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This is Exercise 4.19 of Roman's "Fundamentals of Group Theory: An Advanced Approach". According to Approach0, it is new to MSE.

The Details:

The set product of two subsets $S,T$ of a group $G$ is given by $$ST=\{st\in G\mid s\in S, t\in T\}.$$

Since definitions vary, Roman's book defines normal as follows:

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

The Question:

Let $\mathcal{F}$ be a family of groups with the following properties:

  1. If $G\in\mathcal{F}$ and $H\cong G$, then $H\in\mathcal{F}$

  2. If $G\in\mathcal{F}$ and $N\unlhd G$, then $G/N\in\mathcal{F}$.

  3. If $N\unlhd G$ and if $N\in\mathcal{F}$ and $G/N\in\mathcal{F}$, then $G\in\mathcal{F}$.

Prove that if $N\in\mathcal{F}$ and $K\in\mathcal{F}$ are subgroups of $G$ with $K\unlhd G$, then $NK\in\mathcal{F}$.

Thoughts:

Two previous questions of mine seem related and they're from the same book (and even the same set of exercises):

A guess of mine is that the isomorphism theorems come into play here, especially the second one (since we're aiming for $NK$), except that there's nothing about $N\cap K$ in $\mathcal{F}$.

This seems like it should have a simple answer.

Perhaps a more modest goal would be to show $G\in\mathcal{F}$ if at all possible; it might even be a crucial lemma.

Please help :)

Shaun
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1 Answers1

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Note that $NK$ is a subgroup of $G$.

By the second isomorphism theorem we have $$NK/K \cong N/(N \cap K).$$

Since $N \in \mathcal{F}$ and $N \cap K \trianglelefteq N$, property 2 tells us that $N/(N \cap K) \in \mathcal{F}$.

Property 1 now tells us that $NK/K \in \mathcal{F}$.

Finally, since $K \in \mathcal{F}$ and $NK/K \in \mathcal{F}$, property 3 tells us that $NK \in \mathcal{F}$.

Shaun
  • 47,747