# Changes

## Classification of fully characteristic subgroups in finitely generated Abelian groups

, 21:41, 4 January 2009
no edit summary
where $k_1 \le k_2 \le \dots \le k_r$.
Then a subgroup $H$ of $G$ is a [[characteristic subgroup]] of $G$ if and only if $H$ 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 finite finitely generated groups===
Suppose $G$ is a [[finite finitely generated Abelian group]]. Then, we can write:
$G = \bigoplus_{i=10}^r G_i$
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 [[characteristic subgroup]] if and only if it 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 $G_i$.

===Case of finitely generated groups===

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

$G = A \oplus B$

where $A$ is a torsion-free group. A subgroup $H$ of $G$ is a [[fully characteristic subgroup]] of $G$ if and only if $H = (H \cap A) \oplus (H \cap B)$ and we have that $H \cap A$ is fully characteristic in $A$ and $H \cap B$ is characteristic in $B$.
==Proof==
and is zero elsewhere.
Define $\sigma_{i,j}$ as the sum of the endomorphism $\alpha_{i,j}$ and the identity map on $G$. Clearly, $\sigma_{i,j}$ is an automorphism. 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. Define $\varphi_{j,i}$ as the automorphism obtained as the sum of $\beta_{j,i}$ and the identity map.
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:
$(g_1,g_2, \dots, g_r) \in H$.
We want Since $H$ is fully characteristic, it is invariant under the projections to show that the elements direct factors, which are endomorphisms. Thus, each $(g_10,0, \dots,0, g_i,0, \dots, 0)$ 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 $\alpha_{i, (0j}$,g_2so $l_i \le l_j$.* The endomorphism $\beta_{j,0i}$ sends $H$ to itself: Thus,0$\beta_{j, i}$ induces a surjective endomorphism from $G_j/(H \cap G_j)$ to $G_i/(H \dots 0cap G_i)$, forcing $k_i - l_i \dotsle k_j - l_j$. Now, (0we show that if $H$ satisfies the conditions described above,0then $H$ 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 $\rho(H \cap G_i) \dotsle H$. For this,0in turn,g_rit suffices to show that the $j^{th}$ coordinate of $\rho(H \cap G_i)$ are is contained in $H\cap G_j$.This is easily done in three cases:
If * $j = i$: In this case, it is direct since $H \cap G_i$ is a fully characteristic subgroup of the cyclic group $G_i$ (all except one subgroups of cyclic groups are fully characteristic).* $g_ij > i$ are zero: In this case, then we are done$k_i \le k_j$. 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 $G_j$. OtherwiseBut since $l_i \le l_j$, there exist this subgroup of size $p^{l_i}$ is contained in the subgroup $H \cap G_j$ that has order $p^j</math>.* [itex]j < i < /math>: In this case, [itex]k_j \le k_i$. COnsider the homomorphism from $G_i$ to $G_j$ obtained by composing the $j^{th}$ such that both 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 $g_ik_j - l_j \le k_i - l_i$ and . Thus, it is in $g_jH \cap G_j$ are nonzero.