It is known that any group $G$ can be embedded into a simple group $S$, see, e.g., the discussion at [Can any group be embedded in a simple group?](https://mathoverflow.net/questions/247402/can-any-group-be-embedded-in-a-simple-group) 


My question is whether one can get an embedding such that the ambient group $S$ is of commutator width $1$ (i.e., each element of $S$ can be represented as a single commutator). 



Embedding of a group into a simple group in which every element is a commutator
Here is a construction. Every finite group with $n$ elements embeds into $A\_{2n}$ (and even $A\_{n+2}$) which is simple if $n>2$ and of commutator width 1 (as any other finite simple group by the Ore conjecture proved by Martin W. Liebeck, E. A. O'Brien, Aner Shalev, Pham Huu Tiep, although for $A\_n$ it was probably known to Frobenius and certainly to Ore in 1951). Suppose that the group $G$ is infinite. Using HNN extensions embed your group into a group $S$ where all pairs of elements of the same order are conjugate (this is the standard application of HNN extensions: use HNN extensions with cyclic associated subgroups to make $G<G\_1$ such that all pairs of elements of $G$ of the same order are conjugate in $G\_1$; then do the same to $G\_1$ and obtain $G\_2$,then $G\_3,G\_4,...$; the union $\cup G\_i$ is the desired group $S$). The group $S$ is always simple because the HNN extension with free letter commuting with $1$ in the HNN construction gives a free product with $\mathbb Z$. Hence the normal subgroup generated by any non-trivial element of $S$ has an element of infinite order, hence all elements of infinite orders (see below), hence coincides with $S$. 


Take an element $g\in S$. Then among the HNN extensions used to build $S$ there is one with free letter $t$ such that $tgt^{-1}=g$. That is, $tg=gt$. Then $gt$ and $t$ have infinite order in $S$. So there exists $h\in S$ such that $gt=hth^{-1}$. Therefore $g=[h,t]$, and $g$ is a commutator. This works for groups of any infinite cardinality. If your group is infinite, $S$ will have the same cardinality as your group. 


