Cfrtjryk

1 I'm reading Ash's "Basic Abstract Algebra" and been trying to understand the following: For which I guess the problem lies when we try to take an element $hkin HK$ , it's inverse is $k^{-1}h^{-1}$ but $k^{-1}h^{-1}in HK$ only if $k^{-1}h^{-1}=h^{-1} k^{-1}$ . My problem is the following, taking the definition of subgroup, we have: If $a,bin H$ , then $ab^{-1}in H$ . Then, all the elements $kcdot1=k$ and $1cdot h=h$ are in $HK$ , that is: We have all the elements of both $H,K$ in $HK$ . By the definition, suppose we have $hkin HK$ , then we must have $(hk)(k^{-1})$ and $(hkk^{-1})(h^{-1})$ . Isn't this very similar to having the inverse of $hk$ in $HK$ ? It's a bit odd because it seems we can actually invert $hk$ by performing the two operations I mentioned before ...