The socle of a group $G$ is defined as the subgroup generated by minimal subgroups among normal subgroups of $G$, and it is denoted as $\textrm{Soc}(G)$.

Suppose $A_1,...,A_n$ are finite groups. Is it true that $\textrm{Soc}(A_1 \times ... \times A_n) = \textrm{Soc}(A_1) \times ... \times \textrm{Soc}(A_n) $ ?

The $\supseteq$ inclusion seems to be true, but I'm not sure whether the inclusion $\subseteq$ is generally true (though I would guess it is true in the case of all the groups $A_i$ being abelian, for example - I think this might be true because in a direct product of abelian groups, minimal groups are cyclic groups prime order).


1 Answer 1


I think the equality is true. Here is a sketch proof.

Let $N$ be a minimal normal subgroup of $G = A_1 \times \cdots \times A_n$. If $N \le A_i$ for some $i$ then $N \le {\rm Soc}(A_i)$, so suppose not, and suppose that $N$ projects nontrivially onto $A_1$ and $A_2$, say.

Then $N$ cannot contain any elements $(h,1,\ldots)$ with $h\ne 1$ because the set of such elements would form a smaller normal subgroup than $N$.

But this implies that the projection $N_1$ of $N$ onto $A_1$ must lie in $Z(A_1)$, since otherwise $N$ would contain elements $([g,h],1,\ldots)$ with $h \in N_1$, $g \in A_1$ and $[g,h] \ne 1$.

So $N \le Z(G)$, and $N$ must have prime order, and be a diagonal subgroup of minimal normal subgroups of some of the $A_i$. So $N \le {\rm Soc}(A_1) \times \cdots \times {\rm Soc}(A_n)$


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