SQ-universal Group - Some Elementary Properties of SQ-universal Groups

Some Elementary Properties of SQ-universal Groups

A free group on countably many generators h₁, h₂, ..., h_n, ..., say, must be embeddable in a quotient of an SQ-universal group G. If are chosen such that for all n, then they must freely generate a free subgroup of G. Hence:

Every SQ-universal group has as a subgroup, a free group on countably many generators.

Since every countable group can be embedded in a countable simple group, it is often sufficient to consider embeddings of simple groups. This observation allows us to easily prove some elementary results about SQ-universal groups, for instance:

If G is an SQ-universal group and N is a normal subgroup of G (i.e. ) then either N is SQ-universal or the quotient group G/N is SQ-universal.

To prove this suppose N is not SQ-universal, then there is a countable group K that cannot be embedded into a quotient group of N. Let H be any countable group, then the direct product H × K is also countable and hence can be embedded in a countable simple group S. Now, by hypotheseis, G is SQ-universal so S can be embedded in a quotient group, G/M, say, of G. The second isomorphism theorem tells us:

Now and S is a simple subgroup of G/M so either:

or:

.

The latter cannot be true because it implies K ⊆ H × K ⊆ S ⊆ N/(M ∩ N) contrary to our choice of K. It follows that S can be embedded in (G/M)/(MN/M), which by the third isomorphism theorem is isomorphic to G/MN, which is in turn isomorphic to (G/N)/(MN/N). Thus S has been embedded into a quotient group of G/N, and since H ⊆ S was an arbitrary countable group, it follows that G/N is SQ-universal.

Since every subgroup H of finite index in a group G contains a normal subgroup N also of finite index in G, it easily follows that:

If a group G is SQ-universal then so is any finite index subgroup H of G. The converse of this statement is also true.