2
$\begingroup$

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 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!

1 Answers 1

6

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

  • 1
    I add that the order of $H\cap K$ divides to $|K|=p^d$ but is not equal.2012-01-27