# Embedding of a group into a simple group in which every element is a commutator

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?

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).

• Yes, every infinite group $G$ naturally embeds into the quotient of the symmetric group $S(G)$ by its maximal normal proper subgroup (the subgroup of permutations of support $<|G|$) which is a simple group (Baer) and in which every element is a commutator (because it holds in $S(G)$, in turn because it holds in $S(\omega)$, Ore Proc AMS 1951).
– YCor
Feb 23, 2020 at 21:50
• Note: the above embeds a group of cardinal $\alpha$ into a simple group of cardinal $2^\alpha$. Nevertheless, whenever $G\le H$ with $G$ infinite and $H$ simple, then $G$ is contained in a simple subgroup $L$ of $H$ of cardinal $|G|$ (by an easy argument: add finitely many conjugators for each pair of nontrivial elements and iterate). If $H$ has commutator width 1, do the above at odd times and at even times add for every element 2 elements of which they're commutator. In this way, get $L$ in addition of commutator width 1.
– YCor
Feb 23, 2020 at 22:33
• Note: the fact that every group embeds into a simple group of the same cardinal is asserted (with hints of proof based on the same principle) in: Locally finite groups by Kegel and Wehrfritz, 1973, Remarks p.115. This follows a detailed proof that in an infinite simple group, every countable subset is contained in an infinite countable simple subgroup (Theorem 4.4 therein). In the same remark they also mention that this is a part of general result of universal algebra.
– YCor
Mar 3, 2020 at 9:03
• @YCor: All the comments are greatly appreciated (sorry for unforgivably long delay). Interestingly, similar phenomena take place for associative algebras and Lie algebras (L.A. Bokut', 1960-70's). Mar 14, 2020 at 10:33

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 $$G such that all pairs of elements of $$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.