (0,0, \dots, 0, g_i,0, \dots, 0) is in H. Thus, every element of H can be expressed as a sum of elements in H \cap G_i, and we get that H is a direct sum of the H \cap G_is.

Finally, we need to show the conditions on the l_is. For this, observe that for i < j, we have:

  • The endomorphism \alpha_{i,j} sends H to itself: Thus, H \cap G_i injects into H \cap G_j via \alpha_{i,j}, so l_i \le l_j.
  • The endomorphism \beta_{j,i} sends H to itself: Thus, \beta_{j,i} induces a surjective endomorphism from G_j/(H \cap G_j) to G_i/(H \cap G_i), forcing k_i - l_i \le k_j - l_j.

Now, we show that if H satisfies the conditions described above, then H is fully characteristic in G.

Suppose \rho:G \to G is an endomorphism. Since H is a direct sum of H \cap G_i, it suffices to show that \rho(H \cap G_i) \le H. For this, in turn, it suffices to show that the j^{th} coordinate of \rho(H \cap G_i) is contained in H \cap G_j. This is easily done in three cases:

  • j = i: In this case, it is direct since H \cap G_i is a fully characteristic subgroup of the cyclic group G_i (all subgroups of cyclic groups are fully characteristic).
  • j > i: In this case, k_i \le k_j. Consider the homomorphism from H \cap G_i to G_j obtained by composing the j^{th} projection with \rho. This map must send H \cap G_i to a subgroup of size at most p^{l_i} in G_j. But since l_i \le l_j, this subgroup of size p^{l_i} is contained in the subgroup H \cap G_j that has order p^j.
  • j < i: In this case, k_j \le k_i. COnsider the homomorphism from G_i to G_j obtained by composing the j^{th} projection with \rho. The index of the image of H \cap G_i is at least the index of H \cap G_i in G_i, which is p^{k_i - l_i}. Thus, the image has size at most p^{k_j - (k_i - l_i)} \le p^{l_j} because k_j - l_j \le k_i - l_i. Thus, it is in H \cap G_j.

General case

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