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This is Exercise 3.3.11 of Robinson's "A Course in the Theory of Groups (Second Edition)". According to Approach0 and this search, it is new to MSE.

The Details:

Since definitions vary, on page 15, ibid., paraphrased, it states that

A subgroup $N$ of $G$ is normal in $G$ if one of the following equivalent statements is satisfied:

(i) $xN=Nx$ for all $x\in G$.

(ii) $x^{-1}Nx=N$ for all $x\in G$.

(iii) $x^{-1}nx\in N$ for all $x\in G, n\in N$.

The definition of the socle of $G$, which I denote as ${\rm Soc}(G)$, is on page 87:

The subgroup generated by all minimal normal subgroups of a group $G$ is called the socle: should the group fail to have any minimal normal subgroups, [. . .] the socle of $G$ is defined to be $1$.

On the same page, we find

the concept of a minimal normal subgroup of a group $G$, by which is understood a nontrivial normal subgroup that does not contain a smaller nontrivial normal subgroup of $G$.

The Question:

Identify the socle of an abelian group.

Thoughts:

I haven't quite got the hang of computing socles, so I have used GAP, and, so far, there does not appear to be any obvious pattern to the socles I looked at. For example:

gap> StructureDescription(Socle(CyclicGroup(8)));
"C2"
gap> StructureDescription(Socle(DirectProduct(CyclicGroup(4),CyclicGroup(2))));
"C2 × C2"
gap> StructureDescription(Socle(DirectProduct(CyclicGroup(6),CyclicGroup(3))));
"C6 × C3"

The above is code for showing:

  • ${\rm Soc}(\Bbb Z_8)\cong \Bbb Z_2$.
  • ${\rm Soc}(\Bbb Z_4\times \Bbb Z_2)\cong \Bbb Z_2^2$.
  • ${\rm Soc}(\Bbb Z_6\times \Bbb Z_3)\cong \Bbb Z_6\times \Bbb Z_3$.

Please keep in mind that the Fundamental Theorem of Finitely Generated Abelian Groups has not been proven explicitly in Robinson's book yet.

Nonetheless, if we cheat a little, we could make use of the following:

That is,

$${\rm Soc}(A_1\times\dots\times A_n)\cong {\rm Soc}(A_1)\times \dots\times{\rm Soc}(A_n)$$

for (necessarily finitely generated?) abelian groups $A_i$.

I have no idea how one might tackle, say, $(\Bbb R, +)$, or any other abelian group that is not finitely generated.

Does a classification of socles for abelian groups even exist? A partial answer is here. Have I misinterpreted the question? Because it states "an abelian group", not "each abelian group".


I know that every subgroup of an abelian group $A$ is normal in $A$.


The kind of answer I'm hoping for is a classification of socles of abelian groups with some justification of why it is the way it is. This might be a big ask, so I'm sorry if that's the case.

Please help :)

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    $\begingroup$ In abelian groups, every subgroup is normal. So minimal normal subgroups are just minimal subgroups (smallest nontrivial subgroups). What do minimal subgroups look like? (In any group?) $\endgroup$
    – verret
    Jun 24 at 18:13
  • $\begingroup$ Oh, yeah, I forgot to mention that I knew that each subgroup of each abelian group is normal in that group. Thank you, @verret. Is there a classification of minimal subgroups? If so, I'm not aware of it. $\endgroup$
    – Shaun
    Jun 24 at 18:19
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    $\begingroup$ Say $H$ is a minimal nontrivial subgroup. If $h\in H$, $h\neq 1$, then $\langle h\rangle = H$. So the subgroup must be cyclic of prime order. $\endgroup$ Jun 24 at 19:36
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    $\begingroup$ I guess I didn't explicitly answer the last question, but it follows from I wrote: the socle of an abelian group is a direct sum (possibly empty) of cyclic groups of prime order, and any such sum is the socle of an abelian group (namely, itself). $\endgroup$ Jun 25 at 0:12
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As noted, and as you know, every subgroup of an abelian group is normal; so the socle of an abelian group $G$ is generated by the minimal nontrivial subgroups.

If $H$ is a minimal nontrivial subgroup of $G$, and $h\in H$, $h\neq 1$, then $\langle h\rangle = H$; thus, $H$ is cyclic, and has no nontrivial subgroups. Therefore, $H$ is cyclic of prime order. Conversely, if $H$ is a subgroup of prime order, then it is necessarily minimal nontrivial, since it has no proper nontrivial subgroups.

So the socle must be generated by the elements of prime order. In particular, $(\mathbb{R},+)$, being torsionfree, has trivial socle, as does every torsionfree abelian group.

Thus, if $G$ is abelian, then $\mathrm{socle}(G)=\mathrm{socle}(G_{\rm tor})$, the torsion subgroup of $G$. And $$\mathrm{socle}(G) = \langle x\mid x\text{ has prime order}\rangle.$$

You can get a bit more information (depending on how much Robinson has proven). Since the torsion subgroup is the direct sum of its $p$-parts, you can just look at the separate $p$-parts, and then just look at what some authors denote $G[p]$, the set of elements of order dividing $p$. So $$\mathrm{socle}(G) = \bigoplus_{p\text{ prime}}G[p].$$

(Your final example comes from the fact that $\mathbb{Z}_6\cong\mathbb{Z}_2\times\mathbb{Z}_3$, so you "really" have $\mathbb{Z}_2\times\mathbb{Z}_3^2$...)

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