If $G=⨝_{i\in I} H_i$ and $K\unlhd G$ is perfect, then $K=⨝_{i\in I}(H_i\cap K)$.

Question

This is Exercise 5.11(b) of Roman's "Fundamentals of Group Theory: An Advanced Approach". According to Approach0, it is new to MSE.

The Details:

On page 72 . . .

Definition: If $\mathcal{F}=\{H_i\mid i\in I\}$ is a nonempty family of normal subgroups of a group $G$, we write $$H_{(i)}:=\bigvee \{H_j\mid j\in I, i\neq j\}$$ for the join of all members of $\mathcal{F}$ except $H_i$.

[. . .]

We say $\mathcal{F}$ is strongly disjoint if $$H_i\cap H_{(i)}=\{1\}$$ for all $i\in I$.

I use $⨝$ instead of $\times$ for a reason. It means something specific in Roman's book; namely, on page 153,

Definition: A group $G$ is the (internal) direct sum or (internal) direct product of a family $\mathcal{F}=\{H_i\mid i\in I\}$ of normal subgroups if $\mathcal{F}$ is strongly disjoint and $G=\bigvee\mathcal{F}$. We denote the internal direct product of $\mathcal{F}$ by $$⨝H_i\quad \text{ or }\quad ⨝\mathcal{F}.$$

Since definitions vary, here is a

Definition: A subgroup $H$ of $G$ is normal in $G$, written $H\unlhd G$, if $$aH=Ha$$ for all $a\in G$.

This is on page 65.

Finally,

Definition: A group $G$ is perfect if $G=[G, G].$

This is on page 84.

The Question:

Let $G=⨝_{i\in I}H_i$. Suppose $K\unlhd G$ is perfect. Show that $$K=⨝_{i\in I}(H_i\cap K).$$

Thoughts:

The question concerns an equality, not an isomorphism, so the isomorphism theorems don't help here. (My first attempt, essentially, was to apply the second isomorphism theorem, since intersections are at play.)

One thing I'm not sure of is the definition of $\bigvee \mathcal{S}$ (if it is not just the least upper bound of the family $\mathcal{S}$ with respect to the subgroup relation).

I don't see how to use the fact that $K$ is perfect. Do I need to start with

$$⨝_{i\in I}(H_i\cap [K,K])$$

then manipulate this using properties of the derived subgroup of $K$?

I'm not sure where normality comes in either. I know that $$\mathcal{H}\unlhd \mathcal{G}\quad\iff\quad [\mathcal{H},\mathcal{G}]\le\mathcal{H}.$$

Please help :)

Answer 1

Since $G$ is the internal direct product of the subgroups $H_i$, each element $g \in G$ can be written as $\prod_{i \in I} g_i$ with $g_i \in H_i$, and all but finitely many of the $g_i$ trivial.

Then the commutator of two elements $x = \prod x_i$ and $y = \prod y_i$ of $K$ is $[x,y] = \prod [x_i,y_i]$.

Since $K \unlhd G$, we have $[x,y_i] = [x_i,y_i] \in K$ for each $i \in I$.

Now let $k = \prod k_i \in K$. Since $K$ is perfect, the element $k$ is a product $\prod_{j=1}^n [x(j),y(j)]$ of commutators $[x(j),y(j)]$ of elements $x(j),y(j) \in K$, and then $k_i = \prod _{j=1}^n [x(j)_i,y(j)_i]$ for each $i \in I$.

But we saw above that each commutator $[x(j)_i,y(j)_i] \in K$, and so $k_i \in K$ for all $i \in I$, and the result follows.