Product of a subnormal subgroup and a normal subgroup We all know that
(1) the product of a subgroup and a normal subgroup is a subgroup
(2) the product of two normal subgroups is a normal subgroup
Now my question is: Suppose $G$ is a group, $H$ is a subnormal subgroup of $G$ and $K$ is a normal subgroup of $G$ . Is $HK$ also a normal subgroup of $G$? I have no idea about it.
 A: No, $HK$ need not be normal (though it will be subnormal; see below)
A simple counterexample can be found with $p$-groups, since every subgroup of a $p$-group is subnormal. Let $G$ be the group
$$G = \langle x,y\mid x^{p^2}=y^{p^2}=[y,x]^{p^2}=[y,x,x]=[y,x,y]=e\rangle.$$
Elements of $G$ are of the form $x^ay^b[y,x]^c$, with multiplication given by
$$\left(x^ay^b[y,x]^c\right)\left(x^{\alpha}y^{\beta}[y,x]^{\gamma}\right) = x^{a+\alpha}y^{b+\beta}[y.x]^{c+\gamma+\alpha b}.$$
You can also think of the group as the multiplicative group of $3\times 3$ upper triangular matrices with coefficients in $\mathbb{Z}/p^2\mathbb{Z}$ and ones in the diagonal. The element $x$ corresponds to the matrix that also has a $1$ in the $(2,3)$ entry (and zeroes off the diagonal elsehwere); $y$ to the element that has a $1$ in the $(1,2)$ entry, and $[y,x]$ the element that has a $1$ in the $(1,3)$ entry.
The subgroup $K$ generated by $[y,x]^p$ is central, hence normal. The subgroup $H$ generated by $y$ is subnormal because $G$ is a $p$-group. The subgroup $HK$ consists of elements of the form $y^b[y,x]^{pc}$, $0\leq b\lt p^2$, $0\leq c\lt p$, and is abelian. But this subgroup is not normal, since it contains $y$, but it does not contain $x^{-1}yx = y[y,x]$.

It will, however, be subnormal. If $H\triangleleft N_1\triangleleft N_2\triangleleft\cdots\triangleleft N_k=G$ is a subnormal series for $H$, and $K$ is normal, then $HK\triangleleft N_1K\triangleleft\cdots\triangleleft N_kK=G$ is a subnormal series for $HK$.
To verify this, say $A\triangleleft B$. Then $AK$ and $BK$ are subgroups, and $AK\leq BK$. Given $ax\in AK$, $a\in A$, $x\in K$; and $by\in BK$, $b\in B$, $y\in K$, then $(by)^{-1}ax(by) = y^{-1}(b^{-1}ab)(b^{-1}xb)y$. Now, $b^{-1}xb\in K$, since $K$ is normal; $b^{-1}ab\in A$, since $A\triangleleft B$. So this element lies in $KAK$. But since $K$ is normal, $KA=AK$, so $KAK=AK$. Thus, $(by)^{-1}(ax)(by)\in AK$, proving $AK\triangleleft BK$, as claimed.
