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Let $F_n$ be a free group of rank $n$. For a fixed $k$, consider the subgroup $P_{n,k}$ generated by $k$th powers of elements of $F_n$. This is a characteristic subgroup. What are the quotients $F_n/P_{n,k}$?

I would guess that it's just $C_k^n$ (direct product of $n$ copies of the cyclic group of order $k$), but I'm not sure.

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    ... "*free Burnside groups*" is the magic phrase. Deep mathematic is involved regarding these groups. – 2017-01-09

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The quotient of $F_n$ by the normal subgroup $F_n^k$ generated by all $k$-th powers of elements of $F_n$ is defined to be the free Burnside group $B_n(k)$ of rank $n$ and exponent $k$. In $1902$, W. Burnside asked whether or not $B_n(k)$ has to be finite. This turned out to be a very hard problem, and it is true for some small exponents $k=2,3,4,6$, but not true in general. These groups are infinite in general. In particular, such groups are not the direct product of $n$ copies of $C_k$ in general (but it is true for $k\le 2$).

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    Do you know of any special results in the case $n = 2$? For example, are the quotients finite in this case? Do they have finite rank as free groups? – 2017-01-09
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    $k=1$: cyclic group, $k=2$: Klein four group, $k=3$: elementary abelian group of order 8, $k=4$: elementary abelian group of order 16 ... – 2017-01-09
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    Already for $n=2$, and for any odd period $k ≥ 4381$ the free Burnside group $B(2,k)$ is infinite. For more results see [this thesis](http://digitalcommons.liberty.edu/cgi/viewcontent.cgi?article=1605&context=honors) for example. – 2017-01-09
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    ... what do you mean "finite rank as free groups"? – 2017-01-09
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    @M.U. These burnside groups are subgroups of $F_n$, and hence are free. So I'm asking about their rank. – 2017-01-09
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    That is wrong!! Free Burnside groups are **quotients** of free groups not subgroups. – 2017-01-09
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    @M.U You seem to have mixed up rank and exponent in some of your examples. – 2017-01-09
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    @Tobias Kildetoft: Whups. You are absolutely right :) – 2017-01-09