Question on wreath product Let $H$ be a cyclic group of order n. Let $K$ be a subgroup of $S_n$. Let $K$ acts on $H^n$ as $(h_1, h_2,...h_n)^a=(h_{a^{-1}(1)},.....h_{a^{-1}(n)})$ where $a\in K$. My question is: What is $H^n \rtimes S_2$ in general ? Is $H^2 \rtimes S_2=D_8?$
Fact I know and tried to used in the sum but failed is:
If $N$ is cyclic group of order $n$., and let $P$ be a cyclic  group of order $2$. Let $b$, $a$ generates $N$,$P$ respectively, and $b^a=b^{-1}$ Then the semidirect product of $N$ and $P$ (i.e $N\rtimes P$) is the dihedral group of order $2n$. 
 A: For your second question, a concrete approach is the following. 
You have a group $G = H^2 \rtimes S_2 = \langle a, b, c \rangle$ of order $8$ with $a^2 = b^2 = c^2 = 1$, $ab = ba$ and $a^c = b, b^c=a$.
Consider $x = ac$ and $y = c$. Then $x^2 = a c a c = a a^c = ab$, and $x^4 = 1$, so $x$ has order $4$. Moreover $x^y = (a c)^c = b c = x^{-1}$, as $(a c) (b c) = a b^c = a^2 = 1$. So you see $G$ is dihedral.
For your first question, note first that 
$$
H^n \rtimes S_2 \cong (H^2 \rtimes S_2) \times H^ {n-2},
$$
so I'll just treat the case $G = H^2 \rtimes S_2$, where $H$ is cyclic of order $n$.
If $n$ is odd, then $G \cong D_{2n} \times C_{n}$. This is because you can rewrite $H^{2} = F \times I$, where $F = \{ (x, x) : x \in H \}$ and $I = \{ (x, x^{-1}) : x \in H \}$, where $F$ is clearly central, while $I S_{2}$ is clearly dihedral.
When $n$ is even, this is not so anymore. I will treat only the case when $n = 2^{k}$. The point here is that $I \cap F$ contains $(i,i) \ne (1,1)$, where $i$ is the unique involution of $H$. 
Write $$G = \langle a, b, c : a^n = b^n = c^2 = 1, ab = ba, a^c = b, b^c=a \rangle.$$  We have that $F = \langle a b \rangle$ is the centre of $G$, and $G/F$ is dihedral of order $2^{k+1}$. (This is because modulo the subgroup $F$ we have $a^c = b \equiv a^{-1}$.) But there is no splitting as in the case when $n$ is odd.
