# Socle of a direct product of finite groups.

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

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 $$N \le {\rm Soc}(A_1) \times \cdots \times {\rm Soc}(A_n)$$