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Let $G$ be a group such that for each $n$: $$G^\dot{n}=\{a^n|a\in G\}$$ is a subgroup of $G$. Is $G$ abelian?

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because it's generally false, let it be any n. – user59671 Feb 11 '13 at 1:21
This is not a duplicate of that other question. – user53153 Feb 11 '13 at 2:03
up vote 10 down vote accepted

Let $Q = \{1, -1, i, -i, j, -j, k, -k\}$ be the quaternion group of order 8. Then

$$Q^2 = \{1, -1\}$$ $$Q^3 = Q$$ $$Q^4 = 1$$

So $Q^k$ is a subgroup for all $k$ (because $Q^5 = Q$, $Q^6 = Q^2$, etc.). But $Q$ is not abelian.

I've found that whenever you conjecture that some property implies a group is abelian, it's always useful to check if $Q$ is a counterexample. It's the standard counterexample to show that "every subgroup is normal" does not imply abelian.

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Here it is a large family of those groups. A a finite $p$-group is said to be powerful if $p$ is odd and $[G,G]\leq G^p$ or if $p=2$ and $[G,G]\leq G^4$. If $G$ is a powerful $p$-group then for any integer $n$ we have $G^n=\{g^n:n\in G\}$. This is Theorem 2.7 in the book Analytic pro-$p$ group:

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In the statement of the theorem $n$ is a power of $p$ but it can be extended for any $n$ because if G is any finite p-group and p does not divide n then any element of G is a n-th power – Diego Feb 11 '13 at 1:21
You need $[G,G] \le \{g^p : g \in G\}$, right ? (For every $G$, the commutator subgroup $[G,G]$ is always contained in the subgroup generated by the $p$th powers, since the quotient group is cyclic. You need the commutators to actually be $p$th powers.) – Ted Feb 11 '13 at 1:23
$G/G^p$ need not be abelian. I will come with an example later – Diego Feb 11 '13 at 1:37
@Diego Sorry, you're right, I was confused - $G/G^p$ is only a group of exponent $p$ which need not abelian. – Ted Feb 11 '13 at 1:55

Look at the quaternion group...

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