Socle of abelian divisible periodic group

Question

I'm trying to prove that the socle of a periodic divisible abelian group J is a proper subgroup of J. I know that J is direct sum of quasicyclic groups, say

$$ J={\oplus}_{i\in I} P_i $$

and that the socle of J is

$$ Soc(J)={\oplus}_p J[p] $$

Now, I'm wondering if it is true that

$$J[p]={\oplus}_{i\in I} P_i[p]$$

and that

$$Soc(J)={\oplus}_p\big({\oplus}_{i\in I}P_i[p]\big) ={\oplus}_{i\in I}\big({\oplus}_p P_i[p]\big) $$

Thank you very much!

Answer 1

I take $G[p]$ to mean the subgroup of elements satisfying $x^p=\rm identity$ in an abelian group $G$.

Yes, $\displaystyle\left(\bigoplus G_i\right)[p]=\bigoplus G_i[p]$ holds; this is because an element in the direct sum $\displaystyle\bigoplus G_i$ satisfies the relation $x^p=\rm identity$ if and only if each coordinate also satisfies this relation. So, yes, we have

$${\rm soc}(J)\cong\bigoplus_p J[p]\cong\bigoplus_p \left(\bigoplus_{i\in I} Q_i\right)[p]=\bigoplus_p\bigoplus_{i\in I} Q_i[p]\cong\bigoplus_{i\in I}\left(\bigoplus_p Q_i[p]\right)\cong \bigoplus_{i\in I}{\rm soc}(Q_i)$$

when $\displaystyle J=\bigoplus_{i\in I}Q_i$ (let's avoid overusing the letter 'P') over some index set $I$.

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However, this sort of classification/decomposition theory is not fully necessary. To prove that the socle of a nontrivial torsion abelian divisible group $G$ is proper, show the following:

• $G$ contains a nontrivial element whose order is a perfect square
• A minimal (abelian $\Rightarrow$ normal) subgroup of $G$ has prime order
• The socle of $G$ contains only elements of squarefree order