1
$\begingroup$

Let $G$ be a group with subgroups $P$ and $Q$. Assume $Q$ is normal.

Define $G$ to be the semi direct product of $P$ and $Q$ if $PQ=\{pq: p\in P,q\in Q\}=G$ and $P\cap Q=\{e\}$, where $e$ is the identity element in $G$.

Let $\iota : P\rightarrow G$ be the inclusion map $\iota(p)=p$ and $\phi : G\rightarrow G/Q$ be the natural surjection $\phi(g)=gQ$.

I'm trying to prove that $G$ is the semi direct product of $P$ and $Q$ if and only if the composition $\phi\circ \iota : P\rightarrow G/Q$ is an isomorphism, but I'm struggling to prove that $PQ=G$. Obviously, $PQ\subseteq G$, so all I need to prove is that $G\subseteq PQ$, but I can't see how to do it. I would be very thankful for any tips or hints that would lead me in the right direction.

1 Answers 1

3

Let $g\in G$ be given. Since $\phi\circ \iota$ is an isomorphism, there is some $p\in P$ such that $\phi\circ \iota(p)=gQ$. Thus $g^{-1}p=q\in Q$, and and therefore $g=pq^{-1}\in PQ$.

  • 0
    Thanks so much for the reply! Can you expand a bit on your explanation? I've been trying to write it out for myself, and I'm having a hard time seeing how you got from the second to the third line. – 2017-02-24
  • 0
    @omvendtgribb On my screen I only see two lines :) but I think I know where the confusion lies: $gQ=\phi\circ\iota(p)$ from second sentence, and $\phi\circ\iota(p)=\phi(p)=pQ$. – 2017-02-24
  • 0
    Haha, sorry, I confused line with sentence. I'm not a native English speaker. :) You were right though, I see why your answer is correct now. Thank you very much for helping me understand. – 2017-02-24
  • 0
    @omvendtgribb No apologies necessary. You're welcome! – 2017-02-24