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I am currently working on the following exercise:

Let $G$ be a group of order $pq$, where $p$ and $q$ are primes and $p > q$. Prove that $G = N \rtimes H$ for some subgroups $N$ and $H$ of orders $p$ and $q$, respectively.

Now, I just proved before this that for $G$ where $|G|=pq$ for $p$, $q$ primes and $p > q$, $G$ has a normal subgroup of order $p$.

Also, I know that since I have already shown that $N$ is normal, and the Sylow theorems guarantee that $H \cap N = \{ e \}$, in order to show that $G$ is the semidirect product of $N$ and $H$, all that remains to be shown is that $G = HN$, or that $H \cup N$ generates $G$. However, I am not sure how to do this. Could somebody please help me out with it?

Another possibility is to use a result I have at my disposal that states that "if $H \leq G$ and $N \trianglelefteq G$. Then, $G = N \rtimes H$ if and only if the restriction of the canonical epimorphism $\pi: G \to G/N$ onto $H$ is an isomorphism $H \to G/N$". I am equally unsure of how to apply this result in this situation. Again, help would be greatly appreciated!

Thank you.

1 Answers1

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You can recall that $|HN|=\frac{|H|\cdot|N|}{|H \cap N|}$ (see e.g. Size of the product of two subgroups), and then $|HN|=pq$ implies $G=HN$.

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  • probably a stupid question, but why can $G$ not have a proper subgroup of order $pq$? –  Jan 18 '17 at 05:39
  • Because $G$ itself has order pq. A proper subgroup would necessarily have a smaller order. – math Jan 18 '17 at 05:41
  • told ya it was stupid. This is your brain on no sleep... –  Jan 18 '17 at 05:44
  • more awake now. Isn't this also the Second Isomorphism Theorem? –  Jan 18 '17 at 12:39