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

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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 | + | 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 | + | ===Case of finitely generated groups=== |

− | Suppose <math>G</math> is a [[ | + | Suppose <math>G</math> is a [[finitely generated Abelian group]]. Then, we can write: |

− | <math>G = \bigoplus_{i= | + | <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> | + | 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>. | ||

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==Proof== | ==Proof== | ||

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and is zero elsewhere. | 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. | |

− | |||

− | 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 | ||

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>. |

## Revision as of 21:41, 4 January 2009

## Contents

## Statement

### Case of groups of prime power order

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

where:

.

where .

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

- is the direct sum of the intersections .
- If the orders of are , then and .

### Case of finitely generated groups

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

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

- is the direct sum of the intersections .
- Each is characteristic (equivalently, fully characteristic) in .

## Proof

### Case of prime power order

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

where:

.

where .

A subgroup of .

**To prove**: is characteristic if and only if it is fully characteristic in , if and only if is the direct sum of , and if the orders of the intersections are , then and .

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

that sends the generator on the left to times the generator on the right.

There is also a surjective homomorphism:

that sends the generator to the generator.

Recall that we have:

.

For , define as the endomorphism of that sends to via the map:

.

and is zero elsewhere.

Similarly, define as the endomorphism of that sends to via the map and is zero elsewhere.

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

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

.

Since is fully characteristic, it is invariant under the projections to the direct factors, which are endomorphisms. Thus, each is in . Thus, every element of can be expressed as a sum of elements in , and we get that is a direct sum of the s.

Finally, we need to show the conditions on the s. For this, observe that for , we have:

- The endomorphism sends to itself: Thus, injects into via , so .
- The endomorphism sends to itself: Thus, induces a surjective endomorphism from to , forcing .

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

Suppose is an endomorphism. Since is a direct sum of , it suffices to show that . For this, in turn, it suffices to show that the coordinate of is contained in . This is easily done in three cases:

- : In this case, it is direct since is a fully characteristic subgroup of the cyclic group (all subgroups of cyclic groups are fully characteristic).
- : In this case, . Consider the homomorphism from to obtained by composing the projection with . This map must send to a subgroup of size at most in . But since , this subgroup of size is contained in the subgroup that has order .
- : In this case, . COnsider the homomorphism from to obtained by composing the projection with . The index of the image of is at least the index of in , which is . Thus, the image has size at most because . Thus, it is in .