This is Exercise 7, page 40 from Hungerford's book Algebra.

Let $G$ be a group of order $p^{k}m,$ with $p$ prime and $(p,m)=1.$ Let $H$ be a subgroup of order $p^{k}$ and $K$ a subgroup of order $p^{d}$, with $0<d\leq k$ and $K\nsubseteq H.$ Show that $HK$ is not a subgroup of $G$.

I've started assuming that $HK$ is a subgroup, but it din't help me.

Thanks in advance!


Hint. $\displaystyle |HK|=\frac{|H||K|}{|H\cap K|}$.

  • 1
    $\begingroup$ I add that the order of $H\cap K$ divides to $|K|=p^d$ but is not equal. $\endgroup$ – emiliocba Jan 27 '12 at 22:45

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