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

Revision as of 21:41, 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 $a < b$ , 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 $H \cap G_i$ s.

Finally, we need to show the conditions on the $l_i$ s. 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 $H \cap G_j$ .

@@ Line 13: / Line 13: @@
 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>.
 * 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>.
 * 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==
@@ Line 77: / Line 69: @@
 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.
-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>.
-We want to show that the elements <math>(g_1,0,0, \dots, 0), (0,g_2,0,0, \dots 0), \dots, (0,0,\dots,0,g_r)</math> are in <math>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:
-If all except one <math>g_i</math> are zero, then we are done. Otherwise, there exist <math>i < j</math> such that both <math>g_i</math> and <math>g_j</math> are nonzero.
+* <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>.

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

Revision as of 21:41, 4 January 2009

Contents

Statement

Case of groups of prime power order

Case of finitely generated groups

Proof

Case of prime power order

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