If $G$ is a group and $H$ is a subgroup of $S\_n$ we can form their wreath product $G \wr H = \{(g\_1, ..., g\_n; \pi): g\_i \in G$ and $\pi \in H\}$. I'm wondering whether the following is correct:


1. $<(e, ..., g\_i, ..., e; e)> = \{(g\_1, ..., g\_n; e)\}$
2. $\{(g\_1, ..., g\_n; e)\} \circ \{(e, ..., e; \pi)\} = G \wr H$


Additionally, will all elements of $G \wr H$ be generated distinctly? Or in other words is it possible to do this more efficiently?


Edit: For $(g\_1, ..., g\_n; \pi), (a\_1, ..., a\_n; \phi) \in G \wr H, \circ$ is defined by $(a\_1, ..., a\_n; \phi) \circ (g\_1, ..., g\_n; \pi) = (a\_{\pi(1)}\*g\_1, ..., a\_{\pi(n)}\*g\_n; \phi\circ\pi)$ where $\*$ is the operation for $G$, although for my purposes my $G$ can be thought of as $S\_n$.


You also have $(g\_1, ..., g\_n; \pi)^{-1} = (g^{-1}\_{\pi^{-1}(1)}, ...; \pi^{-1})$


By distinct I mean for distinct $(g\_1, ..., g\_n; e), (a\_1, ..., a\_n; e), (e, ..., e; \pi), (e, ..., e; \phi)$ we don't have $(g\_1, ..., g\_n; e) \circ (e, ..., e; \pi) = (a\_1, ..., a\_n; e) \circ (e, ..., e; \pi)$ or $(g\_1, ..., g\_n; e) \circ (e, ..., e; \pi) = (g\_1, ..., g\_n; e) \circ (e, ..., e; \phi)$.


I'm not so sure about that for the first part with generating $\{(g\_1, ..., g\_n; e)\}$, but by efficiency I mean I want to generate an object much like a wreath product algorithmically so I don't want to do more work than necessary when generating it.



How to efficiently generate a wreath product?
Note that the wreath product is a semidirect product with normal subgroup $G^n$ and complement $H$. A group theory book that defines wreath product will cover this fact.


Part 1. seems to ask whether the normal $G^n$ is generated by elements of the form $(e, \ldots, g\_i, \ldots,e)$, i.e. those which are the identity except in one coordinate. This is true by definition of multiplication in the direct product $G^n$.


Part 2. I interpret to ask whether $G \wr H = G^n H$--this is true because, as noted, the wreath product is a semidirect product of these two groups.


The "distinctness" seems to follow from the existence of inverses. For example, if $(g\_1, \ldots, g\_n; e) \circ (e, \ldots, e; \pi) = (a\_1, \ldots, a\_n, ; e) \circ (e, \ldots, e; \pi)$, we may multiply on the right by $(e, \ldots, e; \pi)^{-1}$ and conclude that $(g\_1, \ldots, g\_n; e) = (a\_1, \ldots, a\_n; e)$. 


There could be an algorithmic question in the original post about generating a wreath product, but it requires a more careful statement.


