# Set consisting of fixed power of elements of a group

Let G be a group. Let n be a fixed natural number. Consider $$S=\{g^{n}|g\in G\}$$. Is this set a subgroup of G? For n=1 it is the trivial subgroup, for $$n\geq 2$$ we need the condition that G is abelian to say that this is a subgroup. But I was unable to find a counter example for non- abelian case. Further in non- abelian case how can I get counter examples for each n?

Also motivated by this question I got one more question in mind. Let G be a group. Let n be a fixed natural number. Consider $$S=\{g|\exists a\in G$$ such that $$g=a^{n}\}$$. Is this set a subgroup of G? Again for n=1 it is the trivial subgroup. For $$n\geq 2$$ we need the condition that G is abelian to say that this is a subgroup. For non- abelian case how do I construct counter examples?

• Does this answer your question? Is the subset of squares of a group a subgroup?
– lulu
Sep 25 at 14:54
• How do I construct for each n? Further what are the counter examples for the second question? Sep 25 at 15:00
• Your two questions are the same. The free group on two letters is a counterexample for all $n$, and it should be possible to construct explicit examples using algebraic groups over finite fields, as in the linked duplicate.
– lulu
Sep 25 at 15:01
• Note that, for $p$-groups it makes sense to consider the subgroup generated by these sets, called the "Agemo" and "Omega" respectively: en.wikipedia.org/wiki/Omega_and_agemo_subgroup Sep 25 at 15:05
• Thanks that was helpful. Sep 25 at 15:08

For $$n\geq 2$$ we need the condition that $$G$$ is abelian to say that this is a subgroup.

That statement seems to be saying that $$\{g^n\mid g\in G\} = G^{\{n\}}$$, with $$n\geq 2$$, is a subgroup only if $$G$$ is abelian. But this is not correct. "$$G$$ is abelian" is sufficient, but not necessary.

For an easy example, consider $$G=S_3$$ and $$n=6$$. Then $$G^{\{6\}} = \{g^6\mid g\in S_3\} = \{e\}$$ is a subgroup, even though $$G$$ is not abelian.

More generally, a group $$G$$ is said to be $$n$$-abelian if and only if $$(ab)^n = a^nb^n$$ for all $$a,b\in G$$. Alperin proved that a group $$G$$ is $$n$$-abelian if and only if it is a quotient of a subgroup of a direct product of a group of exponent $$n-1$$, a group of exponent $$n$$, and an abelian group. (It is easy to verify that if $$G$$ is of exponent $$n-1$$, of exponent $$n$$, or abelian, then it satisfies $$(ab)^n= a^nb^n$$; and that the class of groups that satisfy this condition is closed under products, subgroups, and quotients).

If $$G$$ is $$n$$-abelian, then we have that $$G^{\{n\}}$$ is necessarily a subgroup, since it is trivially closed under inverses, contains $$e=e^n$$, and in the $$n$$-abelian case we also have $$g^nh^n = (gh)^n\in G^{\{n\}}$$.

However, a group need not be $$n$$-abelian for this set to be a subgroup, either. For example, consider the dihedral group of order $$8$$, which is not $$2$$-abelian (a group is $$2$$-abelian if and only if it is abelian). Then $$G^{\{2\}} = \{e,r^2\}$$, since every element other than $$r$$ and $$r^3$$ is of order $$1$$ or $$2$$; this is clearly a subgroup of $$G$$.

Not sure what you mean by "counter-examples for each $$n$$". If you mean, for each $$n$$ find a group in which $$G^{\{n\}}$$ is not a subgroup, then they aren't hard to find, even restricting for finite groups. For odd $$n$$, the set $$(S_n)^{\{n\}}$$ contains all transpositions but no cycle of length $$n$$. I'll let you figure out the even $$n$$ case.