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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:
* <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 finitely generated groups===
Suppose <math>G</math> is a [[finite finitely generated Abelian group]]. Then, we can write:
<math>G = \bigoplus_{i=10}^r G_i</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 [[characteristic subgroup]] if and only if it 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==
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. 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 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 Since <math>H</math> is fully characteristic, it is invariant under the projections to show that the elements direct factors, which are endomorphisms. Thus, each <math>(g_10,0, \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, (0j}</math>,g_2so <math>l_i \le l_j</math>.* The endomorphism <math>\beta_{j,0i}</math> sends <math>H</math> to itself: Thus,0<math>\beta_{j, i}</math> induces a surjective endomorphism from <math>G_j/(H \cap G_j)</math> to <math>G_i/(H \dots 0cap G_i)</math>, forcing <math>k_i - l_i \dotsle k_j - l_j</math>. Now, (0we show that if <math>H</math> satisfies the conditions described above,0then <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) \dotsle H</math>. For this,0in turn,g_rit suffices to show that the <math>j^{th}</math> coordinate of <math>\rho(H \cap G_i)</math> are is contained in <math>H\cap G_j</math>.This is easily done in three cases:
If * <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 except one subgroups of cyclic groups are fully characteristic).* <math>g_ij > i</math> are zero: In this case, then we are done<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>. OtherwiseBut since <math>l_i \le l_j</math>, there exist 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> such that both 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>g_ik_j - l_j \le k_i - l_i</math> and . Thus, it is in <math>g_jH \cap G_j</math> are nonzero.
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